Homework #01, due 1/20/10 = 9.1.2, 9.1.4, 9.1.6, 9.1.8, 9.2.3

Additional problems for study: 9.1.1, 9.1.3, 9.1.5, 9.1.13, 9.2.1,9.2.2, 9.2.4, 9.2.5, 9.2.6, 9.3.2, 9.3.3

9.1.1 (This problem was not assigned except for study, but it’s usefulfor the next problem.) Let p and q be polynomials in Z[x, y, z], where

p = p(x, y, z) = 2x2y − 3xy3z + 4y2z5

q = q(x, y, z) = 7x2 + 5x2y3z4 − 3x2z3

(a) Write each of p and q as a polynomial in x with coefficients inZ[y, z].

p = (2y)x2 − (3y3z)x+ 4y2z5

q = (7 + 5y3z4 − 3z3)x2

(b) Find the degree of each of p and q.

deg(p) = 7 deg(q) = 9

(c) Find the degree of p and q in each of the three variables x, y, andz.

degx(p) = 2 degx(q) = 2

degy(p) = 3 degy(q) = 3

degz(p) = 5 degz(q) = 4

(d) Compute pq and find the degree of pq in each of the three variablesx, y, and z.

pq = (2x2y − 3xy3z + 4y2z5)(7x2 + 5x2y3z4 − 3x2z3)

= 2x2y(7x2 + 5x2y3z4 − 3x2z3)

− 3xy3z(7x2 + 5x2y3z4 − 3x2z3)

+ 4y2z5(7x2 + 5x2y3z4 − 3x2z3)

= 2x2y7x2 + 2x2y5x2y3z4 − 2x2y3x2z3

− 3xy3z7x2 − 3xy3z5x2y3z4 + 3xy3z3x2z3

2

+ 4y2z57x2 + 4y2z55x2y3z4 − 4y2z53x2z3

= 14x4y + 10x4y4z4 − 6x4yz3

− 21x3y3z − 15x3y6z5 + 9x3y3z4

+ 28x2y2z5 + 20x2y5z9 − 12x2y2z8

degx(pq) = 2 + 2 = 4 degy(pq) = 3 + 3 = 6 degz(pq) = 5 + 4 = 9

(e) Write pq as a polynomial in the variable z with coefficients inZ[x, y].

pq = 14x4y − (21x3y3)z − (6x4y)z3 + (10x4y4 + 9x3y3)z4

+ (28x2y2 − 15x3y6)z5 − (12x2y2)z8 + (20x2y5)z9

9.1.2 Repeat the preceding exercise under the assumption that thecoefficients of p and q are in Z/3Z. Let p and q be polynomials inZ/3Z[x, y, z], where

p = p(x, y, z) = 2x2y + y2z5

q = q(x, y, z) = x2 + 2x2y3z4

(a) Write each of p and q as a polynomial in x with coefficients inZ/3Z[y, z].

We need only reduce the coefficients modulo 3, which gives us

p = (2y)x2 + y2z5

q = (1 + 2y3z4)x2

(b) Find the degree of each of p and q.

deg(p) = 7 deg(q) = 9

(c) Find the degree of p and q in each of the three variables x, y, andz.

degx(p) = 2 degx(q) = 2

degy(p) = 2 degy(q) = 3

degz(p) = 5 degz(q) = 4

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(d) Compute pq and find the degree of pq in each of the three variablesx, y, and z.

pq = (2x2y + y2z5)(x2 + 2x2y3z4)

= 2x2y(x2 + 2x2y3z4) + y2z5(x2 + 2x2y3z4)

= 2x2y(x2) + 2x2y(2x2y3z4) + y2z5x2 + y2z5(2x2y3z4)

= 2x4y + x4y4z4 + x2y2z5 + 2x2y5z9

degx(pq) = 2 + 2 = 4 degy(pq) = 3 + 3 = 6 degz(pq) = 5 + 4 = 9

(e) Write pq as a polynomial in the variable z with coefficients inZ/3Z[x, y].

pq = 2x4y + x4y4z4 + x2y2z5 + 2x2y5z9

9.1.4 Prove that the ideals (x) and (x, y) are prime ideals in Q[x, y]but only the latter ideal is a maximal ideal.

To show (x) is a prime ideal of Q[x, y], we must assume pq ∈ (x),where p, q ∈ Q[x, y], and show that either p ∈ (x) or q ∈ (x). Supposep = an(y)x

n+· · ·+a1(y)x+a0(y) and q = bm(y)xm+· · ·+b1(y)x+b0(y),where an(y), . . . , a0(y), bm(y), . . . , b0(y) ∈ Q[y]. Then pq has exactlyone term with x-degree 0, namely the polynomial a0(y)b0(y) ∈ Q[y].

From the assumption pq ∈ (x) we know there is some polynomials ∈ Q[x, y] such that pq = sx. Note that every (non-zero) term in sxhas x as a factor, hence its x-degree is at least 1. Hence every nonzeroterm in pq has degree at least 1. It therefore follows from pq = sx thata0b0 = 0.

Now Q is an integral domain, hence Q[y] is also an integral domainby Prop. 9.2, so from a0b0 = 0 we conclude that either a0 = 0 or b0 = 0.If a0 = 0 then every term of p has x as a factor, hence p is equal to xmultiplied by some polynomial, so p ∈ (x). Similarly, if b0 = 0, thenq ∈ (x). This shows that p or q is in (x), so (x) is prime.

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To show (x, y) is a prime ideal of Q[x, y], we must assume pq ∈ (x, y),where p, q ∈ Q[x, y], and show that either p ∈ (x, y) or q ∈ (x, y). Frompq ∈ (x, y) we get pq = sx + ty for some polynomials s, t ∈ Q[x, y].This means that the constant term of pq is 0, because sx + ty is apolynomial in which every term either has x as a factor or has y as afactor. Let a0 ∈ Q be the constant term of p and let b0 ∈ Q be theconstant term of q. Then a0b0 = 0, so either a0 = 0 or b0 = 0. But ifa0 = 0 then every term of p has either x or y (or both) as a factor, andthis implies that p ∈ (x, y). Similarly, if b0 = 0 then q ∈ (x, y). Thisshows that (x, y) is prime.

Define a function ϕ : Q → Q[x, y]/(x, y) by ϕ(q) = q+(x, y) for everyq ∈ Q. It is easy to show that ϕ preserves + and · and is therefore aring homomorphism. Note that ϕ sends each rational q to the set ofpolynomials in Q[x, y] which have constant term equal to q. Now everypolynomial in Q[x, y] has a constant term, so ϕ is onto. Furthermore,polynomials with distinct constant terms are distinct, so ϕ is injective.These considerations show that Q[x, y]/(x, y) is isomorphic to Q. ButQ is a field, so the ideal (x, y) is a maximal ideal by Prop. 7.12 (whichsays, if R a commutative ring then an ideal M is maximal if and onlyif R/M is a field).

One the other hand, (x) is not a maximal ideal because (as we willshow) it is a proper subset of the maximal ideal (x, y). We have (x) ⊆(x, y) since {x} ⊆ {x, y}. To show the inclusion is proper it is enoughto note that y ∈ (x, y) but y /∈ (x) because the x-degree of y is 0 butthe x-degree of every polynomial in (x) is at least 1 (and not 0).

9.1.6 Prove that (x, y) is not a principal ideal in Q[x, y].

Let us suppose, to the contrary, that (x, y) = (p) for some polynomialp ∈ Q[x, y]. Then, since x, y ∈ (x, y) = (p), there are s, t ∈ Q[x, y] suchthat x = sp and y = tp. Then 0 = degy(x) = degy(s) + degy(p), so0 = degy(p), and 0 = degx(y) = degx(t)+degx(p), so 0 = degx(p). From0 = degy(p) = degx(p) we get deg(p) = 0 and p ∈ Q. But p ∈ (p) =(x, y), so p = fx+gy for some f, g ∈ Q[x, y], so deg(p) = deg(fx+gy) =

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min(deg(f)+deg(x), deg(g)+deg(y)) = min(deg(f)+1, deg(g)+1) ≥ 1,contradicting deg(p) = 0.

9.1.8 Let F be a field and let R = F [x, x2y, x3y2, . . . , xnyn−1, . . . ] bea subring of the polynomial ring F [x, y].

(a) Prove that the field of fractions of R and F [x, y] are the same.

Just as a review, we note that Th. 7.15 says, if R is a commutativering, ∅ 6= D ⊆ R, 0 /∈ D, D is closed under ·, and D contains no0-divisors, then there is a ring of quotients Q = D−1R ⊇ R such that

(1) Q is a commutative ring with 1,(2) D ⊆ Q×,(3) Q = {d−1r|d ∈ D, r ∈ R},(4) if D is a set of units in S ⊇ R, a commutative ring with 1

extending R, then Q is isomorphic to a subring Q′ of S extendingR, i.e., Q ∼= Q′ and R ⊆ Q′ ⊆ S.

If R is an integral domain, then D = R−{0} has the needed propertiesand D−1R is a field, called the field of quotients of R.

If F is a field, then the polynomial ring F [x, y] is an integral domain,so F [x, y] has a field of fractions obtained by choosingD = F [x, y]−{0}.Let Q1 be the field of fractions of F [x, y]. Then

R = F [x, x2y, x3y2, . . . , xnyn−1, . . . ] ⊆ F [x, y] ⊆ Q1 = D−1F [x, y]

By Th. 7.15, there is a subring Q2 of Q1 which is isomorphic to thefield of quotients of R, so

R = F [x, x2y, x3y2, . . . , xnyn−1, . . . ] ⊆ Q2 ⊆ Q1

and we need only show Q1 ⊆ Q2.

Note that x ∈ R so x ∈ Q2. Also, 0 6= x, x2y ∈ R so x, x2y ∈ Q2, butx is a unit of Q2, so y = (x−1)2x2y ∈ Q2. From x, y ∈ Q2 we get

F [x, y] ⊆ Q2.

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Consider an arbitrary element d−1p ∈ Q1, where 0 6= d ∈ F [x, y] andp ∈ F [x, y]. Then d, p ∈ Q2 and d 6= 0 and Q2 is a field, so pd−1 = p

d ∈Q2. This shows Q1 ⊆ Q2, completing the proof that Q1 = Q2.

(b) Prove that R contains an ideal that is not finitely generated.

For every n ∈ Z+ let Gn := {x, · · · , xnyn−1}, and let 〈Gn〉 be theclosure of Gn under multiplication. Then a polynomial f ∈ F [x, y] be-longs to F [x, x2y, x3y2, . . . , xnyn−1] if and only if f is a linear combina-tion of a finite subset of 〈Gn〉, that is, there is a finite set of monomialsM ⊆ 〈Gn〉 and a finite set of coefficients αm ∈ F , m ∈ M , such thatf = Σm∈Mαmm.

For every f ∈ F [x, y] let r(f) = degx(f) − degy(f). Note that ifm ∈ Gn then r(m) = 1. If f, g ∈ F [x, y] then

r(fg) = degx(fg)− degy(fg)

= degx(f) + degx(g)− (degy(f) + degy(g))

= degx(f)− degy(f) + degx(g)− degy(g)

= r(f) + r(g)

Therefore, for every m ∈ 〈Gn〉, either m ∈ Gn and r(m) = 1, or elser(m) > 1 (because m is a nontrivial product of two or more monomialsfrom Gn).

Claim xn+1yn /∈ F [x, · · · , xnyn−1].

Proof Suppose, to the contrary, that xn+1yn ∈ F [x, · · · , xnyn−1].Then there is a finite set of monomials M ⊆ 〈Gn〉 and a finite set ofcoefficients αm ∈ F , m ∈ M , such that xn+1yn = Σm∈Mαmm. Nowtwo polynomials in F [x, y] are equal if and only if they have the samecoefficients. In this case, this means that xn+1yn is one of the monomialsin M , say xn+1yn = m ∈ M , and its coefficient is αm = 1, and all theother coefficients are 0. Now m ∈M ⊆ 〈Gn〉, but r(m) = r(xn+1yn) =1, so in fact m ∈ Gn. However, this is a contradiction because xn+1yn /∈Gn.

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R is an ideal of itself, so by the following claim R contains an ideal(itself) that is not finitely generated.

Claim R is not finitely generated.

Proof Suppose, to the contrary, that G ⊆ R is a finite set of gener-ators of R. Now R is the union of a chain of subrings since

R = F [x, x2y, · · · , xnyn−1, · · · ] =⋃n∈Z+

F [x, · · · , xnyn−1].

Each element of G is in one of the subrings F [x, · · · , xnyn−1] ⊆ R, n =1, 2, 3, · · · . For each element ofG, choose such a subring containing thatelement. There are only finitely many such subrings since G is finite,and one of them, say F [x, · · · , xNyN−1] for some N ∈ Z+, contains allthe others (because every finite subset of a linearly ordered set has amaximum element). ThusG ⊆ F [x, · · · , xNyN−1], so, sinceG generatesR, we have R ⊆ F [x, · · · , xNyN−1], but F [x, · · · , xNyN−1] ⊆ R, soF [x, · · · , xNyN−1] = R. We now have a contradiction since, by theclaim above, xN+1yN /∈ F [x, · · · , xNyN−1].

9.2.3 Let f(x) be a polynomial in F [x]. Prove that F [x]/(f(x)) is afield if and only if f(x) is irreducible. [Use Proposition 7, Section 8.2]

We note that Prop. 8.7 says, “Every nonzero prime ideal in a PIDis maximal”. The proof actually shows that, in an integral domain,prime ideals are maximal among principal ideals. In a PID, all idealsare principal, so prime ideals are maximal.

We assume (although it is not explicitly mentioned in the text of theproblem) that F is a field. Then F [x] is a Euclidean domain, henceF [x] is a PID, so by Prop. 8.7, prime ideals in F [x] are maximal.Prop. 7.14 says that, in a commutative ring, every maximal ideal isprime. Therefore,

(1) in F [x] an ideal is prime if and only if it is maximal.

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Now complete the proof as follows—

F [x]/(f(x)) is a field

⇐⇒ (f(x)) is a maximal ideal of F [x] Prop. 7.12, below

⇐⇒ (f(x)) is a prime ideal of F [x] see (1) above

⇐⇒ f(x) is prime in F [x] Def. (2), p. 284

⇐⇒ f(x) is irreducible in F [x] Prop. 8.11, below

Prop. 7.12 says, “If R is a commutative ring then an ideal M is maxi-mal if and only if R/M is a field.” Prop. 8.11 says, “In a PID a nonzeroelement is a prime if and only if it is irreducible.”

Homework #02, due 1/27/10 = 9.4.1, 9.4.2, 9.4.5, 9.4.6, 9.4.7.

Additional problems recommended for study: 9.4.4, 9.4.9, 9.4.11,9.4.13, 9.4.17

9.4.1 Determine whether the following polynomials are irreduciblein the rings indicated. For those that are reducible, determine theirfactorization into irreducibles. The notation Fp denotes the finite fieldZ/pZ, p a prime.

9.4.1(a) x2 + x+ 1 in F2[x]

By Prop. 9.10 the quadratic polynomial p(x) = x2 + x + 1 is ir-reducible in F2[x] if and only if p(x) has no root in F2. This wecheck by substituting all the elements of the field F2 (there are only0 and 1) into p(x) and calculating the result modulo 2, as follows:p(0) = 02 + 0 + 1 = 1 6= 0 and p(1) = 12 + 1 + 1 = 1 + 1 + 1 = 1 6= 0.Since p(x) has no roots in F2, p(x) is irreducible.

9.4.1(b) x3 + x+ 1 in F3[x].

By Prop. 9.10 the cubic polynomial q(x) = x3+x+1 is irreducible inF3[x] if and only if p(x) has no root in F3. This we check by substitutingall the elements of the field F3 (there are only 0, 1, and 2) into p(x) andcalculating the result modulo 3, as follows: p(0) = 03 + 0 + 1 = 1 6= 0,

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p(1) = 13 + 1 + 1 = 1 + 1 + 1 = 0, and p(2) = 23 + 2 + 1 = 8 + 2 + 1 =11 = 2 6= 0. Since p(x) has root 1 in F2, p(x) has linear factor x − 1and factorization p(x) = (x− 1)(x2 +x− 1) = (x+2)(x2 +x+2) (note−1 = 2 mod 3).

9.4.1(c) x4 + 1 in F5[x].

Let p(x) = x4 + 1. By Prop. 9.10 we can show that p(x) has nolinear factor by substituting the elements 0, 1, 2, 3, 4 of F5 in p(x) andcalculating modulo 5: p(0) = 1 6= 0, p(1) = 14 + 1 = 2 6= 0, p(2) =24 + 1 = 17 = 2 6= 0, p(3) = 34 + 1 = 82 = 2 6= 0, and p(4) = 44 + 1 =256+1 = 257 = 2 6= 0. It remains to consider factorizations into monicquadratic polynomials.

Suppose p(x) = (x2 + ax+ b)(x2 + cx+ d) where a, b, c, d ∈ F5. Then

x4 + 1 = p(x) = x4 + (a+ c)x3 + (b+ ac+ d)x2 + (ad+ bc)x+ bd

so 0 = a+ c, 0 = b+ac+d, 0 = ad+ bc, and 1 = bd. From 0 = a+ c wehave c = −a, so 0 = b + a(−a) + d and 0 = ad + b(−a), so a2 = b + dand ad = ab.

If a 6= 0 then b = d by cancellation from ad = ab, so a2 = 2b andb2 = 1 from a2 = b + d and 1 = bd. Now 12 = 1, 22 = 4, 32 = 9 = 4,42 = 1, so b is 1 or 4. But then either a2 = 2 or a2 = 2(4) = 8 = 3, andneither of these equations has a solution in F5 (since 12 = 1, 22 = 4,32 = 4, 42 = 1). On the other hand, if a = 0 then 02 = 0 = b + d, sob = −d, so b2 = −bd = −1 = 4, so b = 2, d = −2, a = 0, c = 0, givinga factorization

p(x) = x4 + 1 = (x2 + 2)(x2 − 2) = (x2 + 2)(x2 + 3).

Thus p(x) is a reducible polynomial in F5[x].

9.4.1(d) x4 + 10x2 + 1 in Z[x].

This polynomial has no real-valued roots, since x4 ≥ 0 and x2 ≥ 0for all x ∈ R, hence x4 + 10x2 + 1 ≥ 1 for all x ∈ R. This shows, byProp. 9.10, that x4+10x2+1 has no linear factors in Z[x]. We considerfactoring x4 + 10x2 + 1 into quadratic polynomials. If a, b, c, d ∈ Z and

x4 + 10x2 + 1 = (x2 + ax+ b)(x2 + cx+ d)

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= x4 + (a+ c)x3 + (b+ ac+ d)x2 + (ad+ bc)x+ bd

then 0 = a+c, b+ac+d = 10, ad+bc = 0, and bd = 1. From 0 = a+cwe get c = −a, so, by ad+ bc = 0 we have ad− ab = 0, hence a = 0 ord = b.

If a = 0 then c = 0 and the factorization becomes

x4 + 10x2 + 1 = (x2 + b)(x2 + d) = x4 + (b+ d)x2 + bd

so b + d = 10 and bd = 1. These equations have no solution in Z.Indeed, we have either b = d = 1 or b = d = −1 by bd = 1, and neitherchoice satisfies the equation b+ d = 10.

If a 6= 0 then b = d and the factorization becomes

x4 + 10x2 + 1 = (x2 + ax+ b)(x2 − ax+ b)

= x4 + (2b− a2)x2 + b2

where 2b = a2 + 10 and b2 = 1. From 2b = a2 + 10 ≥ 10 we get b ≥ 5,but from b2 = 1 we get b = ±1, a contradiction. Thus there is nofactorization into quadratic factors, and x4 + 10x2 + 1 is irreducibleover Z[x].

9.4.2 Prove that the following polynomials are irreducible in Z[x].

9.4.2(a) x4 − 4x3 + 6

This polynomial is irreducible by Eisenstein’s Criterion, since theprime 2 does not divide the leading coefficient, does divide all thecoefficients of the low order terms, namely −4, 0, 0, and 6, but thesquare of 2 does not divide the constant 6.

9.4.2(b) x6 + 30x5 − 15x3 + 6x− 120

The coefficients of the low order terms are 30, 0, −15, 0, 6, and −120.These numbers are divisible by the prime 3, but 32 = 9 does not divide−120 = −23 · 3 · 5, so the polynomial is irreducible by Eisenstein’sCriterion.

9.4.2(c) x4 + 6x3 + 4x2 + 2x+ 1 [Substitute x− 1 for x.]

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Let p(x) = x4 + 6x3 + 4x2 + 2x + 1. Then to calculate p(x − 1) wefirst note that

(x− 1)4 = x4 − 4x3 + 6x2 − 4x+ 1

6(x− 1)3 = 6x3 − 18x2 + 18x− 6

4(x− 1)2 = 4x2 − 8x+ 4

2(x− 1) = 2x− 2

1 = 1

Then we add these up and get

p(x− 1) = (x− 1)4 + 6(x− 1)3 + 4(x− 1)2 + 2(x− 1) + 1

= x4 + 2x3 − 8x2 + 8x− 2

Now x4 + 2x3 − 8x2 + 8x − 2 is irreducible by Eisenstein’s Criterion,since the prime 2 divides all the lower order coefficients, but 22 =4 does not divide the constant −2. Any factorization of p(x) wouldprovide a factorization of p(x−1) as well, since if p(x) = a(x)b(x) thenx4 +2x3−8x2 +8x−2 = p(x−1) = a(x−1)b(x−1), contradicting theirreducibility of p(x− 1). Therefore p(x) is itself irreducible in Z[x].

9.4.2(d) (x+2)p−2p

x , where p is an odd prime, in Z[x].

To express (x+2)p−2p

x as a polynomial we expand (x+2)p according tothe binomial theorem. The final constant 2p cancels with −2p, so everyremaining term has x as a factor, which we cancel, leaving

xp−1 + 2

(p1

)xp−2 + 22

(p2

)xp−3 + · · ·+ 2p−1

(p

p− 1

)Every lower order coefficient in this polynomial has the form

2k(pk

)xp−k−1 = 2k · p · (p− 1) · · · · · (p− k − 1)

with 0 < k < p. Each lower order coefficient has p as a factor but doesnot have p2 as a factor, so the polynomial is irreducible by Eisenstein’sCriterion.

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9.4.5 Find all the monic irreducible polynomials of degree ≤ 3 inF2[x], and the same in F3[x].

All the monic linear (degree 1) polynomials x − a are irreducible.For F2 these irreducible linear polynomials are x and x − 1(= x + 1).For F3 these irreducible linear polynomials are x, x− 1(= x + 2), andx− 2(= x+ 1).

In F2[x] the only (monic) quadratic polynomials are x2, x2+1, x2+x,and x2 + x + 1. Obviously, x2 and x2 + x are reducible since x2 = xxand x2 + x = x(x + 1). Less obviously, x2 + 1 has 1 as a root, whichleads to the factorization x2 + 1 = (x + 1)(x + 1). Finally, x2 + x + 1has no root in F2 since 12 + 1 + 1 = 1 6= 0 and 02 + 0 + 1 = 1 6= 0, so,because it is quadratic, it is irreducible.

In F3[x] the nine monic quadratic polynomials are x2, x2 + 1, x2 + 2,x2 + x, x2 + 2x, x2 + x + 1, x2 + 2x + 1, x2 + x + 2, and x2 + 2x + 2.Obviously, x2, x2 + x, and x2 + 2x are reducible. Since the remainingpolynomials x2 + 1, x2 + 2, x2 + x + 1, x2 + 2x + 1, x2 + x + 2, andx2 + 2x+ 2 are quadratic, they are reducible if and only if they have aroot in F3, so we check each of them for roots:

x2 + 1 02 + 1 = 1 12 + 1 = 2 22 + 1 = 2x2 + 2 02 + 2 = 2 12 + 2 = 0 22 + 2 = 0x2 + x+ 1 02 + 0 + 1 = 1 12 + 1 + 1 = 0 22 + 2 + 1 = 1x2 + 2x+ 1 02 + 2 · 0 + 1 = 1 12 + 2 · 1 + 1 = 1 22 + 2 · 2 + 1 = 0x2 + x+ 2 02 + 0 + 2 = 2 12 + 1 + 2 = 1 22 + 2 + 2 = 2x2 + 2x+ 2 02 + 2 · 0 + 2 = 2 12 + 2 · 1 + 2 = 2 22 + 2 · 2 + 2 = 1

Since x2 + 2 has 1 and 2 as roots, we get the factorization x2 + 2 =(x− 1)(x− 2) = (x+ 2)(x+ 1). Since x2 + x+ 1 has 1 as a (multiple)root, we get a factorization x2 +x+1 = (x−1)(x−1) = (x+2)(x+2).Finally, x+2x+1 = (x+1)(x+1). The remaining polynomials, whichare the only irreducible quadratic polynomials in F3[x], are x2+1 (usedbelow), x2 + x+ 2, and x2 + 2x+ 2.

There are quite a few monic cubic polynomials, so I thought it best towrite some code in GAP to do the computations. The following GAP

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code computes, counts, and prints the irreducible cubic polynomialsover the field Fp, for p ranging over the first 12 primes.

irrcubic:=function(p)

local f,c,t,roots,irr;

f:=function(c,p)

return (c[4]^3+c[1]*c[4]^2+c[2]*c[4]+c[3]) mod p;

end;

t:=Tuples([0..p-1],4);

roots:=Set(Filtered(t,c->f(c,p)=0),c->c{[1..3]});

irr:=Difference(Tuples([0..p-1],3),roots);

Print("The number of irreducible monic cubic ");

Print("polynomials over F_",p," is\t",Size(irr));

# Print("The irreducible monic cubic ");

# Print("polynomials over F_",p," are\n");

# for c in irr do

# Print("x^3");

# if c[1]=1 then Print(" + x^2");fi;

# if c[1] in [2..p-1] then Print(" + ",c[1],"x^2");fi;

# if c[2]=1 then Print(" + x");fi;

# if c[2] in [2..p-1] then Print(" + ",c[2],"x");fi;

# if c[3]=1 then Print(" + 1");fi;

# if c[3] in [2..p-1] then Print(" + ",c[3]);fi;

# Print("\n");

# od;

Print("\n");

return irr;

end;

LogTo("roots.output");

for p in Primes{[1..12]} do

irrcubic(p);

od;

Running the GAP code without printing the polynomials produces thedata in the following table, which shows the number np of irreducible

14

monic cubic polynomials over Fp for the first 12 primes.

p 2 3 5 7 11 13 17 19 23 29 31 37np 2 8 40 112 440 728 1632 2280 4048 8120 9920 16872

By uncommenting the Print commands we get lists of irreducible moniccubic polynomials. The 2 irreducible monic cubic polynomials in F2[x]are

x3 + x+ 1 x3 + x2 + 1

The 8 irreducible monic cubic polynomials over F3[x] are

x3 + 2x+ 1 x3 + x2 + 2x+ 1

x3 + 2x+ 2 x3 + 2x2 + 1

x3 + x2 + 2 x3 + 2x2 + x+ 1

x3 + x2 + x+ 2 x3 + 2x2 + 2x+ 2

The 40 irreducible monic cubic polynomials over F5 are

x3 + x+ 1 x3 + 2x2 + 2x+ 2

x3 + x+ 4 x3 + 2x2 + 2x+ 3

x3 + 2x+ 1 x3 + 2x2 + 4x+ 2

x3 + 2x+ 4 x3 + 2x2 + 4x+ 4

x3 + 3x+ 2 x3 + 3x2 + 2

x3 + 3x+ 3 x3 + 3x2 + 4

x3 + 4x+ 2 x3 + 3x2 + x+ 1

x3 + 4x+ 3 x3 + 3x2 + x+ 2

x3 + x2 + 1 x3 + 3x2 + 2x+ 2

x3 + x2 + 2 x3 + 3x2 + 2x+ 3

x3 + x2 + x+ 3 x3 + 3x2 + 4x+ 1

x3 + x2 + x+ 4 x3 + 3x2 + 4x+ 3

x3 + x2 + 3x+ 1 x3 + 4x2 + 3

15

x3 + x2 + 3x+ 4 x3 + 4x2 + 4

x3 + x2 + 4x+ 1 x3 + 4x2 + x+ 1

x3 + x2 + 4x+ 3 x3 + 4x2 + x+ 2

x3 + 2x2 + 1 x3 + 4x2 + 3x+ 1

x3 + 2x2 + 3 x3 + 4x2 + 3x+ 4

x3 + 2x2 + x+ 3 x3 + 4x2 + 4x+ 2

x3 + 2x2 + x+ 4 x3 + 4x2 + 4x+ 4

9.4.6 Construct fields of each of the following orders: (a) 9, (b) 49,(c) 8, (d) 81. (you may exhibit these as F [x]/(f(x)) for some F andf). [Use Exercises 2 and 3 in Section 2.]

By Exer. 9.2.3, F [x]/(f(x)) is a field when f ∈ F [x] is irreducible,and by Exer. 9.2.2, F [x]/(f(x)) has qn elements if |F | = q and deg(f) =n.

(a) To get a field of order 9, we choose F = F3 so that q = |F3| = 3,and we choose an irreducible quadratic, so that n = 2 and the orderof the field F [x]/(f(x)) is qn = 32 = 9, as desired. For f we mayuse x2 + 1, which is an irreducible quadratic polynomial in F3[x] byExer. 9.4.5, solved above.

(b) To get a field of order 49, we choose F = F7 so that q = |F7| = 7,and we let f be any irreducible quadratic in F7[x]. Then n = 2 and theorder of the field F [x]/(f(x)) is qn = 72 = 49. One choice that worksis f(x) = x2 + 1. By the way, the quadratic irreducible polynomials inF7[x] are

x2 + 1 x2 + 2 x2 + 4

x2 + x+ 3 x2 + x+ 4 x2 + x+ 6

x2 + 2x+ 2 x2 + 2x+ 3 x2 + 2x+ 5

x2 + 3x+ 1 x2 + 3x+ 5 x2 + 3x+ 6

16

x2 + 4x+ 1 x2 + 4x+ 5 x2 + 4x+ 6

x2 + 5x+ 2 x2 + 5x+ 3 x2 + 5x+ 5

x2 + 6x+ 3 x2 + 6x+ 4 x2 + 6x+ 6

(c) To get a field of order 8, we choose F = F2 so that q = |F2| = 2,and we choose an irreducible cubic for f , so that n = 3 and the orderof the field F [x]/(f(x)) is qn = 23 = 8. For f we may use x3 + x+ 1 orx3 +x2 +1 (the two irreducible monic cubics in F2[x], as shown above).

(d) To get a field of order 81, we choose F = F3 so that q = |F3| = 3,and we choose an irreducible quartic polynomial f in F3[x], so thatn = 4 and the order of the field F [x]/(f(x)) is qn = 34 = 81. Letf(x) = x4 +x+2. Then f has no linear factors because it has no rootsin F3, since 04 + 0 + 2 = 2, 14 + 1 + 2 = 1, and 24 + 2 + 2 = 2.

Suppose f(x) = (x2 + ax+ b)(x2 + cx+ d) where a, b, c, d ∈ F3. Then

x4 + x+ 2 = x4 + (a+ c)x3 + (b+ ac+ d)x2 + (ad+ bc)x+ bd,

which implies that 0 = a+ c, 0 = b+ ac+ d, 1 = ad+ bc, and 2 = bd.From 0 = a+ c we have c = −a. Substituting −a for c in the equations0 = b+ ac+ d and 1 = ad+ bc (and simplifying) yields a2 = b+ d and1 = a(d − b). By the latter equation, a = a2(d − b), so by a2 = b + dwe have a = (d − b)(b + d) = d2 − b2. From bd = 1 we know b 6= 0and d 6= 0. The only nonzero element that is a square in F3 is 1, since12 = 1 and 22 = 1 in F3. Therefore a = b2 − d2 = 1− 1 = 0, but then1 = a(d − b) = 0(d − b) = 0, a contradiction. This shows that f isactually irreducible, so F3[x]/(f) is a field with 81 elements in it.

9.4.7 Show that R[x]/(x2 + 1) is isomorphic to the field of complexnumbers.

Let p(x) = x2 + 1 and let P = (x2 + 1) be the principal ideal ofR[x] generated by p(x). Obviously p(x) has no real roots since, forevery r ∈ R, r2 + 1 ≥ 1. Since p(x) is quadratic and has no realroots, p(x) is irreducible in R[x] by Prop. 9.10. Therefore R[x]/P isa field by Exer. 9.2.3 (or the upcoming Prop. 9.15). The elementx = x + P ∈ R[x]/P is a solution to x2 = −1 and plays the same role

17

as the imaginary i =√−1 in C, because x2 = (x + P )2 = x2 + P =

x2 − (x2 + 1) + P = −1 + P .

The elements of R[x]/P are cosets of the form a+ bx+ P , a, b,∈ R.Define a map ϕ : R[x]/P → C by ϕ(a+ bx+P ) = a+ b

√−1. To show

ϕ is well-defined, let us assume that a+ bx+ P = c+ dx+ P for somea, b, c, d ∈ R. Then a−c+(b−d)x ∈ P , hence a−c+(b−d)x = p(x)q(x)for some q(x) ∈ R[x]. If b− d 6= 0 then

1 = deg(a− c+ (b− d)x) = deg(p(x)q(x))

= deg(p(x)) + deg(q(x)) = 2 + deg(q(x)) ≥ 2,

a contradiction, so b = d. But then a−c ∈ P , hence a−c = p(x)q′(x) forsome q′(x) ∈ R[x]. If a− c 6= 0 then 0 = deg(a− c) = deg(p(x)q′(x)) =deg(p(x)) + deg(q′(x)) = 2 + deg(q′(x)) ≥ 2, a contradiction, so a = c.Thus, each element of R[x]/P has the form a + bx + P for uniquelydetermined reals a, b ∈ R. Next are calculations that show ϕ is a ringhomomorphism because it preserves differences and products.

ϕ((a+ bx+ P )− (c+ dx+ P ))

= ϕ((a− c) + (b− d)x+ P )

= (a− c) + (b− d)√−1

= (a+ b√−1)− (c+ d

√−1)

= ϕ(a+ bx+ P )− ϕ(c+ dx+ P )

ϕ((a+ bx+ P )(c+ dx+ P ))

= ϕ((a+ bx)(c+ dx) + P )

= ϕ(ac+ (ad+ bc)x+ bdx2 + P )

= ϕ(ac+ (ad+ bc)x+ bdx2 + (−bd)(x2 + 1) + P )

= ϕ(ac+ (ad+ bc)x+ bdx2 − bdx2 − bd+ P )

= ϕ(ac− bd+ (ad+ bc)x+ P )

= ac− bd+ (ad+ bc)√−1

= ac+ (ad+ bc)√−1 + bd(

√−1)2

18

= (a+ b√−1)(c+ d

√−1)

= ϕ(a+ bx+ P )ϕ(c+ dx+ P )

Finally, to see that ϕ is injective, we note that kernel of ϕ is trivialsince if ϕ(a + bx + P ) = 0 then a + b

√−1 = 0, hence a = b = 0, so

a+ bx+ P = P = 0 ∈ R[x]/P .

Homework #03, due 2/3/10 = 9.4.3, 9.4.8, 9.4.14, 9.4.18 (use9.4.17), 9.5.2.

Additional problems recommended for study: 9.4.10, 9.4.12, 9.4.16,9.4.20, 9.5.1, 9.5.3, 9.5.4, 9.5.7

9.4.3 Show that the polynomial (x − 1)(x − 2) · · · (x − n) − 1 isirreducible over Z[n] for all n ≥ 1. [If the polynomial factors considerthe values of the factors at x = 1, 2, . . . , n.]

Let p(x) = (x−1)(x−2) · · · (x−n)−1. To show that p(x) is irreduciblein Z[x] for all n ≥ 1, we assume p(x) = a(x)b(x) for some (wlog monic)polynomials a(x), b(x) ∈ Z[x], and we will show this factorization istrivial.

For all k ∈ {1, · · · , n} we have p(k) = −1 = a(k)b(k) so, since a(k)and b(k) are integers, they both must be 1 or −1, and they can’t havethe same sign, so a(x) 6= b(x). Thus we have 1 = (a(k))2 = (b(k))2

for all k ∈ {1, · · · , n}. Since deg(p) = n = deg(a(x)) + deg(b(x)), thedegrees of a(x) and b(x) cannot both be strictly larger than n

2 , so wemay assume one of them has degree no more than n

2 . We assume wlogthat deg(a(x)) ≤ n

2 , hence deg(a2(x)) ≤ n. Therefore p(x) + a2(x) is apolynomial of degree n that has roots 1, · · · , n. These are distinct roots,so by Prop. 9.9 p(x)+a2(x) has n distinct linear factors x−1, x−2, . . . ,x−n. Now p(x)+a2(x) has factor x−1, so p(x)+a2(x) = (x−1)q1(x)for some q1(x) ∈ Z[x], but p(x) + a2(x) has roots 2, · · · , n and thesearen’t roots of x − 1, so they’re roots of q1(x), so q1(x) has a linearfactor x− 2, hence q1(x) = (x− 2)q2(x), and so on. Thus

p(x) + a2(x) = (x− 1)q1(x)

19

= (x− 1)(x− 2)q2(x)

= · · · = (x− 1)(x− 2) · · · (x− n)

= p(x) + 1

Note the final equation gives a2(x) = 1, so the degree of a2(x) is actually0, hence a(x) is a constant in Z whose square is 1, so a(x) = ±1 andp(x) = ∓b(x). Therefore the factorization is trivial.

9.4.8 Prove that K1 = F11[x]/(x2 + 1) and K2 = F11/(y

2 + 2y + 2)are both fields with 121 elements. Prove that the map which sendsthe element p(x) of K1 to the element p(y + 1) of K2 (where p is anypolynomial with coefficients in F11) is well-defined and gives a ring(hence field) isomorphism from K1 onto K2.

First we note that x2 +1 is quadratic and has no roots in the groundfield F11 since, calculating modulo 11, we have

(0)2 + 1 = 1 (1)2 + 1 = 2 (2)2 + 1 = 5

(3)2 + 1 = 10 (4)2 + 1 = 6 (5)2 + 1 = 4

(6)2 + 1 = 4 (7)2 + 1 = 6 (8)2 + 1 = 10

(9)2 + 1 = 5 (10)2 + 1 = 2

It follows by Prop. 9.10 that x2+1 is irreducible in F11[x]. The elementsof the quotient field K1 = F11[x]/(x

2 + 1) are ax + b + P , where P =(x2 + 1) and a, b ∈ F11.

By Exer. 9.2.3, F [x]/(f(x)) is a field when f ∈ F [x] is irreducible,and by Exer. 9.2.2, F [x]/(f(x)) has qn elements if |F | = q and deg(f) =n. In this case we have F = F11 so q = 11, and f(x) = x2 +1 so n = 2,and the number of elements in K1 is therefore qn = 112 = 121. (Part ofthe proof goes like this: if a+bx+P = c+dx+P for some a, b, c, d ∈ F11,then a− c+ (b− d)x ∈ P , hence a− c+ (b− d)x = p(x)q(x) for someq(x) ∈ F11[x]. If b− d 6= 0 then

1 = deg(a− c+ (b− d)x) = deg(p(x)q(x))

= deg(p(x)) + deg(q(x)) = 2 + deg(q(x)) ≥ 2,

20

a contradiction, so b = d. But then a−c ∈ P , hence a−c = p(x)q′(x) forsome q′(x) ∈ F11[x]. If a−c 6= 0 then 0 = deg(a−c) = deg(p(x)q′(x)) =deg(p(x)) + deg(q′(x)) = 2 + deg(q′(x)) ≥ 2, a contradiction, so a = c.Thus, each element of F11[x]/P has the form a + bx + P for uniquelydetermined a, b ∈ F11.)

Define a map ϕ from F11[x] to F11[y] by ϕ(p(x)) = p(y+ 1) for everyp(x) ∈ F11[x]. Then ϕ is a ring homomorphism because

ϕ(p(x)q(x)) = p(y + 1)q(y + 1) = ϕ(p(x))ϕ(q(x))

ϕ(p(x) + q(x)) = p(y + 1) + q(y + 1) = ϕ(p(x)) + ϕ(q(x))

The map ϕ can be described as “substitute y + 1 for x”. This maphas an inverse ϕ−1 which may be described as “substitute x − 1 fory”, that is, for all q(y) ∈ F11[y] we let ψ(q(y)) = q(x − 1), so thatψ(ϕ(p(x)) = ψ(p(y + 1)) = p((x − 1) + 1) = p(x) and ϕ(ψ(q(y)) =ϕ(q(x − 1)) = q((y + 1) − 1) = q(y), so ψ = ϕ−1. This shows thatϕ is actually a ring isomorphism. Note that ϕ(x2 + 1) = (y + 1)2 +1 = y2 + 2y + 2. Corresponding elements in isomorphic rings generatecorresponding ideals, so the image of the principal ideal (x2+1) ⊆ F11[x]under ϕ is the ideal (y2 + 2y+ 1) ⊆ F11[y]. Consequently the quotientsby corresponding ideals are isomorphic, so

K1 = F11[x]/(x2 + 1) ∼= F11/(y

2 + 2y + 2) = K2.

9.4.14 Factor each of the two polynomials x8 − 1 and x6 − 1 intoirreducibles over each of the following rings. (a) Z (b) Z/2Z (c) Z/3Z.

(a) Z: First we note that

x8 − 1 = (x4)2 − 1 = (x4 + 1)(x4 − 1)

= (x4 + 1)(x2 + 1)(x2 − 1)

= (x4 + 1)(x2 + 1)(x+ 1)(x− 1)

Next we show that the factors x4 + 1, x2 + 1, x + 1, and x − 1 areall irreducible in Z[x]. In Example (3), p. 310, following Cor. 14 inSection 9.4, the polynomial p(x) = x4 + 1 ∈ Z[x] was shown to beirreducible in two steps. First, by the Eisenstein Criterion, applied with

21

prime 2 to the polynomial p(x+1) = (x+1)4+1 = x4+4x3+6x2+4x+2,we know that p(x + 1) is irreducible in Z[x]. Second, if there were anontrivial factorization p(x) = a(x)b(x), with a(x), b(x) ∈ Z[x], 1 ≤deg(a(x)) < 4, and 1 ≤ deg(b(x)) < 4, then we would have p(x+ 1) =a(x+1)b(x+1) where 1 ≤ deg(a(x+1)) < 4 and 1 ≤ deg(b(x+1)) < 4,contradicting the irreducibility of p(x + 1). The factor x2 + 1 has noroots in Z and is quadratic, hence is irreducible by Prop. 9.10. Finally,the linear factors x+ 1 and x− 1 are irreducible.

A factorization of x6 − 1 into irreducibles in Z[x] is

x6 − 1 = (x3)2 − 1 = (x3 + 1)(x3 − 1)

= (x2 − x+ 1)(x+ 1)(x2 + x+ 1)(x− 1)

The linear factors x+1 and x−1 are irreducible. Let p(x) = x2−x+1.Then

p(x+ 1) = (x+ 1)2 − (x+ 1) + 1

= x2 + 2x+ 1− x− 1 + 1 = x2 + x+ 1

p(x+ 2) = (x+ 2)2 − (x+ 2) + 1

= x2 + 4x+ 4− x− 2 + 1 = x2 + 3x+ 3

Now x2 +3x+3 is irreducible in Z[x] by the Eisenstein Criterion, sinceprime 3 does not divide the leading coefficient 1, but does divide thelower order coefficients (3 and 3), and its square 32 = 9 does not dividethe constant 3. If p(x) had a nontrivial factorization into noncon-stant polynomials, say p(x) = a(x)b(x), this would yield a nontrivialfactorization p(x+ 2) = a(x+ 2)b(x+ 2), contradicting the irreducibil-ity of p(x + 2), so p(x) is irreducible. A nontrivial factorization ofp(x+1), say p(x+1) = c(x)d(x), would yield a nontrivial factorizationp(x+2) = c(x+1)d(x+1), contradicting the irreducibility of p(x+2),so p(x+1) is also irreducible. This shows that the two quadratic factorsin the factorization of x6 − 1 are irreducible in Z[x].

22

(b): Z/2Z. This ring is the 2-element field F2, whose only elementsare 0 and 1, and in which we have −1 = 1, so the previous factorizationscan be written

x8 − 1 = (x4 + 1)(x2 + 1)(x+ 1)(x+ 1)

x6 − 1 = (x2 + x+ 1)(x+ 1)(x2 + x+ 1)(x+ 1)

Note that (x+1)(x+1) = x2 +2x+1 = x2 +1, and (x2 +1)(x2 +1) =x4 + 2x2 + 1 = x4 + 1, so we can continue to factor x8 − 1, obtaining

x8 − 1 = (x4 + 1)(x2 + 1)(x+ 1)(x+ 1) = (x+ 1)8

On the other hand, x2+x+1 is irreducible in F2[x], so a factorization ofx6− 1 into irreducibles over Z/2Z is the same as it was over Z, namely

x6 − 1 = (x2 + x+ 1)(x+ 1)(x2 + x+ 1)(x+ 1)

GAP that confirms these results:

gap> r:=PolynomialRing(GF(2),["x"]);

GF(2)[x]

gap> x:=IndeterminatesOfPolynomialRing(r)[1];

x

gap> Factors(x^8-1);Factors(x^6-1);

[ x+Z(2)^0, x+Z(2)^0, x+Z(2)^0, x+Z(2)^0,

x+Z(2)^0, x+Z(2)^0, x+Z(2)^0, x+Z(2)^0 ]

[ x+Z(2)^0, x+Z(2)^0, x^2+x+Z(2)^0, x^2+x+Z(2)^0 ]

(c): Z/3Z. This ring is the 3-element field F3, in which −1 = 2. Thefactorizations of x8 − 1 and x6 − 1 into irreducibles are

x8 − 1 = (x+ 1)(x+ 2)(x2 + 1)(x2 + x+ 2)(x2 + 2x+ 2)

x6 − 1 = (x+ 1)(x+ 1)(x+ 1)(x+ 2)(x+ 2)(x+ 2)

These results were obtained by the following calculation in GAP:

gap> r:=PolynomialRing(GF(3),["x"]);

GF(3)[x]

gap> IndeterminatesOfPolynomialRing(r);

[ x ]

23

gap> x:=IndeterminatesOfPolynomialRing(r)[1];

x

gap> Factors(x^8-1);Factors(x^6-1);

[ x+Z(3)^0, x-Z(3)^0, x^2+Z(3)^0, x^2+x-Z(3)^0, x^2-x-Z(3)^0 ]

[ x+Z(3)^0, x+Z(3)^0, x+Z(3)^0, x-Z(3)^0, x-Z(3)^0, x-Z(3)^0 ]

9.4.18 Show that 6x5 +14x3−21x+35 and 18x5−30x2 +120x+360are irreducible in Q[x].

We may use Exer. 9.4.17, which is the extended version of the Eisen-stein Criterion that was proved in class.

Let p(x) = 6x5 +14x3− 21x+35. Then p(x) is irreducible in Z[x] byEisenstein’s Criterion using prime 7, which does not divide the leadingcoefficient 6, does divide the remaining coefficients 14, −21, and 35,but its square 72 = 49 does not divide the constant 35. Suppose p(x)is reducible in Q[x], say p(x) = A(x)B(x) for some A(x), B(x) ∈ Q[x].Note that Q is the field of fractions of Z, so by Prop. 9.5 (Gauss’sLemma) there are rationals r, s ∈ Q such that p(x) = rA(x)sB(x),rA(x) ∈ Z[x], and sB(x) ∈ Z[x], contradicting the irreducibility ofp(x) in Z[x].

Let q(x) = 18x5 − 30x2 + 120x + 360. Then q(x) is irreducible byEisenstein’s Criterion using prime 5, which does not divide the leadingcoefficient 18, does divide the remaining coefficients −30, 120, and 360,but 52 = 25 does not divide the constant 360. By Prop. 9.5 (Gauss’sLemma), q(x) is also irreducible in Q[x], as was argued for p(x) above.

9.5.2 For each of the fields in Exercise 6 of Section 4 exhibit a gen-erator for the (cyclic) multiplicative group of nonzero elements.

Exer. 9.4.6 asks for fields of order (a) 9, (b) 49, (c) 8, and (d) 81,exhibited in the form F [x]/(f(x)) where F is a field and f ∈ F [x].

(a) For a field F [x]/(f(x)) of order 9, let F = F3 and let f be theirreducible quadratic polynomial f(x) = x2 + 1 ∈ F3[x]. The elementsF [x]/(f(x)) have the form ax + b + (f), where a, b ∈ {0, 1, 2}. Themultiplicative group of F [x]/(f(x)) is generated by x + 2 + (f), as isshown in the table of calculations below, where we write simply “ax+b”

24

instead of “ax+b+(f)”. With this notational convention, observe thatin F [x]/(f(x)) we have x2 + 1 = 0 (more precisely, x2 + 1 + (f) = (f)),so x2 = −1 = 2.

(x+ 2)1 = x+ 2

(x+ 2)2 = x2 + 4x+ 4 = 2 + x+ 1 = x

(x+ 2)3 = (x+ 2)x = x2 + 2x = 2x+ 2

(x+ 2)4 = (2x+ 2)(x+ 2) = 2x2 + 6x+ 4 = 2(2) + 1 = 2

(x+ 2)5 = 2(x+ 2) = 2x+ 1

(x+ 2)6 = (2x+ 1)(x+ 2) = 2x2 + 5x+ 2 = 2(2) + 2x+ 2 = 2x

(x+ 2)7 = 2x(x+ 2) = 2x2 + 4x = 2(2) + x = x+ 1

(x+ 2)8 = (x+ 1)(x+ 2) = x2 + 3x+ 2 = 2 + 2 = 1

(b) To get a field F [x]/(f(x)) of order 49, we choose F = F7 so thatq = |F7| = 7, and we let f be any one of the 21 monic irreducible qua-dratic polynomials in F7[x]. Then n = 2 and the order of F [x]/(f(x))is qn = 72 = 49. The 21 monic irreducible quadratic polynomialsin F7[x] are organized in the following table according to the 7 linearsubstitutions, “substitute x+ n for x”.

ϕ(x) x2 + 1 x2 + 2 x2 + 4ϕ(x+ 1) x2 + 2x+ 2 x2 + 2x+ 3 x2 + 2x+ 5ϕ(x+ 2) x2 + 4x+ 5 x2 + 4x+ 6 x2 + 4x+ 1ϕ(x+ 3) x2 + 6x+ 3 x2 + 6x+ 4 x2 + 6x+ 6ϕ(x+ 4) x2 + x+ 3 x2 + x+ 4 x2 + x+ 6ϕ(x+ 5) x2 + 3x+ 5 x2 + 3x+ 6 x2 + 3x+ 1ϕ(x+ 6) x2 + 5x+ 2 x2 + 5x+ 3 x2 + 5x+ 5

Claim: F7/(x2 + 1) is generated by x+ 3 = x + 3 + (x2 + 1). From

x2 + 1 = 0 we have x2 = −1 = 6, so we can derive a rule for raisingx+ 3 to powers in F7/(x

2 + 1). If ax+ b is a power of x+ 3, we get thenext power as follows:

(ax+ b)(x+ 3) = ax2 + (3a+ b)x+ 3b = (3a+ b)x+ (6a+ 3b)

25

Using GAP, I computed the powers of x+3, and obtained these results,which prove that x+ 3 is indeed a generator.

(x+ 3)1 = x+ 3 (x+ 3)2 = 6x+ 1

(x+ 3)3 = 5x+ 4 (x+ 3)4 = 5x

(x+ 3)5 = x+ 2 (x+ 3)6 = 5x+ 5

(x+ 3)7 = 6x+ 3 (x+ 3)8 = 3

(x+ 3)9 = 3x+ 2 (x+ 3)10 = 4x+ 3

(x+ 3)11 = x+ 5 (x+ 3)12 = x

(x+ 3)13 = 3x+ 6 (x+ 3)14 = x+ 1

(x+ 3)15 = 4x+ 2 (x+ 3)16 = 2

(x+ 3)17 = 2x+ 6 (x+ 3)18 = 5x+ 2

(x+ 3)19 = 3x+ 1 (x+ 3)20 = 3x

(x+ 3)21 = 2x+ 4 (x+ 3)22 = 3x+ 3

(x+ 3)23 = 5x+ 6 (x+ 3)24 = 6

(x+ 3)25 = 6x+ 4 (x+ 3)26 = x+ 6

(x+ 3)27 = 2x+ 3 (x+ 3)28 = 2x

(x+ 3)29 = 6x+ 5 (x+ 3)30 = 2x+ 2

(x+ 3)31 = x+ 4 (x+ 3)32 = 4

(x+ 3)33 = 4x+ 5 (x+ 3)34 = 3x+ 4

(x+ 3)35 = 6x+ 2 (x+ 3)36 = 6x

(x+ 3)37 = 4x+ 1 (x+ 3)38 = 6x+ 6

(x+ 3)39 = 3x+ 5 (x+ 3)40 = 5

(x+ 3)41 = 5x+ 1 (x+ 3)42 = 2x+ 5

(x+ 3)43 = 4x+ 6 (x+ 3)44 = 4x

(x+ 3)45 = 5x+ 3 (x+ 3)46 = 4x+ 4

26

(x+ 3)47 = 2x+ 1 (x+ 3)48 = 1

(x+ 3)49 = x+ 3

(c) For a field F [x]/(f(x)) of order 8, let F = F2 and let f bethe irreducible cubic polynomial f(x) = x3 + x + 1 ∈ F2[x]. Themultiplicative group of F3[x]/(x

3 +x+1) has 7 elements, and since 7 isprime it follows that every non-zero element of the field will generate themultiplicative group. In particular, x generates, as is explicitly shownby the following calculations. First note that since x3 + x+ 1 = 0, weget x3 = −x− 1 = x+ 1 (all modulo 2), so

x1 = x

x2 = x2

x3 = x+ 1

x4 = x2 + x

x5 = x3 + x2 = x2 + x+ 1

x6 = x3 + x2 + x = x2 + 1

x7 = x3 + x2 = x+ 1 + x = 1

(d) For a field F [x]/(f(x)) of order 81, let F = F3 and let f be theirreducible quartic polynomial f(x) = x4 + x + 2 ∈ F [x] (shown tobe irreducible in Exer. 9.4.6(d). Let P = (f) ⊆ F [x]. The elementsF [x]/(f(x)) have the form ax3 + bx2 + cx + d + P , where a, b, c, d ∈{0, 1, 2}. The multiplicative group of F [x]/(f(x)) is generated by x+P .The 80 powers of x + P are shown in the table below, which has only“ax3 + bx2 + cx+ d” instead of “ax3 + bx2 + cx+ d+ P”.

x1 = x x41 = 2x

x2 = x2 x42 = 2x2

x3 = x3 x43 = 2x3

x4 = 2x+ 1 x44 = x+ 2

27

x5 = 2x2 + x x45 = x2 + 2x

x6 = 2x3 + x2 x46 = x3 + 2x2

x7 = x3 + x+ 2 x47 = 2x3 + 2x+ 1

x8 = x2 + x+ 1 x48 = 2x2 + 2x+ 2

x9 = x3 + x2 + x x49 = 2x3 + 2x2 + 2x

x10 = x3 + x2 + 2x+ 1 x50 = 2x3 + 2x2 + x+ 2

x11 = x3 + 2x2 + 1 x51 = 2x3 + x2 + 2

x12 = 2x3 + 1 x52 = x3 + 2

x13 = 2x+ 2 x53 = x+ 1

x14 = 2x2 + 2x x54 = x2 + x

x15 = 2x3 + 2x2 x55 = x3 + x2

x16 = 2x3 + x+ 2 x56 = x3 + 2x+ 1

x17 = x2 + 2 x57 = 2x2 + 1

x18 = x3 + 2x x58 = 2x3 + x

x19 = 2x2 + 2x+ 1 x59 = x2 + x+ 2

x20 = 2x3 + 2x2 + x x60 = x3 + x2 + 2x

x21 = 2x3 + x2 + x+ 2 x61 = x3 + 2x2 + 2x+ 1

x22 = x3 + x2 + 2 x62 = 2x3 + 2x2 + 1

x23 = x3 + x+ 1 x63 = 2x3 + 2x+ 2

x24 = x2 + 1 x64 = 2x2 + 2

x25 = x3 + x x65 = 2x3 + 2x

x26 = x2 + 2x+ 1 x66 = 2x2 + x+ 2

x27 = x3 + 2x2 + x x67 = 2x3 + x2 + 2x

x28 = 2x3 + x2 + 2x+ 1 x68 = x3 + 2x2 + x+ 2

x29 = x3 + 2x2 + 2x+ 2 x69 = 2x3 + x2 + x+ 1

28

x30 = 2x3 + 2x2 + x+ 1 x70 = x3 + x2 + 2x+ 2

x31 = 2x3 + x2 + 2x+ 2 x71 = x3 + 2x2 + x+ 1

x32 = x3 + 2x2 + 2 x72 = 2x3 + x2 + 1

x33 = 2x3 + x+ 1 x73 = x3 + 2x+ 2

x34 = x2 + 2x+ 2 x74 = 2x2 + x+ 1

x35 = x3 + 2x2 + 2x x75 = 2x3 + x2 + x

x36 = 2x3 + 2x2 + 2x+ 1 x76 = x3 + x2 + x+ 2

x37 = 2x3 + 2x2 + 2x+ 2 x77 = x3 + x2 + x+ 1

x38 = 2x3 + 2x2 + 2 x78 = x3 + x2 + 1

x39 = 2x3 + 2 x79 = x3 + 1

x40 = 2 x80 = 1

Homework #04, due 2/10/10 = 10.1.1, 10.1.8, 10.1.9, 10.1.10.

Additional problems recommended for study: 10.1.2, 10.1.3, 10.1.4,10.1.5, 10.1.6, 10.1.7, 10.1.11, 10.1.12, 10.1.15

10.1.1 Assume R is a ring with 1 and M is a left R-module. Provethat 0m = 0 and (−1)m = −m for all m ∈M .

Let m ∈ M . Since M is a group with identity element 0, we have0 = 0+0 and 0+0m = 0m. Then 0+0m = 0m = (0+0)m = 0m+0mby the module axiom (r+ s)m = rm+ sm. Since M is a group, we cancancel 0m from both sides, leaving 0 = 0m.

Since R is a ring with 1, we have 0 = 1+(−1), so 0m = (1+(−1))m =1m + (−1)m, but 1m = m by one of the module axioms and 0m = 0by the first part of this problem, so 0 = m + (−1)m. By adding −mto both sides we get −m = (−1)m.

10.1.8 Assume R is a ring with 1 and M is a left R-module. Anelement m of the R-module M is called a torsion element if rm = 0 for

29

some nonzero element r ∈ R. The set of torsion elements is denotedby

Tor(M) := {m ∈M |rm = 0 for some nonzero r ∈ R}

(a) Prove that if R is an integral domain then Tor(M) is a submoduleof M (called the torsion submodule of M).

Note that, by hypothesis, R is commutative, has 1, and has no zerodivisors. Also, for every r ∈ R we have 0+r0 = r0 = r(0+0) = r0+r0,so, cancelling r0 from both sides leaves 0 = r0.

Let m,n ∈ Tor(M). Then there are nonzero r, s ∈ R such thatrm = 0 = sn. Since R is an integral domain and neither r nor s iszero, we conclude that rs 6= 0. Then m+ n ∈ Tor(M) because

rs(m+ n) = (rs)m+ (rs)n

= (sr)m+ (rs)n R is commutative

= s(rm) + r(sn) by a module axiom

= s0 + r0 since rm = 0 = sn

= 0 + 0 proved above

= 0

This shows Tor(M) is closed under +. Also, for every t ∈ R we have

r(tm) = (rt)m by a module axiom

= (tr)m R is commutative

= t(rm)

= t0 since rm = 0

= 0 proved above

so tm ∈ Tor(M). Since Tor(M) is closed under + and the action of R,it is a submodule.

(b) Give an example of a ring R and an R-module M such thatTor(M) is not a submodule of M . [Consider the torsion elements inthe R-module R.]

30

Let R = Z/6Z and n = n+ 6Z for every n ∈ Z. Then

2 · 3 = (2 + 6Z)(3 + 6Z) = 2 · 3 + 6Z = 6 + 6Z = 6Z = 0.

This equation implies that if we consider R as a module over itself, wehave 3 ∈ Tor(R), but we also have 3 · 2 = 0, so 2 ∈ Tor(M) as well.Note that 2+3 = 5 and 5 /∈ Tor(M) because 1 ·5 = 5 6= 0, 2 ·5 = 4 6= 0,3 · 5 = 3 6= 0, 4 · 5 = 2 6= 0, and 5 · 5 = 1 6= 0. Thus 2 and 3 are inTor(M), but their sum in not in Tor(M). This shows that Tor(M) isnot a submodule.

(c) If R has zero divisors show that every nonzero R-module hasnonzero torsion elements.

Assume R is a commutative ring with 1 and with zero divisors, sayrs = 0 where r, s ∈ R and r 6= 0 6= s. Assume M is a nonzero R-module. Since M is not the zero module, we may choose some m ∈Mwith m 6= 0. If sm = 0, then m ∈ Tor(M) since s 6= 0, and we’re done,so assume sm 6= 0. Then

r(sm) = (rs)m module axiom

= 0m since rs = 0

= 0 by 10.1.1 above

Since r 6= 0, this shows that sm ∈ Tor(M). Either way, we have anonzero element in Tor(M).

10.1.9 Assume R is a ring with 1 and M is a left R-module. IfN is a submodule of M , the annihilator of N in R is defined to be{r ∈ R|rn = 0 for all n ∈ N}. Prove that the annihilator of N in R isa 2-sided ideal of R.

Let AnnR(N) ⊆ R be the annihilator ofN inR. Assume r ∈ AnnR(N)and let s ∈ R. We wish to show sr ∈ AnnR(N) and rs ∈ AnnR(N).For every n ∈ N we have (sr)n = s(rn) = s · 0 = 0, so sr ∈ AnnR(N).Also, for every n ∈ N , sn ∈ N since n ∈ N and N is a submodule,so (rs)n = r(sn) = 0 since r ∈ AnnR(N). So far we have shownthat AnnR(N) is closed under multiplication by elements of R, bothon the left and on the right. To complete the proof we need to show

31

AnnR(N) is closed under +. Let r, s ∈ AnnR(N). Then for all n ∈ N ,we have rn = sn = 0, so (r + s)n = rn + sn = 0 + 0 = 0. Thereforer + s ∈ AnnR(N).

10.1.10 Assume R is a ring with 1 and M is a left R-module. If I isan ideal of R, the annihilator of I in M is defined to be {m ∈M |am =0 for all a ∈ I}. Prove that the annihilator of I in M is a submoduleof M .

Let AnnM(I) be the annihilator of I in M . We will show AnnM(I)is closed under − and closed under the action of elements in R. Letm,n ∈ AnnM(I). Then, for all a ∈ I, am = an = 0, so a(m − n) =am− an = 0− 0 = 0. This shows m− n ∈ AnnM(I). Let r ∈ R. Thenrm ∈ AnnM(I) because, for every a ∈ I, we have ar ∈ I since I is anideal, so (ar)m = 0 since m ∈ AnnM(I), hence a(rm) = 0 by a moduleaxiom.

Homework #05, due 2/17/10 = 10.3.1, 10.3.4, 10.3.5, 10.3.7,10.3.15

Additional problems recommended for study: 10.2.1, 10.2.2, 10.2.3,10.2.5, 10.2.6, 10.2.10, 10.2.11, 10.3.2, 10.3.9, 10.3.12, 10.3.13,10.3.14.

10.3.1 Assume R is a ring with 1 and M is a left R-module. Provethat if A and B are sets with the same cardinality, then the free modulesF (A) and F (B) are isomorphic.

Since A and B have the same cardinality there exists a bijectionf : A → B. Since f is a bijection its inverse is also a bijection andf−1 : B → A. By Th. 10.6 the free modules F (A) and F (B) havethe universal mapping property. Let us apply this property to f . Notethat the range of f is B since f is surjective, and B is a subset of thefree module F (B). Thus we have another function (which we call g)that maps A into F (B) and is equal to f on all elements of A, that is,g : A→ F (B) and g(a) = f(a) for all a ∈ A. By the universal mappingproperty there is a unique R-module homomorphism ϕ : F (A) → F (B)

32

which extends g (and f), that is, f(a) = g(a) = ϕ(a) for all a ∈ A. Bythe same reasoning, with A and B interchanged and f replaced by f−1,we get another R-module homomorphism ψ : F (B) → F (A) extendingf−1. The composition of R-module homomorphisms is again an R-module homomorphism, so we obtain the R-module homomorphismψ◦ϕ : F (A) → F (A). For every a ∈ A, (ψ◦ϕ)(a) = ψ(ϕ(a)) = ψ(f(a)),but f(a) ∈ B and ψ extends f−1, so ψ(f(a)) = f−1(f(a)) = a. Thisshows that ψ◦ϕ is an extension of the identity map ι : A→ F (A) whichsends every element of A to itself, that is, ι(a) = a for every a ∈ A.Now, by the universal mapping property, ι has a unique extension to anR-module homomorphism from F (A) to itself. The identity map fromF (A) to F (A) is an R-module homomorphism of F (A) onto itself, andby the universal mapping property of F (A) it is the only R-modulehomomorphism of F (A) onto itself. However, we found that ψ ◦ ϕ isa homomorphism of F (A) onto itself that extends the identity map onA, so ψ ◦ ϕ must therefore be the identity map from F (A) to F (A).Similarly, ϕ◦ψ is the identity map from F (B) to F (B). Thus ϕ and ψare R-module homomorphisms between F (A) and F (B), and they areinverses of each other, so they are both injective and surjective, and aretherefore isomorphisms between F (A) and F (B). Thus F (A) ∼= F (B)whenever |A| = |B|.

10.3.4 An R-module M is called a torsion module if for each m ∈Mthere is a nonzero element r ∈ R such that rm = 0, where r may dependon m (i.e., M = Tor(M) in the notation of Exercise 8 of Section 10.1).Prove that every finite abelian group is a torsion Z-module. Give anexample of an infinite abelian group that is a torsion Z-module.

Let M be a finite abelian group, so that M is also a Z-module. Let nbe the order of M , that is, n = |M | ∈ Z+. Then, for every a ∈M , theorder of a divides the order n of the abelian group M , so na = 0. Sincen ∈ Z and na is the result of the action of n on a in the module M ,we have a ∈ Tor(M) for every a ∈ M , so M ⊆ Tor(M). The oppositeinclusion holds trivially, so M = Tor(M).

33

Let M =∏

i∈Z+ Z/2Z. Thus M is the direct product of countablymany copies of the 2-element cyclic group Z/2Z. M is an infiniteabelian group whose cardinality is the same as the set of real numbers.Every element of M has order 2, so 2 · m = 0 for every m ∈ M , soTor(M) = M .

10.3.5 Let R be an integral domain. Prove that every finitely gener-ated torsion R-module has a nonzero annihilator, i.e., there is a nonzeror ∈ R such that rm = 0 for all m ∈M—here r does not depend on m(the annihilator of a module was defined in Exercise 9 of Section 10.1).Give an example of a torsion R-module whose annihilator is the zeroideal.

Assume M is a finitely generated torsion R-module. Then there isa finite set A ⊆ M of nonzero elements such that M = RA. LetA = {a1, · · · , an}, n ∈ ω. Since M is a torsion module, there existnonzero elements r1, · · · , rn ∈ R such that r1a1 = 0, r2a2 = 0, . . . ,rnan = 0. Let q = r1 · . . . · rn. Consider an arbitrary element m ∈ M .Since M = RA, there are ring elements s1, · · · , sn ∈ R such thatm = s1a1 + · · · + snan. Let 1 ≤ i ≤ n. Since R is an integral domain,R is commutative, so q = pri where p is the product of all the otherfactors rj with j 6= i. Then qai = (pri)ai = p(riai) = p0 = 0. Thisshows qa1 = qa2 = · · · = qan = 0, so we have

qm = q(s1a1 + · · ·+ snan)

= q(s1a1) + · · ·+ q(snan) module axiom

= (qs1)a1 + · · ·+ (qsn)an module axiom

= (s1q)a1 + · · ·+ (snq)an since R is commutative

= s1(qa1) + · · ·+ sn(qan) module axiom

= s1(0) + · · ·+ sn(0) shown above

= 0 + · · ·+ 0

= 0

Since m was an arbitrary element of M , we have shown that qm = 0for every m ∈M . Finally, we observe that q is nonzero because ri 6= 0

34

and the product on nonzero elements in the integral domain R cannotbe zero.

For an example of a torsion R-module whose annihilator is the zeroideal, let R = Z, and let M be the direct sum of the finite cyclic groupsZ/nZ with 2 ≤ n ∈ Z+, so

M =⊕

2≤n∈Z+

Z/nZ.

Let m ∈ M . Then there are k ∈ Z+ and a1, . . . , ak ∈ Z such thatm = (a1 + 2Z, · · · , ak + kZ, 0, 0, 0, · · · ), so k! ·m = 0, so m ∈ Tor(M).Then M is a torsion module, but no element of Z annihilates everyelement of M , for if k ∈ Z+ then k · (. . . , 1 + (k + 1)Z, 0, 0, . . . ) 6= 0.

10.3.7 Assume R is a ring with 1 and M is a left R-module. LetN be a submodule of M . Prove that if both M/N and N are finitelygenerated, then so is M .

N is a finitely generated R-module, so there is a finite subset A ⊆ Nsuch that N = RA. M/N is also a finitely generated R-module. Theelements of M/N have the form b + N with b ∈ M , so there is afinite subset B ⊆ M such that M/N = R{b + N |b ∈ B}. Let A ={a1, · · · , an} and B = {b1, · · · , bk}. Then

N = {r1a1 + · · ·+ rnan|r1, · · · , rn ∈ R}M/N = {s1(b1 +N) + · · ·+ sk(bk +N)|s1, · · · , sk ∈ R}

= {s1b1 + · · ·+ skbk +N |s1, · · · , sk ∈ R}

We will show M = R(A∪B). Let m ∈M . Then m+N ∈M/N , so bythe second equation above there exist s1, · · · , sk ∈ R such thatm+N =s1b1 + · · ·+ skbk +N . Now m ∈ m+N , so m ∈ s1b1 + · · ·+ skbk +N ,so by the first equation above there exist r1, · · · , rn ∈ R such thatm = s1b1 + · · · + skbk + r1a1 + · · · + rnan ∈ R(A ∪ B), as was to beshown. Now A ∪ B is finite because both A and B are finite, so M istherefore finitely generated by the finite set A ∪B.

35

10.3.15 An element e ∈ R is called a central idempotent if e2 = eand er = re for all r ∈ R. If e a central idempotent in R, prove thatM = eM ⊕ (1− e)M . [Recall Exercise 14 in Section 10.1.]

We wish to show that eM and (1− e)M are submodules and that Mis the internal direct sum of the two submodules eM and (1− e)M .

To see that eM is a submodule, first note that em1 + em2 = e(m1 +m2) ∈ eM for all m1,m2 ∈ M , so eM is closed under addition. Tosee that eM is closed under the action of R, let m ∈ M and r ∈ R.Then, using the fact that e is in the center and hence er = re, we haver(em) = (re)m = (er)m = e(rm) ∈ eM .

The proof that (1 − e)M is also closed under addition is essentiallythe same, namely, (1−e)m1+(1−e)m2 = (1−e)(m1+m2) ∈ (1− e)Mfor all m1,m2 ∈ M . Next we show that (1 − e)M is closed under theaction of R. Let m ∈ M and r ∈ R. Then, using the fact that e is inthe center and hence er = re, we have

r((1− e)m) = (r(1− e))m

= (r1− re)m

= (1r − er)m

= ((1− e)r)m

= (1− e)(rm) ∈ (1− e)M

Next we show eM ∩ (1− e)M = {0}. Let m ∈ eM ∩ (1− e)M . Thenthere are m1,m2 ∈M such that m = em1 = (1− e)m2, so

m = em1

= e2m1 e = e2

= (ee)m1

= e(em1)

= e((1− e)m2) em1 = (1− e)m2

= (e(1− e))m2

= (e1− e2)m2

36

= (e− e)m2 e = e2

= 0m2

= 0.

It therefore follows by Prop. 10.5 that M ∼= eM ⊕ (1− e)M .

Homework #06, due 2/24/10 = 10.2.9, 10.2.10, 10.3.12, 10.3.13,10.3.14

Additional problems recommended for study: 10.3.18, 10.3.19, 10.3.20,10.3.21, 10.3.22, 10.3.23, 10.3.24

10.2.9 Let R be a commutative ring. Prove that HomR(R,M) andM are isomorphic as left R-modules. [Show that each element ofHomR(R,M) is determined by its value on the identity of R.]

Let 1 be the identity of R. (The hint suggests that such an elementexists, so we assume R is a commutative ring with 1.) First note thatHomR(R,M) is indeed an R-module by Prop. 10.2(2), for which weneed the commutativity of R. We wish to show HomR(R,M) ∼= M .

For every ϕ ∈ HomR(R,M), let Ψ(ϕ) = ϕ(1). This defines a map Ψfrom HomR(R,M) to M . For Ψ to be an R-module homomorphism,we must check Ψ(ϕ + ψ) = Ψ(ϕ) + Ψ(ψ) and Ψ(rϕ) = rΨ(ϕ) for allϕ, ψ ∈ HomR(R,M) and all r ∈ R.

Ψ(ϕ+ ψ) = (ϕ+ ψ)(1) def of Ψ

= ϕ(1) + ψ(1) def of + in HomR(R,M)

= Ψ(ϕ) + Ψ(ψ) def of Ψ

Ψ(rϕ) = (rϕ)(1) def of Ψ

= r(ϕ(1)) def of action of R on HomR(R,M)

= rΨ(ϕ) def of Ψ

If Ψ(ϕ) = Ψ(ψ), then ϕ(1) = ψ(1) by the definition of Ψ, so, for allr ∈ R, ϕ(r) = ϕ(r1) = rϕ(1) = rψ(1) = ψ(r1) = ψ(r), which showsϕ = ψ. Thus Ψ is injective.

37

For surjectivity of Ψ, start with an arbitrary m ∈M . Define a map ξmby ξm(r) = rm for all r ∈ R. Then ξm is an R-module homomorphismof R to M because

ξm(r + s) = (r + s)m def of ξm

= rm+ sm module axiom

= ξm(r) + ξm(s) def of ξm

ξm(rs) = (rs)m def of ξm

= r(sm) module axiom

= rξm(s) def of ξm

Finally, we have Ψ(ξm) = ξm(1) = 1m = m, and this holds for allm ∈M , so Ψ is surjective.

10.2.10 Let R be a commutative ring. Prove that HomR(R,R) and Rare isomorphic as rings.

As in the previous problem, we assumeR is a ring with 1. By Exer. 10.2.9above, HomR(R,R) and R are isomorphic as left R-modules via the iso-morphism Ψ, which is defined for every ϕ ∈ HomR(R,R) by Ψ(ϕ) =ϕ(1). HomR(R,R) is closed under composition, and is a ring under the+ of HomR(R,R) together with ◦. To show HomR(R,R) and R areisomorphic as rings we must show that composition in HomR(R,R) ismapped to multiplication in R.

Define Ψ and ξ as above, but with R in place of M , so the definitionsare Ψ(ϕ) = ϕ(1) and ξm(r) = rm with ϕ ∈ HomR(R,R) and r,m ∈ R.It was shown above that Ψ is a bijection, hence its inverse is surjective.It was proved above that the inverse of Ψ is the map ξ that sendsm ∈ R to ξm ∈ HomR(R,R), so ϕ = ξϕ(1) and ψ = ξψ(1). Then

Ψ(ϕ ◦ ψ) = Ψ(ξϕ(1) ◦ ξψ(1)) Ψ and ξ are inverses

= (ξϕ(1) ◦ ξψ(1))(1) def of Ψ

= ξϕ(1)(ξψ(1)(1)) def of ◦= ξϕ(1)(1ψ(1)) def of ξψ(1)

= (1ψ(1))ϕ(1) def of ξϕ(1)

38

= ψ(1)ϕ(1)

= Ψ(ψ)Ψ(ϕ) def of Ψ

= Ψ(ϕ)Ψ(ψ) R is commutative

10.3.12 Let R be a commutative ring and let A, B and M be R-modules. Prove the following isomorphisms of R-modules.

(a) HomR(A×B,M) ∼= HomR(A,M)× HomR(B,M)

(b) HomR(M,A×B) ∼= HomR(M,A)× HomR(M,B)

First some observations on both parts. Since R is a commutative ring,it follows by Prop. 10.2 that HomR(A,M), HomR(B,M), HomR(M,A),HomR(M,B), HomR(A×B,M), and HomR(M,A×B) are R-modules,so HomR(A,M) × HomR(B,M) and HomR(M,A) × HomR(M,B) arealso R-modules (since they are direct products of R-modules).

(a) An arbitrary element of HomR(A,M)×HomR(B,M) is an orderedpair (α1, β1) consisting of α1 ∈ HomR(A,M) and β1 ∈ HomR(B,M).Given an arbitrary pair of module homomorphisms

(α1, β1) ∈ HomR(A,M)× HomR(B,M),

define a map Ψ(α1, β1) : A×B →M by Ψ(α1, β1)(a, b) = α1(a)+β1(b)for all a ∈ A and all b ∈ B. We show now that Ψ(α1, β1) ∈ HomR(A×B,M). For all a1, a2 ∈ A, b1, b2 ∈ B, and r ∈ R we have

Ψ(α1, β1)((a1, b1) + (a2, b2))

= Ψ(α1, β1)(a1 + a2, b1 + b2) def of + in A×B

= α1(a1 + a2) + β1(b1 + b2) def of Ψ

= (α1(a1) + α1(a2)) + (β1(b1) + β1(b2)) α1, β1 are R-module homs

= α1(a1) + β1(b1) + α1(a2) + β1(b2) properties of +

= Ψ(α1, β1)(a1, b1) + Ψ(α1, β1)(a2, b2) def of Ψ

Ψ(α1, β1)(r(a1, b1))

= Ψ(α1, β1)(ra1, rb1) def of action of R on A×B

39

= α1(ra1) + β1(rb1) def of Ψ

= rα1(a1) + rβ1(b1) α1, β1 are R-module homs

= r(α1(a1) + β1(b1)) module axiom

= rΨ(α1, β1)(a1, b1) def of Ψ

This shows that Ψ : HomR(A,M)×HomR(B,M) → HomR(A×B,M).

Now we must show that Ψ is itself an R-module homomorphism fromthe R-module HomR(A,M)×HomR(B,M) to the R-module HomR(A×B,M). Let α1, α2 ∈ HomR(A,M) and β2, β1 ∈ HomR(A,M), so thatwe have

(α1, β1), (α2, β2) ∈ HomR(A,M)× HomR(B,M)

We wish to show

(1) Ψ((α1, β1) + (α2, β2)) = Ψ(α1, β1) + Ψ(α2, β2)

and

(2) Ψ(r(α1, β1)) = rΨ(α1, β1)

For all a ∈ A, b ∈ B, and r ∈ R we have

Ψ((α1, β1) + (α2, β2))(a, b)

= Ψ(α1 + α2, β1 + β2)(a, b) def of + in HomR(A,M)× HomR(B,M)

= ((α1 + α2)(a), (β1 + β2)(b)) def of Ψ

= (α1(a) + α2(a), β1(b) + β2(b)) def of + in HomR(A,M) and HomR(B,M)

= (α1(a), β1(b)) + (α2(a), β2(b)) def of + in A×B

= Ψ(α1, β1)(a, b) + Ψ(α2, β2)(a, b) def of Ψ

This calculation shows that (1) holds. For (2) we have

Ψ(r(α1, β1))(a, b)

= Ψ(rα1, rβ1)(a, b) def of R-action on HomR(A,M)× HomR(B,M)

= ((rα1)(a), (rβ1)(b)) def of Ψ

= (r(α1(a)), r(β1(b))) def of R-action on HomR(A,M) and on HomR(B,M)

= r(α1(a), β1(b)) def of R-action on A×B

40

= rΨ(α1, β1)(a, b) def of Ψ

Finally we must show Ψ is a bijection.

To show Ψ is injective, assume Ψ(α1, β1) = Ψ(α2, β2) where α1, α2 ∈HomR(A,M) and β2, β1 ∈ HomR(A,M). We must show (α1, β1) =(α2, β2). First note that for all a ∈ A and b ∈ B, we have Ψ(α1, β1)(a, b) =Ψ(α2, β2)(a, b), hence α1(a) + β1(b) = α2(a) + β2(b) by the definition ofΨ. In particular, for every a ∈ A we have α1(a)+β1(0) = α2(a)+β2(0),but β1(0) = β2(0) = 0 since β1 and β2 are module homomorphisms, sowe have α1(a) = α2(a) for all a ∈ A, which tells us that α1 = α2. Sim-ilarly, for every b ∈ B we have β1(b) = α1(0) + β1(b) = α2(0) + β2(b) =β2(b), so β1 = β2. From α1 = α2 and β1 = β2 we get (α1, β1) = (α2, β2).

To show Ψ is surjective, consider an arbitrary ϕ ∈ HomR(A × B,M).Define ϕ1 : A → M by ϕ1(a) = ϕ(a, 0) for all a ∈ A, and defineϕ2 : B → M by ϕ2(b) = ϕ(0, b). Then ϕ1 and ϕ2 are R-modulehomomorphisms, so ϕ1 ∈ HomR(A,M) and ϕ2 ∈ HomR(B,M). For alla ∈ A and b ∈ B,

Ψ(ϕ1, ϕ2)(a, b)

= ϕ1(a) + ϕ2(b) def of Ψ

= ϕ(a, 0) + ϕ(0, b) def of ϕ1 and ϕ2

= ϕ((a, 0) + (0, b)) ϕ is a module homomorphism

= ϕ(a, b) def of + in A×B

This calculation shows that Ψ(ϕ1, ϕ2) = ϕ.

(b) We will define a map

Φ : HomR(M,A)× HomR(M,B) → HomR(M,A×B)

as follows. First, for an arbitrary pair (α1, β1) ∈ HomR(M,A) ×HomR(M,B), where α1 ∈ HomR(M,A) and β1 ∈ HomR(M,B), letΦ(α1, β1) be the map from M to A × B defined for all x ∈ M byΦ(α1, β1)(x) = (α1(x), β1(x)). We show now that Φ(α1, β1) is actuallyan R-module homomorphism from M to the R-module A×B. For all

41

x, y ∈M and r ∈ R we have

Φ(α1, β1)(x+ y)

= (α1(x+ y), β1(x+ y)) def of Φ

= (α1(x) + α1(y), β1(x) + β1(y)) α1, β1 are R-module homs

= (α1(x), β1(x)) + (α1(y), β1(y)) def + of in A×B

= Φ(α1, β1)(x) + Φ(α1, β1)(y) def of Φ

Φ(α1, β1)(rx)

= (α1(rx), β1(rx)) def of Φ

= (rα1(x), rβ1(x)) α1, β1 are R-module homs

= r(α1(x), β1(x)) def of action of R on A×B

= rΦ(α1, β1)(x) def of Φ

so Φ(α1, β1) is an R-module homomorphism from M into A×B.

Now we must show that Φ is itself an R-module homomorphism fromtheR-module HomR(M,A)×HomR(M,B) to theR-module HomR(M,A×B). Let α1, α2 ∈ HomR(M,A) and β2, β1 ∈ HomR(M,B), so that wehave

(α1, β1), (α2, β2) ∈ HomR(M,A)× HomR(M,B)

We wish to show

Φ((α1, β1) + (α2, β2)) = Φ(α1, β1) + Φ(α2, β2)

and

Φ(r(α1, β1)) = rΦ(α1, β1)

For every x ∈M and r ∈ R we have

Φ((α1, β1) + (α2, β2))(x)

= Φ(α1 + α2, β1 + β2)(x) def of + in HomR(M,A)× HomR(M,B)

= ((α1 + α2)(x), (β1 + β2)(x)) def of Φ

= (α1(x) + α2(x), β1(x) + β2(x)) def of + in HomR(M,A) and HomR(M,B)

= (α1(x), β1(x)) + (α2(x), β2(x)) def of + in A×B

42

= Φ(α1, β1)(x) + Φ(α2, β2)(x) def of Φ

Φ(r(α1, β1))(x)

= Φ(rα1, rβ1)(x) def of R-action on HomR(M,A)× HomR(M,B)

= ((rα1)(x), (rβ1)(x)) def of Φ

= (r(α1(x)), r(β1(x))) def of R-action on HomR(M,A), and HomR(M,B)

= r(α1(x), β1(x)) def of R-action on A×B

= rΦ(α1, β1)(x) def of Φ

Finally, we show Φ is bijective.

To show Φ is injective, assume Φ(α1, β1) = Φ(α2, β2) where α1, α2 ∈HomR(A,M) and β2, β1 ∈ HomR(B,M). Then for all x ∈ M we haveΦ(α1, β1)(x) = Φ(α2, β2)(x), hence (α1(x), β1(x)) = (α2(x), β2(x)), henceα1(x) = α2(x) and β1(x) = β2(x). This shows α1 = α2 and β1 = β2, so(α1, α2) = (β1, β2).

To show Ψ is surjective, consider an arbitrary ϕ ∈ HomR(M,A × B).Let π1 and π2 be the R-module homomorphisms that project A × Bonto A, and B, respectively: π1 : A × B → A and π2 : A × B → B.Then for all x ∈M we have

Ψ(π1 ◦ ϕ, π2 ◦ ϕ)(x) = ((π1 ◦ ϕ)(x), (π2 ◦ ϕ)(x)) def of Φ

= (π1(ϕ(x)), π2(ϕ(x)))) def of ◦= ϕ(x) def of π1 and π2

so Ψ(π1 ◦ ϕ, π2 ◦ ϕ) = ϕ.

10.3.13 Let R be a commutative ring and let F be a free R-module offinite rank. Prove the following isomorphism of R-modules:

HomR(F,R) ∼= F

AssumeM is free module of finite rank n < ω and letX = {x1, · · · , xn} ⊆F be a free basis for F . EveryR-module homomorphism ϕ ∈ HomR(F,R)obviously determines a map f : X → R, where f(x) = ϕ(x) ∈ Rfor every x ∈ X. Conversely, since X is a free basis, every function

43

f : X → R determines a unique R-module homomorphism ϕ : F → Rextending f .

Let Ψ be the map that sends each R-module homomorphism ϕ ∈HomR(F,R) to ϕ(x1)x1 + · · ·+ ϕ(xn)xn ∈ F , so

Ψ(ϕ) = ϕ(x1)x1 + · · ·+ ϕ(xn)xn

To see that Ψ is surjective, consider an arbitrary element of F , whichhas the form a1x1 + · · · + anxn for some a1, . . . , an ∈ R. Define f :X → R by setting f(xi) = ai for 1 ≤ i ≤ n, and let ϕ be the uniqueR-module homomorphism extending f . Then

Ψ(ϕ) = ϕ(x1)x1 + · · ·+ ϕ(xn)xn

= f(x1)x1 + · · ·+ f(xn)xn

= a1x1 + · · ·+ anxn.

To see that Ψ is injective, suppose Ψ(ϕ) = Ψ(ψ). Then ϕ(x1)x1 +· · · + ϕ(xn)xn = ψ(x1)x1 + · · · + ψ(xn)xn, but X is a free basis, sorepresentations of elements as linear combinations of elements of X areunique, hence ϕ(x1) = ψ(x1), · · · , ϕ(xn) = ψ(xn), so ϕ = ψ becauseR-module homomorphisms that agree on the free basis X must be thesame by uniqueness. Finally, the following calculations show Ψ is ahomomorphism.

Ψ(ϕ) + Ψ(ψ) =(ϕ(x1)x1 + · · ·+ ϕ(xn)xn

)+

(ψ(x1)x1 + · · ·+ ψ(xn)xn

)=

(ϕ(x1)x1 + ψ(x1)x1

)+ · · ·+

(ϕ(xn)xn + ψ(xn)xn

)=

(ϕ(x1) + ψ(x1)

)x1 + · · ·+

(ϕ(xn) + ψ(xn)

)xn

=((ϕ+ ψ)(x1)

)x1 + · · ·+

((ϕ+ ψ)(xn)

)xn

= Ψ(ϕ+ ψ)

Ψ(rϕ) = ((rϕ)(x1))x1 + · · ·+ ((rϕ)(xn))xn def of Ψ

= (rϕ(x1))x1 + · · ·+ (rϕ(xn))xn def of action of R on Hom

= r(ϕ(x1)x1) + · · ·+ r(ϕ(xn)xn) module axiom

44

= r(ϕ(x1)x1 + · · ·+ ϕ(xn)xn) module axiom

= r(Ψ(ϕ)) def of Ψ

10.3.14 Let R be a commutative ring and let F be the free R-moduleof rank n. Prove that HomR(F,M) ∼= M × · · · ×M (n times). [UseExercise 9 in Section 10.2 and Exercise 12.]

By (repeated or inductive use of) Exer. 10.3.12(a) above,

HomR(

n︷ ︸︸ ︷R× · · · ×R,M) ∼=

n︷ ︸︸ ︷HomR(R,M)× · · · × HomR(R,M)

and by Exer. 10.2.9 above, HomR(R,M) and M are isomorphic as leftR-modules, so

HomR(Rn,M) ∼=n︷ ︸︸ ︷

M × · · · ×M

However, Rn is the free R-module of rank n, so F = Rn, hence

HomR(F,M) ∼=n︷ ︸︸ ︷

M × · · · ×M

Homework #07, due 3/3/10 = 12.1.1, 12.1.2, 12.1.3, 10.3.18, 12.1.10

Additional problems recommended for study: 12.1.4, 12.1.5, 12.1.6,12.1.7, 12.1.8, 12.1.9

12.1.1 Let M be a module over the integral domain R.

(a) Suppose x is a nonzero torsion element in M . Show that x and 0are “linearly dependent.” Conclude that the rank of Tor(M) is 0, sothat in particular any torsion R-module has rank 0.

(b) Show that the rank of M is the same as the rank of the (torsionfree) quotient M/Tor(M).

(a) Since x is a nonzero torsion element ofM , there exists some nonzeror ∈ R such that rx = 0. The rank of a module over an integraldomain is the maximum number of linearly independent elements. Aset of elements of a module is linearly dependent if some linearcombination of them is 0, and yet not all of the coefficients are 0. The

45

equation rx = 0 is just such an example; a linear combination of theelements in {x} is 0, and yet not all coefficients are 0 (because r 6= 0),so it shows that {x} is not a linear independent set whenever x is atorsion element. Every element in a torsion module is a torsion element,so a torsion module cannot have a linearly independent 1-element set,so the largest set of linearly independent elements is the empty set, sothe rank of a torsion module is 0. In particular, the rank of the torsionsubmodule Tor(M) of M is 0.

(b) Let N = Tor(M) be the torsion submodule of M . Suppose the rankof M is n and {x1, · · · , xn} is a linearly independent set of elements ofM . Suppose {x1+N, x2+N, · · · , xn+N} is not linearly independent inthe quotient module M/N . Then there are coefficients r1, · · · , rn ∈ R,not all zero, such that r1(x1 +N)+ · · ·+ rn(xn+N) = N . This impliesr1x1 + · · ·+ rnxn ∈ N , so r1x1 + · · ·+ rnxn is a torsion element. Thereis some nonzero s ∈ R such that s(r1x1 + · · ·+rnxn) = 0, hence sr1x1 +· · ·+ srnxn = 0. But, since there is at least one nonzero coefficient, sayri 6= 0, and s 6= 0, we get sri 6= 0 because R is an integral domain. Sincenot all the coefficients sr1, · · · , srn are zero and sr1x1 + · · ·+srnxn = 0,we conclude that {x1, · · · , xn} is linearly dependent, contradicting thehypothesis. This shows that {x1 + N, x2 + N, · · · , xn + N} is linearlyindependent, so the rank of the quotient M/N is at least n.

The rank of M/N cannot be more than n, because if we consider anarbitrary set of n+1 elements of M/N , say y1 +N , . . . , yn+1 +N withy1, · · · , yn+1 ∈ M , then {y1, · · · , yn+1} is linearly dependent since Mhas rank n, so r1y1 + · · ·+ rn+1yn+1 = 0 for some r1, · · · , rn+1 ∈ R, notall zero, hence

N = 0 +N

= r1y1 + · · ·+ rn+1yn+1 +N

= r1y1 +N + · · ·+ rn+1yn+1 +N

= r1(y1 +N) + · · ·+ rn+1(yn+1 +N)

which shows that {y1+N, · · · , yn+1+N} is linearly dependent in M/N .

Therefore the ranks of M and M/Tor(M) are the same.

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12.1.2 Let M be a module over the integral domain R.

(a) Suppose that M has rank n and x1, x2, · · · , xn is any maximal setof linearly independent elements of M . Let N = Rx1 + · · · + Rxn bethe submodule generated by x1, x2, · · · , xn. Prove that N is isomorphicto Rn and that the quotient M/N is a torsion R-module (equivalently,the elements x1, · · · , xn are linearly independent and for any y ∈ Mthere is a nonzero element r ∈ R such that ry can be written as a linearcombination r1x1 + · · ·+ rnxn of the xi).

(b) Prove conversely that if M contains a submodule N that is freeof rank n (i.e., N ∼= Rn) such that the quotient M/N is a torsion R-module then M has rank n. [Let y1, y2, · · · , yn+1 be any n+1 elementsof M . Use the fact that M/N is torsion to write riyi as a linear com-bination of a basis for N for some nonzero elements r1, · · · , rn+1 of R.Use an argument as in the proof of Proposition 3 to see that the riyi,and hence also the yi, are linearly dependent.]

(a) This proof proceeds by a direct construction of an isomorphismfrom Rn to N (rather than an appeal to theorems of Chapter 10).

Define ϕ : Rn → N by ϕ(r1, · · · , rn) = r1x1 + · · · + rnxn for allr1, · · · , rn ∈ R. Then ϕ is obviously surjective, just by the defini-tion of N . To show ϕ is injective, assume there are two elements of Rn,(r1, · · · , rn) and (s1, · · · , sn), such that ϕ(r1, · · · , rn) = ϕ(s1, · · · , sn).Then r1x1 + · · · + rnxn = s1x1 + · · · + snxn by the definition of ϕ, so(r1 − s1)x1 + · · · + (rn − sn)xn = 0 by module arithmetic. But thenr1−s1 = 0, · · · , rn−sn = 0 by the linear independence of {x1, · · · , xn},so r1 = s1, · · · , rn = sn, hence (r1, · · · , rn) = (s1, · · · , sn). Thus ϕ isinjective.

To check ϕ is a module homomorphism, consider two arbitrary elements(r1, · · · , rn), (s1, · · · , sn) ∈ Rn. Then

ϕ((r1, · · · , rn) + (s1, · · · , sn)

)= ϕ(r1 + s1, · · · , rn + sn) def of + in Rn

47

= (r1 + s1)x1 + · · ·+ (rn + sn)xn def of ϕ

= (r1x1 + · · ·+ rnxn) + (s1x1 + · · ·+ snxn) module arithmetic

= ϕ(r1, · · · , rn) + ϕ(s1, · · · , sn) def of ϕ

and, for all r ∈ R,

ϕ(r(r1, · · · , rn)

)= ϕ(rr1, · · · , rrn) def of action of R on Rn

= (rr1)x1 + · · ·+ (rrn)xn def of ϕ

= r(r1x1 + · · ·+ rnxn) module arithmetic

= rϕ(r1, · · · , xn) def of ϕ

Now we show that the quotient M/N is a torsion R-module. We needto show every nonzero element of the quotient is torsion because it canbe annihilated by a nonzero element of R. Let y + N ∈ M/N wherey ∈ M and y /∈ N , so that y + N is nonzero in M/N . Note that, inparticular, y /∈ {x1, · · · , xn}. Since {x1, · · · , xn} is a maximal linearindependent set, the strictly larger set {y, x1, · · · , xn} must be linearlydependent, so there are coefficients s, r1, · · · , rn ∈ R, not all zero, suchthat sy + r1x1 + · · · + rnxn = 0. We cannot have s = 0, otherwiser1x1 + · · · + rnxn = 0 and one of the coefficients r1, · · · , rn is nonzero,contradicting the linear independence of {x1, · · · , xn}. But then wehave sy = −r1x1−· · ·− rnxn ∈ N , hence sy+N = s(y+N) = N , i.e.,s(y +N) is zero in the quotient M/N .

(b) Assume M contains a submodule N that is free of rank n (i.e.,N ∼= Rn) and the quotient M/N is a torsion R-module. We will showthat M has rank n.

Since N has rank n, it contains a maximal linearly independent subsetX = {x1, · · · , xn} ⊆ N . Then X is also a linearly independent subsetof M with cardinality n, so the rank of M has to be at least n.

We need only show that the rank of M can be at most n. To thatend, let y1, y2, · · · , yn+1 be any n+ 1 elements of M . We want to showthese elements are linearly dependent. The images y1, y2, · · · , yn+1 in

48

the quotient module M/N are

y1 +N, y2 +N, · · · , yn+1 +N.

Since M/N is torsion, each of these elements has a nonzero element ofR that annihilates it, that is, there are nonzero r1, · · · , rn+1 ∈ R suchthat

r1y1 +N = r2y2 +N = · · · = rn+1yn+1 +N = N

Note that if yi is already in N , we can use any nonzero ri ∈ R, and if yiis not in N , we use the fact that yi +N is torsion in M/N . Thereforewe have {r1y1, r2y2, · · · , rn+1yn+1} ⊆ N . Let

Y := {r1y1, r2y2, · · · , rn+1yn+1}.By a theorem proved in class1, the submodule of N generated by Yis free and has rank no more than the rank n of N . Since Y hasmore than n elements, Y is linearly dependent, so there are coefficientss1, s2, . . . , sn+1 ∈ R, not all zero, such that

s1(r1y1) + s2(r2y2) + · · ·+ sn+1(rn+1yn+1) = 0

By module arithmetic, this equation says

(s1r1)y1 + (s2r2)y2 + · · ·+ (sn+1rn+1)yn+1 = 0

Note that not all the coefficients s1r1, s2r2, . . . , sn+1rn+1 are zero,because we know that some si is not zero and that all the ri are notzero, hence siri 6= 0 because R is an integral domain. This shows thatany set Y ⊆ M with n + 1 elements is linearly dependent, completingthe proof that the rank of M is n.

12.1.3 Let R be an integral domain and let A and B be R-modules ofranks m and n, respectively. Prove that the rank of A ⊕ B is m + n.[Use the previous exercise.]

Let M = A⊕B. According to the meaning of ⊕ in the text, this couldmean thatM is the external direct product ofA andB, i.e.,M = A×B.However, I prefer to assume that A and B are submodules of M and

1Th. Assume R is a P.I.D., M is a free module over R with finite rank n, and N is a submoduleof M . Then N is a free submodule of M with rank no more than n.

49

M is the internal direct product of A and B. This is also expressed byM = A⊕B. This assumption implies that A ∩B = {0}.Since the rank of A is m, there is a maximal (in the submodule A)linearly independent subset X = {x1, · · · , xm} ⊆ A. Similarly, sincethe rank B is n, there is a maximal (in B) linearly independent subsetY = {y1, · · · , yn} ⊆ B. LetA1 = Rx1+· · ·Rxm andB1 = Ry1+· · ·Ryn.Then A/A1 and B/B1 are torsion modules by Exer. 12.1.2(a).

Claim 1. X ∪ Y is linearly independent in M .

Proof: Suppose some linear combination of elements of X ∪Y is zero,say

(3) r1x1 + · · ·+ rmxm + s1y1 + · · ·+ snyn = 0

where r1, · · · , rm, s1, · · · , sn ∈ R. Let z = r1x1 + · · · + rmxm. Thenz ∈ A1 ⊆ A, and z = −s1y1 − · · · − snyn by (3), so z ∈ B1 ⊆ B aswell. Then z ∈ A∩B = {0}, so z = 0. Therefore r1 = · · · = rm = 0 bythe linear independence of {x1, · · · , xm}, and s1 = · · · = sn = 0 by thelinear independence of {y1, · · · , yn}. Thus all the coefficients in (3) arezero. This completes the proof of the Claim.

Let M1 = A1⊕B1. Then M1 is isomorphic to Rm+n by Exer. 12.1.2(a),and is thus a free R-module of rank m+ n.

Claim 2. M/M1 is torsion.

Proof: Assume 0 6= w ∈ M . Then, since M = A ⊕ B, there areunique a ∈ A and b ∈ B such that w = a + b. Now A/A1 is torsion,so some nonzero r ∈ R annihilates a + A1, that is, r(a + A1) = A1, sora ∈ A1. Similarly, B/B1 is torsion, so there is some nonzero s ∈ Rsuch that sb ∈ B1. Note that rs = sr since R is an integral domain andhence is commutative. Then we have rsz = rs(a + b) = rsa + rsb =s(ra) + r(sb) ∈ A1 +B1 = M1. Note that rs is nonzero since R is anintegral domain and s 6= 0 6= r. Thus rs(z+M1) = rsz+M1 = M1, soz +M1 is annihilated by the nonzero rs ∈ R.

It follows from Claims 1 and 2 by Exer. 12.1.2(b) that the rank of Mis m+ n.

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10.3.18 Let R be a Principal Ideal Domain and let M be an R-module that is annihilated by the proper, nonzero ideal (a). Leta = pα1

1 pα22 · · · pαk

k be the unique factorization of a into distinct primepowers in R. Let Mi be the annihilator of pαi

i , i.e., Mi is the set{m ∈ M |pαi

i m = 0} — called the pi-primary component of M . Provethat

M = M1 ⊕M2 ⊕ · · · ⊕Mk

This problem, assigned after covering Section 12.1, can be solved by thePrimary Decomposition Theorem 12.7. By the way, for a prime p ∈ Z,the p-primary component of a finite abelian group is the subgroup ofelements whose order is a power of p (p. 142). In Exer. 10.3.22, if pis a prime in the PID R, the p-primary component of a torsion R-module is the set of elements annihilated by some positive power of p.Exer. 10.3.18 follows directly as a special case of part of Th. 12.7, (thePrimary Decomposition Theorem), which states that if R is a PID,M isa nonzero torsion R-module with nonzero annihilator a = upα1

1 · · · pαnn ,

where u ∈ R× is a unit of R and p1, · · · , pn ∈ R are distinct primes ofR, and Ni = {x ∈ M |pαi

i x = 0}, 1 ≤ i ≤ n, then Ni is a submoduleof M with annihilator pαi

i , Ni is the pi-primary component of M , andM = N1 ⊕ · · · ⊕ Nn. The remaining part of Th. 12.7 is that if Mis finitely generated then Ni is the direct sum of finitely many cyclicmodules whose annihilators are divisors of pαi

i .

12.1.10 For p a prime in the P.I.D. R and N an R-module prove thatthe p-primary component of N is a submodule of N and prove thatN is the direct sum of its p-primary components (there need not befinitely many of them).

There may be infinitely many p-primary components, perhaps one forevery prime in R. The proof cannot be completed unless N is assumedto be a torsion module.

N may not be finitely generated, so we have a situation in which variouselements could be annihilated by arbitrarily large powers of a givenprime p ∈ R. Let Np be the p-primary component of M , i.e., the set ofelements annihilated by a positive power of p. Suppose x, y ∈ Np. Then

51

for some positive powers pm and pn, with m,n ∈ Z+, we have pmx =0 = pny. Then pm+n(x−y) = pm+nx−pm+ny = pn(pmx)−pn(pmy) = 0,so x − y ∈ Np. If r ∈ R, then pm(rx) = r(pmx) = r0 = 0 since R iscommutative, soNp is closed under the action of R. Thus the p-primarycomponent Np of N is a submodule of N . To show

(4) N =⊕

prime p∈R

Np

we must show, for an arbitrary prime q ∈ R, that

(5) Nq ∩⊕

prime p∈R, p 6=q

Np = {0}

and that every element of N is the sum of elements in p-primary com-ponents (for which we need to assume N is torsion).

For (5), suppose x ∈ Nq and

(6) x ∈⊕

prime p∈R, p 6=q

Np.

Then 0 = qnx for some n ∈ Z+, and x is a finite sum of elements, eachof which is annihilated by a positive power of a prime of R distinctfrom q. Let r be the product of these prime power annihilators. Bysome simple module computations (actually carried out in the solutionto Exer. 10.3.5 above), we get rx = 0. Since all the prime divisors of rare distinct from q, the greatest common divisor of qn and r is 1. SinceR is a PID, there exist some s, t ∈ R such that 1 = sqn + tr. Hencex = 1x = (sqn + tr)x = sqnx + trx = s(qnx) + t(rx) = s0 + t0 = 0, asdesired.

Assume N is torsion, and let x be an arbitrary element of N . Then xis annihilated by some nonzero r ∈ R, so rx = 0. Since R is a PID,it is also a UFD, so r has a unique factorization into a finite productof finite powers of primes of R, say r = pa1

1 · · · · · pann , where p1, · · · , pn

are distinct primes and a1, . . . , an are non-negative integers. We shownext by induction on n that

(7) x ∈ Np1⊕ · · · ⊕Npn

52

If n = 1 then x is annihilated by a power of p1 since r = pa11 and

0 = rx = pa11 x, so x ∈ Np1

. Now we assume (7) holds for n, and wewill prove it for n + 1. Suppose rx = 0 where r = pa1

1 · · · · · pann p

an+1

n+1 ,where p1, · · · , pn+1 are distinct primes in R and a1, . . . , an+1 are non-negative integers. Then 0 = rx = pa1

1 · · · pann (p

an+1

n+1 x) so by the inductivehypothesis we have

(8) pan+1

n+1 x ∈ Np1⊕ · · · ⊕Npn

We also have 0 = rx = pan+1

n+1 (pa11 · · · · · pan

n x) since R is commutative, so

(9) pa11 · · · pan

n x ∈ Npn+1

Notice that pan+1

n+1 and pa11 · · · pan

n have no common factors and theirgreatest common divisor is 1. Hence there are s, t ∈ R such that1 = sp

an+1

n+1 + tpa11 · · · pan

n . Multiplying both sides by x, we get

x = span+1

n+1 + tpa11 x · · · pan

n x ∈ (Np1⊕ · · · ⊕Npn

)⊕Npn+1

by (8) and(9). This completes the proof of (7), which implies that x isin

⊕prime p∈RNp, as desired.

Homework #08, due 3/10/10 = 12.1.4, 12.1.5, 12.1.6, 12.1.7, 12.1.8,12.1.9

12.1.4 Let R be an integral domain, let M be an R-module and let Nbe a submodule of M . Suppose M has rank n, N has rank r and thequotient M/N has rank s. Prove that n = r+ s. [Let x1, x2, · · · , xs beelements of M whose images in M/N are a maximal set of independentelements and let xs+1, xs+2, · · · , xs+r be a maximal set of independentelements in N . Prove that x1, x2, · · · , xs+r are linearly independentin M and that for any elements y ∈ M there is a nonzero elementr ∈ R such that ry is a linear combination of these elements. Then useExercise 12.1.2.]

Note first that the authors are using “r” with two quite different mean-ings, both as an integer and as an element of R. Following the hint,

53

let x1, x2, · · · , xs ∈ M be elements whose images in M/N are a maxi-mal set of independent elements and let xs+1, xs+2, · · · , xs+r ∈ N be amaximal set of independent elements in N .

Claim: x1, x2, · · · , xs+r are linearly independent in M .

Suppose a1, · · · , as+r ∈ R and

(10) a1x1 + · · ·+ asxs + as+1xs+1 + · · ·+ as+rxs+r = 0.

Note that xs+1, xs+2, · · · , xs+r ∈ N , hence

(11) as+1xs+1 + · · ·+ as+rxs+r +N = N,

so, applying to both sides of (10) the canonical module homomorphismthat sends y ∈M to the coset y +N , we obtain

(12) a1x1 +N + · · ·+ asxs +N = N,

hence

(13) a1(x1 +N) + · · ·+ as(xs +N) = N,

but we chose x1, x2, · · · , xs ∈M so that x1 +N, x2 +N, · · · , xs +N ∈M/N are independent elements, so from (13) we conclude that

(14) a1 = · · · = as = 0.

Substituting these values back into (10) produces

(15) as+1xs+1 + · · ·+ as+rxs+r = 0.

Since we chose independent elements xs+1, xs+2, · · · , xs+r ∈ N , it fol-lows that

(16) as+1 = · · · = as+r = 0.

We have proved that (14) and (16) follow from (10), completing theproof of the Claim.

Let y ∈M . Then y +N ∈M/N . We chose x1, x2, · · · , xs ∈M so thatx1 +N, x2 +N, · · · , xs +N ∈M/N is a maximal set of s independentelements, so the s+1 elements y+N, x1+N, x2+N, · · · , xs+N ∈M/N

54

cannot be independent. There are coefficients b, a1, · · · , as ∈ R, not allzero, such that

(17) b(y +N) + a1(x1 +N) + · · ·+ as(xs +N) = N,

which implies

(18) z := by + a1x1 + · · ·+ asxs ∈ N.

Now if b = 0 then

(19) a1(x1 +N) + · · ·+ as(xs +N) = N,

hence a1 = · · · = as = 0 by the independence of x1 + N, · · · , xs + N ,contradicting our assumption that not all coefficients are zero, so b 6= 0.(We could have used Exer. 12.1.2(a) here, as is done in the nextparagraph.)

Let L := Rxs+1 + Rxs+2 + · · · + Rxs+r. Then L is a submodule of N .Since {xs+1, xs+2, · · · , xs+r} ⊆ N is a maximal set of independent ele-ments in N , by Exer. 12.1.2(a) the quotient N/L is torsion. Applyingthis to z ∈ N , we conclude there is some nonzero c ∈ R such thatc(z+L) = L, hence cz ∈ L. By the definitions of z and L this gives us

cz = c(by + a1x1 + · · ·+ asxs)

= cby + ca1x1 + · · ·+ casxs

∈ L = Rxs+1 +Rxs+2 + · · ·+Rxs+r

which implies the existence of coefficients as+1, · · · , as+r ∈ R such that

(20) cby + ca1x1 + · · ·+ casxs = as+1xs+1 + · · ·+ as+rxs+r.

Let r = cb. Then r 6= 0 because c 6= 0, b 6= 0, and R is an integraldomain. From (20) we get

ry = −ca1x1 − · · · − casxs + as+1xs+1 + · · ·+ as+rxs+r,

which implies

(21) ry ∈ Rx1 + · · ·+Rxs +Rxs+1 + · · ·+Rxs+r.

55

We have shown that for every y ∈M there is some nonzero r ∈ R suchthat (21), which says that the quotient module

M/(Rx1 + · · ·+Rxs +Rxs+1 + · · ·+Rxs+r)

is torsion. It follows by Exer. 12.1.2(b) that M has rank s + r. Byassumption, M has rank n, so n = s+ r.

12.1.5 Let R = Z[x] and let M = (2, x) be the ideal generated by2 and x, considered as a submodule of R. Show that {2, x} is not abasis of M . [Find a nontrivial R-linear dependence between these twoelements.] Show that the rank of M is 1 but that M is not free of rank1 (cf. Exercise 12.1.2).

Since R contains (a copy of) Z, the submodule M of R generated byx and 2 contains all multiples of x, and all sums of such multiples, soit contains all polynomials with constant 0. But M also contains allmultiples of 2, hence all even integers, and these can be constants inpolynomials in M , so M consists of those polynomials in Z[x] whoseconstant is even.

The two elements 2 and x of the R-module Z[x] satisfy the relationx·2+(−2)x = 0, where x and −2 are coefficients in the ring Z[x], so theset {2, x} is not independent. In fact, no 2-element subset {p, q} ⊆Mcan be independent, since q · p + (−p)q = 0. Note that p and q can’tboth be zero, otherwise {p, q} has only one element, so the coefficientsin the relation q · p+(−p)q = 0 are not both zero. Thus the rank of Mcannot be 2. Now the singleton subset {2} of M is independent, for ifr ∈ R and r · 2 = 0, then r = 0. So the rank of M is at least 1. Thisshows the rank of M is 1.

To show M is not free of rank 1, we need assume otherwise and geta contradiction. Suppose there is a single polynomial p(x) ∈ M suchthat M = Rp(x). Note that Rp(x) is simply the ideal (p(x)) of Z[x] =R generated by p(x). Then (2, x) = (p(x)) so there are polynomialsq(x), s(x) ∈ Z[x] = R such that 2 = p(x)q(x) and x = p(x)s(x). From2 = p(x)q(x) it follows that these two polynomials are actually bothconstants, i.e., p(x) = p ∈ Z and q(x) = q ∈ Z, hence 2 = pq and

56

x = ps(x). From 2 = pq we get {p, q} ⊆ {±1,±2}. If p were ±1 thenevery polynomial would be a multiple of p, contrary to the assumptionthat (p) = (2, x) 6= Z[x] = R. Therefore p is ±2, so by x = ps(x) wehave x = ±2s(x), hence the coefficient of x (which is 1) is equal tothe coefficient of x in the polynomial ±2s(x), but that coefficient is amultiple of 2, so again we have a contradiction. (This is the proof that(2, x) is a non-principal ideal of Z[x].) This shows that M cannot befree of rank 1.

12.1.6 Show that if R is an integral domain and M is any nonprincipalideal of R then M is torsion free of rank 1 but is not a free R-module.

For every nonzero m ∈ M ⊆ R, there is no nonzero r ∈ R such thatrm = 0, because R is an integral domain and has no zero divisors. Thisshows that every submodule M of R (treated as R-modules) is torsionfree.

Every submodule M ⊆ R has rank at least 1, for if m ∈M , then {m}is an independent set because if rm = 0 with r ∈ R, then r = 0, aswas just observed.

No submodule M ⊆ R has rank 2, for if nonzero m1,m2 ∈ M aredistinct, then the equation m1m2 +(−m2)m1 = 0 shows that {m1,m2}is not independent.

If a submodule M is free of rank 1, then there is some generator g ∈ Rsuch that M = Rg = (g), hence M is a principal ideal of R. Therefore,if M is a nonprincipal ideal of R, then M is a torsion free submoduleof R with rank 1, but M is not free of rank 1.

12.1.7 Let R be any ring, let A1, A2, · · · , Am be R-modules and let Bi

be a submodule Ai, 1 ≤ i ≤ m. Prove that

(A1 ⊕ A2 ⊕ · · · ⊕ Am)/(B1 ⊕B2 ⊕ · · · ⊕Bm)∼= (A1/B1)⊕ (A2/B2)⊕ · · · ⊕ (Am/Bm)

No solution given here.

12.1.8 Let R be a P.I.D., let B be a torsion R-module, and let p bea prime in R. Prove that if pb = 0 for some nonzero b ∈ B, thenAnn(B) ⊆ (p).

57

We assume R is a PID, B is a torsion R-module, p is prime in R,pb = 0, and 0 6= b ∈ B, and will show Ann(B) ⊆ (p). Let r ∈ Ann(B).This means that ∀a ∈ B, ra = 0. In particular, rb = 0. Since R is aPID, there is some d ∈ R such that (r, p) = (d). Then p ∈ (r, p) = (d),so d|p, so there is some q ∈ R such that dq = p. But p is prime in R,so either p|d or p|q. Case 1. Suppose first that p|d. Then (d) ⊆ (p),but we already have (p) ⊆ (r, p) = (d), so (p) = (d). In this case wehave r ∈ (r, p) = (d) = (p) so p|r and r ∈ (p), as desired. Case 2Suppose p|q. Then there is some u ∈ R such that pu = q. But we sawearlier that dq = p, so we now have p = dq = d(pu) = p(du), hence1 = du by cancellation. (This implies that d and u are units of R.)Now d ∈ (r, p), so there are s, t ∈ R such that d = rs+ pt. Then

(22) db = (rs+ pt)b = s(rb) + t(pb) = s0 + t0 = 0 + 0 = 0,

so b = 1b = (du)b = u(db) = u0 = 0, contradicting b 6= 0. This meansthat the second case cannot occur, so we’re done.

12.1.9 Give an example of an integral domain R and a nonzero torsionR-module M such that Ann(M) = 0. Prove that if N is a finitelygenerated torsion R-module then Ann(N) 6= 0.

Exer. 10.3.5 is essentially the same as this problem. Here’s a copy ofmy earlier solution to that problem, edited a little: 10.3.5 says, “LetR be an integral domain. Prove that every finitely generated torsionR-module has a nonzero annihilator, i.e., there is a nonzero r ∈ Rsuch that rm = 0 for all m ∈ M—here r does not depend on m (theannihilator of a module was defined in Exer. 10.1.9). Give an exampleof a torsion R-module whose annihilator is the zero ideal.”

Solution: Assume N is a finitely generated torsion R-module. Thenthere is a finite set A ⊆ N of nonzero elements such that N = RA.Let A = {a1, · · · , ak}, k ∈ ω. Since N is a torsion module, thereexist nonzero elements r1, · · · , rk ∈ R such that r1a1 = 0, r2a2 = 0,. . . , rkak = 0. Let q = r1 · . . . · rk. Consider an arbitrary elementn ∈ N . Since N = RA, there are ring elements s1, · · · , sk ∈ R suchthat n = s1a1 + · · · + skak. Let 1 ≤ i ≤ k. Since R is an integraldomain, R is commutative, so q = pri where p is the product of all the

58

other factors rj with j 6= i. Then qai = (pri)ai = p(riai) = p0 = 0.This shows qa1 = qa2 = · · · = qak = 0, so we have

qn = q(s1a1 + · · ·+ skak)

= q(s1a1) + · · ·+ q(skak) module axiom

= (qs1)a1 + · · ·+ (qsk)ak module axiom

= (s1q)a1 + · · ·+ (skq)ak since R is commutative

= s1(qa1) + · · ·+ sk(qak) module axiom

= s1(0) + · · ·+ sk(0) shown above

= 0 + · · ·+ 0

= 0

Since n was an arbitrary element of N , we have shown that qn = 0 forevery n ∈ N . Finally, we observe that q is nonzero because ri 6= 0 andthe product of nonzero elements in the integral domain R cannot bezero.

For an example of a torsion R-module whose annihilator is the zeroideal, let R = Z, and let M be the direct sum of the finite cyclic groupsZ/kZ with 2 ≤ k ∈ Z+, so

M =⊕

2≤k∈Z+

Z/kZ.

Let m ∈ M . Then there are k ∈ Z+ and a1, . . . , ak ∈ Z such thatm = (a1 + 2Z, · · · , ak + kZ, 0, 0, 0, · · · ), so k! ·m = 0, so m ∈ Tor(M).Then M is a torsion module, but no element of Z annihilates everyelement of M , for if k ∈ Z+ then k · (. . . , 1 + (k + 1)Z, 0, 0, . . . ) 6= 0.

Homework #09, due 3/24/10 = 12.2.2, 12.2.3, 12.2.4, 12.2.7, 12.2.8

12.2.2 Let M be as is Lemma 19. Prove that the minimal polynomialof M is the least common multiple of the minimal polynomials of A1,. . . , Ak.

59

In Lemma 19, F is a field and M is the block diagonal matrix

M =

A1 0 · · · 00 A2 · · · 0...

... . . . ...0 0 · · · Ak

where A1, · · · , Ak are square matrices with entries in F . It followsfrom the definition of matrix multiplication that for any positive integern ∈ Z+, we have

Mn =

An

1 0 · · · 00 An

2 · · · 0...

... . . . ...0 0 · · · An

k

and for any scalar a ∈ F , we have

aM =

aA1 0 · · · 00 aA2 · · · 0...

... . . . ...0 0 · · · aAk

so for any polynomial p(x) ∈ F [x],

p(M) =

p(A1) 0 · · · 0

0 p(A2) · · · 0...

... . . . ...0 0 · · · p(Ak)

Therefore p(x) annihilates M if and only if p(x) annihilates all the ma-trices A1, · · · , Ak. In particular, the minimum polynomial mM(x) anni-hilates M , hence annihilates all the matrices A1, · · · , Ak, and thereforeis a multiple of each of the minimum polynomials mA1

(x), · · · , mAk(x).

(The minimum polynomial of a matrix A divides every polynomial thatannihilates A, since otherwise the nonzero remainder upon division bythe minimum polynomial mA(x) would be a polynomial annihilatingA but with smaller degree than mA(x), contradicting the definition ofmA(x).) Therefore, to obtain mM(x), we simply choose, among allthe common multiples of the minimum polynomials of A1, · · · , Ak, the

60

one with smallest degree, but that is, by definition, the least commonmultiple of mA1

(x), · · · ,mAk(x).

12.2.3 Prove that two 2 × 2 matrices over F which are not scalarmatrices are similar if and only if they have the same characteristicpolynomial.

Let A be a 2×2 matrix over F . Then A determines a linear transforma-tion T of the vector space V = F 2, and T determines an F [x]-modulecalled in class “V [T ]” (vs. just “V ”), as a reminder that multiplicationby x acts the same as applying T .

The characteristic polynomial cA(x) is quadratic (since A is 2×2), andis the product of the invariant factors of V [T ]. The invariant factorshave degree 1 or more and the largest invariant factor is the minimalpolynomial mA(x).

One possibility is that the only invariant factor is cA(x) = mA(x) = x2+

a1x+ a0. In this case A is similar to the companion matrix

(0 −a01 −a1

).

The only other possibility is that cA(x) is not the minimal polynomial,but instead is the product of the minimal polynomial with the smallerinvariant factors. Since cA(x) is quadratic, the only way this can hap-pen is that there are two linear invariant factors. Since the invariantfactors are linear and monic, the only way one of them can divideanother is that they are equal to each other, so mA(x) = x − b andcA(x) = (x− b)2. The matrix A is similar to the block diagonal matrixwhose blocks are the companion matrices for the invariant factors x−band x − b, namely

(b)

and(b), so A is similar to the scalar matrix

S =

(b 00 b

)= bI. But for scalar matrices, similarity is equality, since

for any 2×2 matrix P we have PSP−1 = PbIP−1 = bPIP−1 = bI = S.Thus A is a scalar matrix.

Suppose A and B are 2×2 matrices but are not scalar matrices. Then,as was shown above, the rational canonical form of A is the companion

61

matrix

(0 −a01 −a1

), where cA(x) = mA(x) = x2 + a1x + a0 and the

rational canonical form of B is the companion matrix

(0 −b01 −b1

), where

cB(x) = mB(x) = x2 + b1x + b0. Now A and B are similar if and onlyif they have the same rational canonical form, so A and B are similarif and only if (

0 −a01 −a1

)=

(0 −b01 −b1

),

which is obviously equivalent to the equality of their characteristic poly-nomials.

12.2.4 Prove that two 3× 3 matrices over F are similar if and only ifthey have the same characteristic and the same minimal polynomials.Give an explicit counterexample to this assertion for 4× 4 matrices.

Let A be a 3×3 matrix over F . Then A determines a linear transforma-tion T of the vector space V = F 3, and T determines an F [x]-modulecalled V [T ] (see comments above on this notation).

The characteristic polynomial cA(x) is cubic (since A is 3 × 3), andis the product of the invariant factors of V [T ]. The invariant factorshave degree 1 or more and the largest invariant factor is the minimalpolynomial mA(x). The degree of the minimal polynomial is 1, 2 or 3.

Case 1. mA(x) has degree 3. Since cA(x) is cubic and is divisible by thecubic invariant factor mA(x), there can be no other invariant factors,hence the only invariant factor is cA(x) = mA(x) = x3+a2x

2+a1x+a0.

In this case A is similar to the companion matrix

0 0 −a01 0 −a10 1 −a2

.

Case 2. mA(x) has degree 1, say mA(x) = x− b. The other invariantfactors must be at least linear and must divide mA(x), so they areall linear. Each invariant factor divides the next and all of them aremonic, so they are all the same. Hence cA(x) = (x− b)3. The rationalcanonical form of A is the block diagonal matrix with three blocks of

62

the form (b), namely b 0 00 b 00 0 b

Case 3. mA(x) has degree 2. In this case there must be one otherlinear invariant factor

a1(x) = x− b,

and the minimal polynomial must be the second invariant factor a2(x) =mA(x), but this second factor is divisible by a1(x), so we have

a2(x) = mA(x) = (x− b)(x− c) = x2 − (b+ c)x+ bc,

and finally

cA(x) = a1(x)a2(x) = (x− b)2(x− c).

The rational canonical form in this case has a block of the form (b),and then a block that is the companion matrix of mA(x), namelyb 0 0

0 0 −bc0 1 b+ c

If two 3×3 matrices A and B have the same minimal and characteristicpolynomials, then they both fall into exactly one of the cases examinedabove, and in each case the rational canonical forms of A and B aredetermined by the coefficients of their minimal and characteristic poly-nomials, so that A and B will have the same rational canonical forms(and hence are similar) if and only if their characteristic and minimalpolynomials are the same.

For an explicit counterexample among 4× 4 matrices we look for ma-trices A and B such that A has invariant factors x − 1, x − 1, and(x−1)2 = mA(x), hence cA(x) = (x−1)4, while B has invariant factors

63

(x− 1)2 and (x− 1)2 = mB(x), with cB(x) = (x− 1)4. For example,

A =

1 0 0 00 1 0 00 0 0 −10 0 1 2

B =

0 −1 0 01 2 0 00 0 0 −10 0 1 2

Then A and B have the same characteristic and minimal polynomials,and yet are not similar because they have distinct rational canonicalforms (and are already in rational canonical form).

12.2.7 Determine the eigenvalues of the matrix

A =

0 1 0 00 0 1 00 0 0 11 0 0 0

We calculate the characteristic polynomial directly,

det(xI − A) =

∣∣∣∣∣∣∣∣x −1 0 00 x −1 00 0 x −1−1 0 0 x

∣∣∣∣∣∣∣∣= x

∣∣∣∣∣∣x −1 00 x −10 0 x

∣∣∣∣∣∣− (−1)

∣∣∣∣∣∣0 −1 00 x −1−1 0 x

∣∣∣∣∣∣= x

∣∣∣∣∣∣x −1 00 x −10 0 x

∣∣∣∣∣∣ +

∣∣∣∣∣∣0 −1 00 x −1−1 0 x

∣∣∣∣∣∣= x4 − 1 = (x2 + 1)(x− 1)(x+ 1)

which shows that the eigenvalues are ±1,±√−1.

64

12.2.8 Verify that the characteristic polynomial of the companion ma-trix

C =

0 0 0 · · · 0 −a01 0 0 · · · 0 −a10 1 0 · · · 0 −a2...

......

......

0 0 0 · · · 1 −an−1

is xn + an−1x

n−1 · · ·+ a1x+ a0.

The characteristic polynomial of C is the determinant of the matrix

xI − C =

x 0 0 · · · 0 a0−1 x 0 · · · 0 a10 −1 x · · · 0 a2...

......

......

0 0 0 · · · −1 x+ an−1

The determinant of xI − C is unchanged if we add multiples of rowsin xI − C to other rows in xI − C (see Prop. 11.22). Here is xI − Cwith the second row multiplied by x, the third row multiplied by x2,. . . , and row n multiplied by xn−1:

x 0 0 · · · 0 a0−x x2 0 · · · 0 a1x0 −x2 x3 · · · 0 a2x

2

......

......

...0 0 0 · · · −xn−1 xn + an−1x

n−1

Let p(x) = xn + an−1x

n−1 · · · + a1x + a0, and add rows 2 through n ofthe matrix above to the first row of matrix xI − C, obtaining

D =

0 0 0 · · · 0 p(x)−1 x 0 · · · 0 a10 −1 x · · · 0 a2...

......

......

0 0 0 · · · x an−20 0 0 · · · −1 x+ an−1

65

Now xI − C and D have the same determinant. If we compute thedeterminant of D by expanding along the first row, we get

(23) detD = (−1)n−1p(x) detE

where E is the square matrix with n− 1 rows and columns, −1 on thediagonal, and x on the superdiagonal:

E =

−1 x 0 · · · 00 −1 x · · · 0...

......

...0 0 0 · · · x0 0 0 · · · −1

The determinant of E is unchanged by adding x times the first columnto the second column. This operation eliminates the x in the secondcolumn, and the new second column can now be used to eliminate thex in the third column, and so on. The result is that the determinantof E is the same as the determinant of a (n − 1) × (n − 1) matrixwith −1 on the diagonal, whose determinant is (−1)n−1. Thus we havedetE = (−1)n−1, so by (23) we have

detD = (−1)n−1p(x) detE = (−1)n−1p(x)(−1)n−1 = p(x),

as desired.

Homework #10, due 3/31/10 = 12.2.9, 12.2.10, 12.2.11, 12.3.1,

Additional problems recommended for study: 12.2.1, 12.2.5, 12.2.6,12.2.12-18, 12.2.21-26

12.2.9 Find the rational canonical forms of 0 −1 −10 0 0−1 0 0

,

c c −10 c 1−1 1 c

, and

422 465 15 −30−420 −462 −15 30840 930 32 −60−140 −155 −5 12

.

66

12.2.9(a) Let

A =

0 −1 −10 0 0−1 0 0

Compute the characteristic polynomial cA(x) ∈ F [x] by expansiondown the first column:

cA(x) = det (xI − A) = det

x 1 10 x 01 0 x

= x det

(x 00 x

)+ det

(1 1x 0

)= x3 − x = x(x− 1)(x+ 1)

The characteristic polynomial is the product of the invariant factors ofA, by Prop. 12.20. Suppose A has more than one invariant factor, saya1(x), . . . , am(x) ∈ Q[x], where a1(x)| . . . |am(x) and m ≥ 2. If a linearpolynomial x + b divides one of the invariant factors besides the lastone, say x+ b divides ai(x) with i < m, then (x+ b)2 must divide cA(x)since ai(x)|am(x), so, for some p(x), q(x) ∈ F [x],

cA(x) = a1(x) · · · ai(x) · · · am(x)

= a1(x) · · · p(x)(x+ b) · · · q(x)(x+ b)

In this case the characteristic polynomial is x(x− 1)(x+ 1), which hasonly linear factors, so none of them can divide any “earlier” invari-ant factors, i.e., there is only one invariant factor, and it is both thecharacteristic polynomial and minimal polynomial of A. Thus

cA(x) = mA(x) = x(x− 1)(x+ 1) = x3 − x,

and the rational canonical form of A is the companion matrix of x3−x,namely 0 0 0

1 0 10 1 0

67

12.2.9(b) Let

A =

c c −10 c 1−1 1 c

Then the characteristic polynomial of A can be factored into distinctlinear factors:

cA(x) = det (xI − A)

= det

x− c 0 10 x− c −11 −1 x− c

= (x− c) det

(x− c −1−1 x− c

)+ det

(0 1

x− c −1

)= (x− c)((x− c)2 − 1)− (x− c)

= (x− c)((x− c)2 − 2)

= (x− c)(x− c+√

2)(x− c−√

2)

= x3 − 3cx2 + (3c2 − 2)x− c3 + 2c

As argued above, since the characteristic polynomial is the product ofdistinct linear polynomials, the minimal and characteristic polynomi-als are the same and are the sole invariant factor of A. So the rationalcanonical form of A is the companion matrix of the characteristic poly-nomial cA(x) of A, namely0 0 c3 − 2c

1 0 2− 3c2

0 1 3c

12.2.9(c) Let

A =

422 465 15 −30−420 −462 −15 30840 930 32 −60−140 −155 −5 12

68

The Smith normal form is1 0 0 00 x− 2 0 00 0 x− 2 00 0 0 (x− 2)(x+ 3)

so the invariant factors are a1 = x−2, a2 = x−2, and a2 = (x−2)(x+3) = x2 +x−6. The companion matrices for these polynomials are (2),

(2), and

(0 61 −1

), respectively, so the rational canonical form for A is

2 0 0 00 2 0 00 0 0 60 0 1 −1

12.2.10 Find all similarity classes of 6×6 matrices over Q with minimalpolynomial (x+2)2(x−1) (it suffices to give all lists of invariant factorsand write out some of their corresponding matrices.)

First we review some facts that put limits on the possible lists of invari-ant factors for A. The characteristic polynomial cA(x) of a 6×6 matrixA has degree 6 and cA(x) is divisible by the cubic minimal polynomialmA(x) = (x+2)2(x−1), which is the “last” and largest invariant factor.Now cA(x) is the product of the invariant factors, and the remaininginvariant factors (besides mA(x)) must divide mA(x) and have degreeat least 1, so cA(x) is the product of mA(x) and three linear factorsthat divide mA(x). The possible lists of invariant factors for A are:

(x+ 2)2(x− 1) (x+ 2)2(x− 1)(24)

x+ 2 (x+ 2)2 (x+ 2)2(x− 1)(25)

x+ 2 (x+ 2)(x− 1) (x+ 2)2(x− 1)(26)

x− 1 (x+ 2)(x− 1) (x+ 2)2(x− 1)(27)

x+ 2 x+ 2 x+ 2 (x+ 2)2(x− 1)(28)

x− 1 x− 1 x− 1 (x+ 2)2(x− 1)(29)

69

The corresponding characteristic polynomials are

(x+ 2)4(x− 1)2 in case (24)

(x+ 2)5(x− 1) in case (25)

(x+ 2)4(x− 1)2 in case (26)

(x+ 2)3(x− 1)3 in case (27)

(x+ 2)5(x− 1) in case (28)

(x+ 2)2(x− 1)4 in case (29)

Note that in the pairs of cases (24)(27) and (25)(28), there exist pairsof dissimilar 6 × 6 matrices with the same minimal and characteristicpolynomials. The polynomials that occur in the lists above as potentialinvariant factors, together with their companion matrices, are listednext.

(x+ 2)2(x− 1) = x3 + 3x2 − 4

0 0 41 0 00 1 −3

(x+ 2)2 = x2 + 4x+ 4

(0 −41 −4

)(x+ 2)(x− 1) = x2 + x− 2

(0 21 −1

)x+ 2

(−2

)x− 1

(1)

Finally, rational canonical forms for the six cases:

0 0 4 0 0 01 0 0 0 0 00 1 −3 0 0 00 0 0 0 0 40 0 0 1 0 00 0 0 0 1 −3

case (24)

70 −2 0 0 0 0 00 0 −4 0 0 00 1 −4 0 0 00 0 0 0 0 40 0 0 1 0 00 0 0 0 1 −3

case (25)

−2 0 0 0 0 00 0 2 0 0 00 1 −1 0 0 00 0 0 0 0 40 0 0 1 0 00 0 0 0 1 −3

case (26)

1 0 0 0 0 00 0 2 0 0 00 1 −1 0 0 00 0 0 0 0 40 0 0 1 0 00 0 0 0 1 −3

case (27)

−2 0 0 0 0 00 −2 0 0 0 00 0 −2 0 0 00 0 0 0 0 40 0 0 1 0 00 0 0 0 1 −3

case (28)

1 0 0 0 0 00 1 0 0 0 00 0 1 0 0 00 0 0 0 0 40 0 0 1 0 00 0 0 0 1 −3

case (29)

12.2.11 Find all similarity classes of 6× 6 matrices over C with char-acteristic polynomial (x4 − 1)(x2 − 1).

71

First we factor the characteristic polynomial over C.

cA(x) = (x4 − 1)(x2 − 1)

= (x2 − 1)(x2 + 1)(x− 1)(x+ 1)

= (x− 1)(x+ 1)(x−√−1)(x+

√−1)(x− 1)(x+ 1)

= (x− 1)2(x+ 1)2(x−√−1)(x+

√−1)

The problem is then to give all lists of invariant factors whose productis cA(x). Every linear polynomial that divides cA(x) (or any of theinvariant factors) must also divide the minimum polynomial mA(x).The various possibilities for mA(x) arise from linear factors that appearin cA(x) with powers higher than 1. So mA(x) and the remaininginvariant factors are one of the following possibilities:

(x− 1)2(x+ 1)2(x−√−1)(x+

√−1)(30)

(x− 1)(x+ 1)2(x−√−1)(x+

√−1) x− 1(31)

(x− 1)2(x+ 1)(x−√−1)(x+

√−1) x+ 1(32)

(x− 1)(x+ 1)(x−√−1)(x+

√−1) (x− 1)(x+ 1)(33)

For each of these cases we compute polynomials and companion matri-ces.

Case (30) The polynomial

cA(x) = (x− 1)2(x+ 1)2(x−√−1)(x+

√−1)

= (x4 − 1)(x2 − 1)

= x6 − x4 − x2 + 1

has companion matrix0 0 0 0 0 −11 0 0 0 0 00 1 0 0 0 10 0 1 0 0 00 0 0 1 0 10 0 0 0 1 0

72

and this matrix is a representative of the similarity class.

Case (31) The polynomial

(x− 1)(x+ 1)2(x−√−1)(x+

√−1)

= (x2 − 1)(x+ 1)(x2 + 1)

= (x4 − 1)(x+ 1)

= x5 + x4 − x− 1

has companion matrix0 0 0 0 11 0 0 0 10 1 0 0 00 0 1 0 00 0 0 1 −1

and the other invariant factor is x − 1, which has companion matrix

(1), so a representative in rational canonical form of the similarity classin this case is

1 0 0 0 0 00 0 0 0 0 10 1 0 0 0 10 0 1 0 0 00 0 0 1 0 00 0 0 0 1 −1

Case (32) The polynomial

(x− 1)2(x+ 1)(x−√−1)(x+

√−1)

= (x2 − 1)(x− 1)(x2 + 1)

= (x4 − 1)(x− 1)

= x5 − x4 − x+ 1

73

has companion matrix0 0 0 0 −11 0 0 0 10 1 0 0 00 0 1 0 00 0 0 1 1

and the other invariant factor is x + 1, which has companion matrix

(−1), so a representative in rational canonical form of the similarityclass in this case is

−1 0 0 0 0 00 0 0 0 0 −10 1 0 0 0 10 0 1 0 0 00 0 0 1 0 00 0 0 0 1 1

Case (33) The polynomial

(x− 1)(x+ 1)(x−√−1)(x+

√−1)

= (x2 − 1)(x2 + 1)

= x4 − 1

has companion matrix0 0 0 11 0 0 00 1 0 00 0 1 0

and (x− 1)(x+ 1) = x2 − 1 has companion matrix(

0 11 0

)

74

so a representative in rational canonical form of the similarity class inthis case is

0 1 0 0 0 01 0 0 0 0 00 0 0 0 0 10 0 1 0 0 00 0 0 1 0 00 0 0 0 1 0

12.3.1 Suppose the vector space V is the direct sum of cyclic F [x]-modules whose annihilators are (x + 1)2, (x − 1)(x2 + 1)2, (x4 − 1),and (x + 1)(x2 − 1). Determine the invariant factors and elementarydivisors for V .

Let us first assume that F = Q. That way the polynomial x2 + 1 isprime/irreducible. (Otherwise we could consider doing this problemover C, but then x2 + 1 is the product of two linear factors.) V isassumed to be the direct sum of four cyclic F [x]-modules, three ofwhich may be further decomposed into sums as follows:

F [x]/((x− 1)(x2 + 1)2) ∼= F [x]/(x− 1)⊕ F [x]/((x2 + 1)2)(34)

F [x]/(x4 − 1) ∼= F [x]/(x− 1)⊕ F [x]/(x+ 1)⊕ F [x]/(x2 + 1)(35)

F [x]/((x+ 1)(x2 − 1)) ∼= F [x]/((x+ 1)2)⊕ F [x]/(x− 1)(36)

so the list of elementary divisors of V is x − 1, x − 1, x − 1, x + 1,(x+ 1)2, (x+ 1)2, x2 + 1, (x2 + 1)2. The minimal polynomial must bethe product of the elementary divisors with the highest powers, so theinvariant factors are

mA(x) = (x− 1)(x+ 1)2(x2 + 1)2

a2 = (x− 1)(x+ 1)2(x2 + 1)

a1 = (x− 1)(x+ 1)

Now we solve the problem again, under the assumption on F (madeat the beginning of Section 12.2) that all the polynomials involved canbe factored into linear polynomials. This is certainly true if we assume

75

F = C. Two of the decompositions given above can be extended usingthe factorization x2 +1 = (x−

√−1)(x+

√−1). First, instead of (34),

we note that F [x]/((x− 1)(x2 + 1)2) is isomorphic to

F [x]/(x− 1)⊕ F [x]/((x+√−1)2)⊕ F [x]/((x−

√−1)2)

and instead of (35), we note that F [x]/(x4 − 1) is isomorphic to

F [x]/(x− 1)⊕ F [x]/(x+ 1)⊕ F [x]/(x+√−1)⊕ F [x]/(x−

√−1)

so the list of elementary divisors of V is

x− 1 x− 1 x− 1

x+ 1 (x+ 1)2 (x+ 1)2

x+√−1 (x+

√−1)2

x−√−1 (x−

√−1)2

and the invariant factors are

mA(x) = (x− 1)(x+ 1)2(x−√−1)2(x+

√−1)2

a2 = (x− 1)(x+ 1)2(x+√−1)(x−

√−1)

a1 = (x− 1)(x+ 1)

Homework #11, due 4/7/10 = 13.1.1, 13.1.2, 13.1.3, 13.2.1, 13.2.2

Additional problems recommended for study: 13.1.4, 13.1.5, 13.1.6,13.1.7, 13.1.8

13.1.1 Show that p(x) = x3 + 9x+ 6 is irreducible in Q[x]. Let θ be aroot of p(x). Find the inverse of 1 + θ in Q(θ).

The prime 3 does not divide the leading term of the polynomial x3 +9x+6, does divide the remaining coefficients 0, 9, and 6, but its square32 does not divide the constant term 6. Therefore p(x) is irreducible inQ[x] by the Eisenstein criterion.

By Th. 13.6, Q(θ) ∼= Q[x]/(p), so to find the inverse of 1 + θ in Q(θ)it is enough to compute the inverse of 1 + x in the quotient Q[x]/(p).

76

To do this, use the Euclidean algorithm to find a linear combination ofp(x) and 1 + x that equals 1. First divide x + 1 into x3 + 9x + 6, getquotient x2 − x+ 10 and remainder −4, so that

x3 + 9x+ 6 = (x+ 1)(x2 − x+ 10) + (−4)

It follows that

1 = −14(x

3 + 9x+ 6) + 14(x

2 − x+ 10)(x+ 1)

By the hypothesis on θ, θ3 + 9θ + 6 = 0, so substituting θ for x in theprevious equation gives

1 = 14(θ

2 − θ + 10)(θ + 1)

hence

(θ + 1)−1 = 14(θ

2 − θ + 10) ∈ Q(θ).

13.1.2 Show that x3− 2x− 2 is irreducible over Q and let θ be a root.Compute (1 + θ)(1 + θ + θ2) and 1+θ

1+θ+θ2 in Q(θ).

First observe that the prime 2 does not divide the leading coefficient ofx3−2x−2 (which is 1), does divide all the other coefficients (which are0, −2, and −2), but 22 does not divide the constant −2 of x3− 2x− 2,so x3 − 2x− 2 is irreducible over Q by the Eisenstein criterion.

Since θ is a root of x3 − 2x − 2, we know θ3 − 2θ − 2 = 0, henceθ3 = 2θ + 2, so

(1 + θ)(1 + θ + θ2) = (1 + θ + θ2) + (θ + θ2 + θ3)

= 1 + 2θ + 2θ2 + θ3

= 1 + 2θ + 2θ2 + (2θ + 2)

= 3 + 4θ + 2θ2

We will compute 1+θ1+θ+θ2 by first computing the inverse of 1+θ+θ2. By

Th. 13.6, Q(θ) ∼= Q[x]/(x3−2x−2), so to find the inverse of 1+θ+θ2 in

Q(θ) it is enough to compute the inverse of 1 + x+ x2 in the quotientQ[x]/(x3 − 2x − 2). To do this, use the Euclidean algorithm to find a

77

linear combination of x3− 2x− 2 and 1+x+x2 that equals 1; we needto find polynomials p(x), q(x) ∈ Q[x] such that

(37) (x3 − 2x− 2)p(x) + (x2 + x+ 1)q(x) = 1.

Then (x2 + x+ 1)q(x) = 1 in Q[x]/(x3 − 2x − 2), so the inverse of1 + θ + θ2 is q(θ).

Now such polynomials exist because x3 − 2x − 2 is irreducible, henceprime, and 1 + x+ x2 has smaller degree (2 vs. 3), so x3 − 2x− 2 and1 + x+ x2 are relatively prime, and Q[x] is a Euclidean domain, so thegreatest common divisor of x3 − 2x− 2 and 1 + x+ x2, namely, 1, is alinear combination of these two polynomials.

Use the Euclidean algorithm: divide x2 + x + 1 into x3 − 2x − 2, getquotient x− 1 and remainder −2x− 1, so that

(38) x3 − 2x− 2 = (x2 + x+ 1)(x− 1) + (−2x− 1),

which implies

(39) −2x− 1 = x3 − 2x− 2− (x2 + x+ 1)(x− 1).

Next we divide x2 +x+1 by the remainder −2x−1, obtaining quotient−1

2x−14 and remainder 3

4 , so that

(40) x2 + x+ 1 = (−2x− 1)(−12x−

14) + 3

4 ,

From (40) we get

34 = x2 + x+ 1−

(− 2x− 1

)(−1

2x−14)

so by substituting according to (39) and simplifying we get

34 = x2 + x+ 1−

(x3 − 2x− 2− (x2 + x+ 1)(x− 1)

)(−1

2x−14)

= x2 + x+ 1− (x3 − 2x− 2)(−12x−

14) + (x2 + x+ 1)(x− 1)(−1

2x−14)

= x2 + x+ 1 + (x3 − 2x− 2)(12x+ 1

4)− (x2 + x+ 1)(x− 1)(12x+ 1

4)

= (x3 − 2x− 2)(12x+ 1

4) + (x2 + x+ 1)(1− (x− 1)(12x+ 1

4))

Multiplying both sides by 4/3 gives

1 = 43(x

3 − 2x− 2)(12x+ 1

4) + (x2 + x+ 1)(1− (x− 1)(12x+ 1

4))

78

= (x3 − 2x− 2)13(2x+ 1) + (x2 + x+ 1)1

3(−2x2 + x+ 5)

Substitute θ for x, note that θ3−2θ−2 = 0 since θ is a root of x3−2x−2,and get

1 = (θ2 + θ + 1)13(−2θ2 + θ + 5)

Therefore (1+θ+θ2)−1 = 13(−2θ2+θ+5). To compute 1+θ

1+θ+θ2 , multiplythis answer by 1 + θ:

(1 + θ)(1 + θ + θ2)−1 = (1 + θ)(13(−2θ2 + θ + 5))

= 13(−2θ2 + θ + 5 + (−2θ2 + θ + 5)θ)

= 13(−2θ2 + θ + 5− 2θ3 + θ2 + 5θ)

= 13(−2θ3 − θ2 + 6θ + 5)

13.1.3 Show that x3 + x+ 1 is irreducible over F2 and let θ be a root.Compute the powers of θ in F2(θ).

Since x3 + x + 1 is cubic, it would have a linear factor (and hence aroot) if it were reducible, so to show it is irreducible it is enough toshow it has no roots in F2, but this is an easy calculation ( where =2is equality modulo 2): 03 + 0 + 1 = 1 6= 0, 13 + 1 + 1 = 3 =2 1 6= 0, and23 + 2 + 1 = 11 =2 1 6= 0.

Since θ is a root of x3 + x+ 1 (in some extension field of F2), we haveSince θ3 + θ+1 = 0, so θ3 = −1− θ =2 1+ θ. Then the powers of θ are

θ0 = 1

θ1 = θ

θ2

θ3 = 1 + θ

θ4 = θ + θ2

θ5 = θ2 + θ3 = 1 + θ + θ2

θ6 = θ + θ2 + θ3 = θ + θ2 + 1 + θ = 1 + θ2

θ7 = θ + θ3 = θ + 1 + θ = 1

79

13.2.1 Let F be a finite field of characteristic p. Prove that |F| = pn

for some positive integer n.

F has ground field generated by the unit element 1F. Since the char-acteristic is p, this ground field is (isomorphic to) Fp = Z/pZ. Since Fis an extension field of Fp, it is a vector space over Fp, but it is finite,so it must be a finite dimensional vector space, therefore isomorphicto a direct power of the ground field F, which has cardinality p, so thecardinality of F must therefore be a power of p.

13.2.2 Let g(x) = x2 +x−1 and let h(x) = x3−x+1. Obtain fields of4, 8, 9, and 27 elements by adjoining a root of f(x) to the field F wheref(x) = g(x) or h(x) and F = F2 or F3. Write down the multiplicationtables for the fields with 4 and 9 elements and show that the nonzeroelements form a cyclic group.

The polynomials g(x) and h(x) have low enough degree (2 or 3) that ifone of them were reducible it would have to have a linear factor, hencea root in the ground field. We prove that they are both irreducible overboth F2 and F3 by checking that neither of them has a root in eitherF2 or F3. In F2,

g(0) = 02 + 0− 1 = 1 6= 0 h(0) = 03 − 0 + 1 = 1 6= 0

g(1) = 12 + 1− 1 = 1 6= 0 h(1) = 13 − 1 + 1 = 1 6= 0

while in F3,

g(0) = 02 + 0− 1 = 2 6= 0 h(0) = 03 − 0 + 1 = 1 6= 0

g(1) = 12 + 1− 1 = 1 6= 0 h(1) = 13 − 1 + 1 = 1 6= 0

g(2) = 22 + 2− 1 = 2 6= 0 h(2) = 23 − 2 + 1 = 1 6= 0

4-element field. We get a field K with 4 elements by adding a rootof the quadratic polynomial g(x) = x2 + x − 1 to the field F2. Thisgives an extension of degree 2, so the size of the extension field is|F2|2 = 22 = 4. For an extension K/F2 in which g(x) has root α, we letK = F [x]/(g(x)) and α = x = x+(g(x)). Note that α2 = 1−α = 1+α(because the ground field of K is F2). The nonzero elements of K forma cyclic group generated by α, whose elements are α, α2 = 1 + α, and

80

α3 = α+ α2 = α+ 1 + α = 1. The multiplication table for elements ofK is

0 1 α 1 + α0 0 0 0 01 0 1 α 1 + αα 0 α 1 + α 11 + α 0 1 + α 1 α

8-element field. We get a field K with 8 elements by adding a rootof the cubic polynomial h(x) = x3−x+1 to the field F2. This gives anextension of degree 3, so the size of the extension field is |F2|3 = 23 = 8.Let K = F2[x]/(h(x)) and θ = x = x+ (h(x)). Then θ3 − θ + 1 = 0 soθ3 = θ − 1 = 1 + θ (because the ground field of K is F2). The nonzeroelements of K form a cyclic group generated by θ, whose elements areθ, θ2, θ3 = 1+θ, θ4 = θ+θ2, θ5 = θ2+θ3 = 1+θ+θ2, θ6 = θ+θ2+θ3 =θ+θ2+1+θ = 1+θ2, and θ7 = θ+θ3 = θ+1+θ = 1. The multiplicationtable for elements of K is

0 1 θ θ2 1 + θ θ + θ2 1 + θ + θ2 1 + θ2

0 0 0 0 0 0 0 0 01 0 1 θ θ2 1 + θ θ + θ2 1 + θ + θ2 1 + θ2

θ 0 θ θ2 1 + θ θ + θ2 1 + θ + θ2 1 + θ2 1θ2 0 θ2 1 + θ θ + θ2 1 + θ + θ2 1 + θ2 1 θ1 + θ 0 1 + θ θ + θ2 1 + θ + θ2 1 + θ2 1 θ θ2

θ + θ2 0 θ + θ2 1 + θ + θ2 1 + θ2 1 θ θ2 1 + θ1 + θ + θ2 0 1 + θ + θ2 1 + θ2 1 θ θ2 1 + θ θ + θ2

1 + θ2 0 1 + θ2 1 θ θ2 1 + θ θ + θ2 1 + θ + θ2

9-element field. We get a field K with 9 elements by adding a rootof the quadratic polynomial g(x) = x2 + x − 1 to the field F3. Thisgives an extension of degree 2, so the size of the extension field is|F3|2 = 32 = 9. For an extension K/F3 in which g(x) has root θ, we letK = F3[x]/(g(x)) and θ = x = x+(g(x)). Note that θ2 = 1−θ = 1+2θ(because the ground field of K is F3). The nonzero elements of Kform a cyclic group generated by θ, whose elements are the powersof θ, namely, θ, θ2 = 1 + 2θ, θ3 = θ + 2θ2 = θ + 2 + 4θ = 2 + 2θ,θ4 = 2θ + 2θ2 = θ + 2 + 4θ = 2, θ5 = 2θ, θ6 = 2θ2 = 2 + 4θ = 2 + θ,θ7 = 2θ + θ2 = 2θ + 1 + 2θ = 1 + θ, and θ8 = θ + θ2 = θ + 1 + 2θ = 1.

81

The multiplication table for elements of K is

0 1 θ 1 + 2θ 2 + 2θ 2 2θ 2 + θ 1 + θ0 0 0 0 0 0 0 0 0 01 0 1 θ 1 + 2θ 2 + 2θ 2 2θ 2 + θ 1 + θθ 0 θ 1 + 2θ 2 + 2θ 2 2θ 2 + θ 1 + θ 11 + 2θ 0 1 + 2θ 2 + 2θ 2 2θ 2 + θ 1 + θ 1 θ2 + 2θ 0 2 + 2θ 2 2θ 2 + θ 1 + θ 1 θ 1 + 2θ2 0 2 2θ 2 + θ 1 + θ 1 θ 1 + 2θ 2 + 2θ2θ 0 2θ 2 + θ 1 + θ 1 θ 1 + 2θ 2 + 2θ 22 + θ 0 2 + θ 1 + θ 1 θ 1 + 2θ 2 + 2θ 2 2θ1 + θ 0 1 + θ 1 θ 1 + 2θ 2 + 2θ 2 2θ 2 + θ

27-element field. We get a field K with 27 elements by adding a rootof the cubic polynomial h(x) = x3−x+1 to the field F3. This gives anextension of degree 3, so the size of the extension field is |F3|3 = 33 = 27.Let K = F3[x]/(h(x)) and θ = x = x+ (h(x)). Then θ3 − θ + 1 = 0 soθ3 = θ − 1 = 2 + θ (because the ground field of K is F3). The nonzeroelements of K form a cyclic group of order 26 generated by θ, whoseelements are

θ,

θ2,

θ3 = 2 + θ,

θ4 = 2θ + θ2,

θ5 = 2θ2 + θ3 = 2 + θ + 2θ2,

θ6 = 2θ + θ2 + 2θ3 = 2θ + θ2 + 4 + 2θ = 1 + θ + θ2,

θ7 = θ + θ2 + θ3 = θ + θ2 + 2 + θ = 2 + 2θ + θ2,

θ8 = 2θ + 2θ2 + θ3 = 2θ + 2θ2 + 2 + θ = 2 + 2θ2,

θ9 = 2θ + 2θ3 = 2θ + 4 + 2θ = 1 + θ,

θ10 = θ + θ2,

θ11 = θ2 + θ3 = 2 + θ + θ2,

θ12 = 2θ + θ2 + θ3 = 2θ + θ2 + 2 + θ = 2 + θ2,

82

θ13 = 2θ + θ3 = 2θ + 2 + θ = 2,

θ14 = 2θ,

θ15 = 2θ2,

θ16 = 2θ3 = 4 + 2θ = 1 + 2θ,

θ17 = θ + 2θ2,

θ18 = θ2 + 2θ3 = θ2 + 4 + 2θ = 1 + 2θ + θ2,

θ19 = θ + 2θ2 + θ3 = θ + 2θ2 + 2 + θ = 2 + 2θ + 2θ2,

θ20 = 2θ + 2θ2 + 2θ3 = 2θ + 2θ2 + 4 + 2θ = 1 + θ + 2θ2,

θ21 = θ + θ2 + 2θ3 = θ + θ2 + 4 + 2θ = 1 + θ2,

θ22 = θ + θ3 = θ + 2 + θ = 2 + 2θ,

θ23 = 2θ + 2θ2,

θ24 = 2θ2 + 2θ3 = 2θ2 + 4 + 2θ = 1 + 2θ + 2θ2,

θ25 = θ + 2θ2 + 2θ3 = θ + 2θ2 + 4 + 2θ = 1 + 2θ2,

θ26 = θ + 2θ3 = θ + 4 + 2θ = 1.

This list of powers of θ is a form of multiplication table for K. Forexample, to multiply 1 + θ and 2 + 2θ + 2θ2 we note that θ9 = 1 + θand θ19 = 2 + 2θ + 2θ2, hence

(1 + θ)(2 + 2θ + 2θ2) = θ9θ19 = θ28 = θ2.

Homework #12, due 4/14/10 = 13.2.3, 13.2.4, 13.2.5, 13.2.7, 13.2.11

Additional problems recommended for study: 13.2.8, 13.2.10, 13.2.13,13.2.14, 13.2.16, 13.2.17, 13.2.19, 13.2.20, 13.2.21

13.2.3 Determine the minimal polynomial over Q for the element 1+ i.

Let α = 1 + i. Noting i2 = −1, we have

α2 = (1 + i)2 = 12 + i2 + 2 · 1 · i = 1− 1 + 2i = 2i = 2(α− 1)

83

so α is a root of x2 − 2x + 2 ∈ Q[x]. This polynomial is irreducibleby the Eisenstein criterion using prime 2, and it is monic, so it is theminimal polynomial of its roots, one of which is α.

13.2.4 Determine the degree over Q of 2 +√

3 and of 1 + 3√

2 + 3√

4.

First we note that Q(2 +√

3) = Q(√

3) and√

3 /∈ Q so the degree ofthe extension Q(2+

√3) cannot be 1, hence is 2 or more. But it cannot

be more than 2, because 2 +√

3 satisfies the equation (x− 2)2 = 3 andis therefore a root of the monic quadratic polynomial x2− 4x+ 1. Thedegree of Q(2+

√3) over Q is therefore 2, and the minimal polynomial

of 2 +√

3 over Q is x2 − 4x+ 2.

Let β = 1 + 3√

2 + 3√

4 and α = 3√

2. Notice that β ∈ Q(α) becauseβ = 1 + α + α2, so we have Q ⊆ Q(β) ⊆ Q(α). By “degree multiply”we have [Q(α) : Q(β)][Q(β) : Q] = [Q(α) : Q]. But [Q(α) : Q] = 3since α is a root of the monic irreducible polynomial x3 − 2 ∈ Q[x], so[Q(β) : Q] divides 3. Therefore [Q(β) : Q] must be either 1 or 3. Toshow it is 3, assume, to the contrary, that [Q(β) : Q] = 1. Then β ∈ Q,and

β2 = (1 + α+ α2)2 β = 1 + α+ α2

= 1 + α2 + α4 + 2α+ 2α2 + 2α3

= 1 + α2 + 2α+ 2α+ 2α2 + 4 α3 = 2

= 5 + 4α+ 3α2

so

β2 − 3β = 5 + 4α+ 3α2 − 3(1 + α+ α2)

= 2 + α

hence

α = β2 − 3β − 2 ∈ Q(β) = Qwhich contradicts the fact that α /∈ Q. Of course, these calculationsalso allow us to proceed slightly differently. We have [Q(α) : Q] = 3so we need only show Q(α) = Q(β), but this follows from equations

84

derived above, namely, β = 1 + α+ α2 ∈ Q(α) and α = β2 − 3β − 2 ∈Q(β).

13.2.5 Let F = Q(i). Prove that x3− 2 and x3− 3 are irreducible overF .

[A more general theorem was proved in class that includes this problemas a special case. The problem could be solved by an appeal to thistheorem, but the solution presented below essentially repeats the proofof the general theorem.]

Let α = 3√

2. Then α /∈ Q and α is a root of x3 − 2 ∈ Q[x]. Sincethe monic irreducible cubic polynomial x3 − 2 ∈ Q[x] is irreducible byEisenstein (via prime 2), it is the minimal polynomial of α over Q.Therefore

(41) [Q(α) : Q] = 3

Also, F is a quadratic extension of Q since i /∈ Q and the degree of i is2, because i is the root of the monic irreducible quadratic polynomialx2 + 1 (so x2 + 1 is the minimal polynomial of i over Q). Thus

(42) [F : Q] = 2

Consider the extension field K = Q(α, i) of Q. Now K has subfield F ,so by Corollary 13.15 and (42), [K : Q] is divisible by 2. Similarly, Khas subfield Q(α), so by Corollary 13.15 and (41), [K : Q] is divisibleby 3. Since [K : Q] is divisible by both 2 and 3, and these two numbersare relatively prime, we have

(43) [K : Q] ≥ 6

Now Q is contained in the extension field F , so by Corollary 13.10 theminimal polynomial of α over F must divide the minimal polynomial ofα over Q, hence mα,F |mα,Q. From above we know that mα,Q = x3 − 2,so

(44) deg(mα,F ) ≤ deg(mα,Q) = 3.

85

The degree of F (α) over F is the degree of mα,F , and K = F (α), so by(43) and Th. 13.14 (“degrees multiply”), we have

6 ≤ [K : Q] = deg(mα,F )[F : Q] = 2 deg(mα,F ),

hence

(45) 3 ≤ deg(mα,F ).

From (44) and (46) we have

(46) 3 = deg(mα,F ) = [K : F ] = F (α) : F ].

By (46) and the fact, established above, that mα,F |x3 − 2, we havemα,F = x3 − 2. Minimal polynomials are irreducible, so x3 − 2 isirreducible over F .

Exactly the same argument also shows that x3 − 3 is irreducible overF . This time, let β = 3

√3 /∈ Q. Then β is a root of x3 − 3 ∈ Q[x].

The monic irreducible cubic polynomial x3 − 3 ∈ Q[x] is irreducibleover Q by Eisenstein, this time using prime 3, so x3− 3 is the minimalpolynomial of β over Q, [Q(β) : Q] = 3, etc.

13.2.7 Prove that Q(√

2+√

3) = Q(√

2,√

3) [one inclusion is obvious;for the other consider (

√2+

√3)2, etc.] Find an irreducible polynomial

satisfied by√

2 +√

3.

Clearly,√

2 +√

3 ∈ Q(√

2,√

3), so Q(√

2 +√

3) ⊆ Q(√

2,√

3). Letα =

√2+

√3. Then α2 = 2+3+2

√2√

3, so√

2√

3 = 12(α

2−5) ∈ Q(α).

It follows that α√

2√

3− 2α ∈ Q(α), but

α√

2√

3− 2α = (√

2 +√

3)√

2√

3

= 2√

3 + 3√

2− 2√

2− 2√

3

=√

2

so√

2 ∈ Q(α), which implies that√

3 = α −√

2 ∈ Q(α). From theselast two statements we get Q(

√2,√

3) ⊆ Q(α), as desired.

13.2.11(a) Let√

3 + 4i denote the square root of the complex num-ber 3 + 4i that lies in the first quadrant and let

√3− 4i denote the

86

square root of 3 − 4i that lies in the fourth quadrant. Prove that[Q(

√3 + 4i +

√3− 4i) : Q] = 1. (b) Determine the degree of the

extension Q(√

1 +√−3 +

√1−

√−3) over Q.

(a) Conjugation is an isomorphism of the complex field C onto itself.The conjugate of a+ bi is a− bi. This map takes numbers in the firstquadrant to numbers in the fourth, and numbers in the fourth quadrantto numbers in the first. It is a field isomorphism, and therefore takessquare roots to square roots. In particular, since

√3 + 4i is a square

root of 3 + 4i, its conjugate is a square root of the conjugate 3− 4i of3 + 4i. Furthermore, since

√3 + 4i is the square root of 3 + 4i in the

first quadrant, its conjugate is the square root of 3 − 4i in the fourthquadrant, which is, by definition,

√3− 4i. Thus

√3 + 4i and

√3− 4i

are conjugates of each other. The sum of a complex number and itsconjugate is a real number, so

√3 + 4i+

√3− 4i ∈ R. Next, note that

(√

3 + 4i+√

3− 4i)2 = (3 + 4i) + (3− 4i) + 2√

3 + 4i√

3− 4i

= 6 + 2√

(3 + 4i)(3− 4i)

= 6 + 2√

32 − 42i2 + 12i− 12i

= 6 + 2√

9 + 16

= 6 + 2√

25

= 16

so√

3 + 4i+√

3− 4i = 4 ∈ Q. It follows that Q(√

3 + 4i+√

3− 4i) =Q, so [Q(

√3 + 4i+

√3− 4i) : Q] = 1.

13.2.11(b) Repeating the opening remarks of part (a), we conclude

that√

1 +√−3 +

√1−

√−3 ∈ R. Now calculate:(√

1 +√−3 +

√1−

√−3

)2

= (1 +√−3) + (1−

√−3) + 2

√1 +

√−3

√1−

√−3

= 2 + 2

√(1 +

√−3)(1−

√−3)

87

= 2 + 2

√12 −

√−3

2

= 2 + 2√

1 + 3

= 2 + 2√

4

= 6

so √1 +

√−3 +

√1−

√−3 = ±

√6

hence

Q(√

1 +√−3 +

√1−

√−3

)= Q

(√6)

and the degree of Q(√

6)

over Q is 2, so the degree of the extension

Q(√

1 +√−3 +

√1−

√−3

)over Q is also 2.

Homework #13, due 4/21/10 = 13.2.12, 13.3.1, 13.3.2, 13.3.4, 13.4.1,13.4.2

Additional problems recommended for study: 13.3.3, 13.3.5, 13.4.3,13.4.4, 13.4.5, 13.4.6

13.2.12 Suppose the degree of the extension K/F is a prime p. Showthat any subfield E of K containing F is either K or F .

By the theorem that “degrees multiply”, we have

(47) p = [K : F ] = [K : E][E : F ]

so [E : F ] is a positive integer that divides the prime p, and thereforemust be equal to 1 or p. If [E : F ] = 1 then E = F . If [E : F ] = pthen [K : E] = 1 by (47), so K = F . Thus either K = E or F = E.

13.3.1 Prove that it is impossible to construct the regular 9-gon.

The angles in a regular 9-gon are all equal to 3609◦

= 40◦. Anglescan be bisected with compass and straightedge, so if a 9-gon could beconstructed, then one of its angles could be bisected, producing a 20◦

angle, from which it would be possible to construct cos(20◦). However,

88

by the proof of Th. 13.24(II), cos(20◦) is not constructible. In fact,20◦ has degree 3 over Q because 2 cos(20◦) satisfies the irreducible cubicpolynomial x3 − 3x− 1.

13.3.2 Prove that Archimedes’ construction actually trisects the angleθ. [Note that isosceles triangles in Figure 5 to prove that β = γ = 2α.]

Refer to Figure 5 in the text. Let O be the center of the circle of radius1. Let P be the point outside the circle and lying on a diameter of thecircle. Let X and Y be the points on the circle such that P , X, and Yare collinear, and X is between P and Y . By hypothesis, PX = 1. Letα = ∠XPO, β = ∠Y XO, γ = ∠XY O, and let θ be the angle betweenthe diameter PO and the radius OY , such that θ is an exterior angleof 4PY O.

Then, by the pons asinorum, α = ∠XPO = ∠XOP because 4PXOis isosceles with base PO. Note that β is an exterior angle of 4PXOand is therefore equal to the sum of the two remote interior angles,which are ∠XPO and ∠XOP . These two angles are both equal to α,so β = 2α.

By the pons asinorum, β = ∠Y XO = ∠XY O = γ because 4XOYis isosceles with base XY . Note that since θ is an exterior angle of4PY O, we have θ = γ + α = β + α = 2α+ α = 3α.

13.3.4 The construction of the regular 7-gon amounts to the con-structibility of cos(2π/7). We shall see later (Section 14.5 and Ex-ercise 2 of Section 14.7) that α = cos(2π/7) satisfies the equationx3 + x2 − 2x − 1 = 0. Use this to prove that the regular 7-gon isnot constructible by straightedge and compass.

If p(x) = x3+x2−2x−1 can be factored over Q, then by Gauss’s Lemmait can be factored over Z, and hence has a root in Z which must divide1. But p(1) = −1 6= 0 and p(−1) = 1 6= 0, so p(x) = x3 + x2 − 2x− 1is irreducible over Q. Consequently its roots (including α) have degree3 over Q and therefore do not lie in an finite dimensional extension ofQ whose dimension is a power of 2, and hence α is not constructible.

13.4.1 Determine the splitting field and its degree over Q for x4 − 2.

89

To find the roots of x4 − 2, factor it:

x4 − 2 = (x2 +√

2)(x2 −√

2) = (x+4√

2i)(x− 4√

2i)(x+4√

2)(x− 4√

2)

The roots of x4 − 2 are therefore − 4√

2i, 4√

2i, − 4√

2, and 4√

2, so thesplitting field is

F := Q(4√

2i,− 4√

2i,4√

2,− 4√

2)

Obviously

F = Q(4√

2i,4√

2) = Q(i,4√

2)

Note that i is not in Q( 4√

2) because the latter field in contained in thereal numbers R, but i /∈ R. So the extension obtained by adding i toQ( 4√

2) is a proper extension and therefore has degree at least 2. On theother hand, i is a root of the quadratic polynomial x2 + 1 ∈ Q( 4

√2)[x],

so the degree can also not be any more than 2, and therefore is exactly2. Thus

[Q(i,4√

2) : Q(4√

2)] = 2.

We also know that[Q(

4√

2) : Q] = 4

because 4√

2 is a root of the polynomial x4− 2 which is irreducible overQ by Eisenstein, using prime 2. We therefore have (by the “degreesmultiply” theorem)

[F : Q] = [Q(i,4√

2) : Q(4√

2)][Q(4√

2) : Q] = 2 · 4 = 8

so the degree of the splitting field F over Q is 8.

13.4.2 Determine the splitting field and its degree over Q for x4 + 2.

The roots of x4 + 2 are ±1±i4√

2, so the splitting field of x4 + 2 is

F := Q(

1 + i4√

2,1− i

4√

2,−1 + i

4√

2,−1− i

4√

2

).

We must compute its degree over Q. Let α = 1+i4√

2. Then, using the first

two of the roots, we get 4√

2 ∈ F because

4√

2 =

(1

2

(24√

2

))−1

=

(1

2

(1 + i

4√

2+

1− i4√

2

))−1

∈ F.

90

We then conclude that i ∈ F because

i = α4√

2− 1 ∈ F.Therefore we now know that Q( 4

√2, i) ⊆ F , but the opposite inclusion

is obvious, so

F = Q(4√

2, i),

and [F : Q] = 8, as was shown in the previous problem.

Homework #14, due 4/28/10 = 14.1.1, 14.1.2, 14.1.3, 14.1.4, 14.2.1,14.2.2,

Additional problems recommended for study: 13.4.3, 13.4.4, 13.4.5,13.4.6, 14.1.5, 14.1.6, 14.1.7, 14.1.8, 14.1.9, 14.2.4, 14.2.5, 14.2.6,14.2.7, 14.2.8, 14.2.10, 14.2.12, 14.2.14

14.1.1(a) Show that if the field K is generated over F by the elementsα1, · · · , αn then an automorphism σ of K fixing F is uniquely deter-mined by σ(α1), · · · , σ(αn). In particular show that an automorphismfixes K if and only if it fixes a [i.e., every] set of generators for K.

Except for some minor notational changes, the following proof worksfor an arbitrary (and possibly infinite) set of generators of K over F .Assume K = F (α1, · · · , αn) and σ is an automorphism of K fixing F .To show σ is uniquely determined by σ(α1), · · · , σ(αn), assume that τis some other automorphism of K that fixes F and agrees with σ onthe generators, i.e.

(48) σ(α1) = τ(α1), · · · , σ(αn) = τ(αn).

Let E := {β ∈ K|σ(β) = τ(β)} be the set of elements of K on whichτ and σ agree. We show that E is a field containing F and every αi.First we note that α1, · · · , αn ∈ E by (48), and F ⊆ E because τ andσ are both the identity function on elements of F , so if β ∈ F thenβ = τ(β) = σ(β). Suppose β, γ ∈ E. Then βγ−1 ∈ E because

σ(βγ−1) = σ(β)σ(γ)−1 σ is an automorphism

= τ(β)τ(γ)−1 β, γ ∈ E

91

= τ(βγ−1) τ is an automorphism

and β ± γ ∈ E because

σ(β ± γ) = σ(β)± σ(γ) σ is an automorphism

= τ(β)± τ(γ) β, γ ∈ E= τ(β ± γ) τ is an automorphism

Since E is a field containing F and α1, · · · , αn, we haveK = F (α1, · · · , αn) ⊆E, but by definition we have E ⊆ K, so E = K. This means that σand τ agree on every element in their domain K, and are therefore thesame function.

If an automorphism fixes K then it obviously fixes every subfield F ofK and every set of generators of K over F .

Conversely, for every set A ⊆ K of generators and every subfield F , ifan automorphism σ fixes F and A then it agrees on A with the identityfunction on K, which is an automorphism of K fixing F . By what wasproved above, it follows that the σ and the identity function on K mustbe the same.

14.1.1(b) Let G ≤ Gal(K/F ) be a subgroup of the Galois group ofthe extension K/F and suppose σ1, · · · , σk are generators for G. Showthat the subfield E/F is fixed by G if and only if it is fixed by thegenerators σ1, · · · , σk.One direction is trivial, for if G fixes E/F then all its elements fix E/F ,including the generators σ1, · · · , σk.We therefore need only assume E/F is fixed by the generators σ1, · · · , σkand show that E/F is fixed by G. Let H be the set of elements of Gthat fix E/F , that is,

H = {g ∈ G|∀β ∈ E(g(β) = β)}.Then the generators are inH because, by hypothesis, they fix E/F . Weshow now that H is closed under the group operations. Let g, h ∈ H,that is, g and h fix E/F . Then, for every β ∈ E,

(g ◦ h)(β) = g(h(β))

92

= g(β) since h fixes E/F

= β since g fixes E/F

so g ◦ h ∈ E, and g−1 ∈ H because

g−1(β) = g−1(g(β)) since g fixes E/F

= (g−1 ◦ g)(β) = 1(β) = β

We have shown that H is a subgroup of G that contains the generatorsof G, so G = H. By the definition of H, we conclude G fixes E/F .

14.1.2 Let τ be the map τ : C → C defined by τ(a + bi) = a − bi(complex conjugation). Prove that τ is an automorphism of C.

First note that τ behaves properly with respect to addition and sub-traction because

τ((a+ bi) + (c+ di)) = τ(a+ c+ (b+ d)i)

= a+ c− (b+ d)i

= a− bi+ c− di

= τ(c+ di) + τ(c+ di)

and

τ((a+ bi)− (c+ di)) = τ(a+ c− (b+ d)i)

= a+ c+ (b+ d)i

= a+ bi+ c+ di

= τ(a− bi) + τ(c− di)

For multiplication, we have

τ((a+ bi)(c+ di)) = τ(ac− bd+ (bc+ ad)i)

= ac− bd− (bc+ ad)i

= (a− bi)(c− di)

= τ(a+ bi)τ(c+ di)

93

For quotients, first note that

τ

(a+ bi

c+ di

)= τ

((a+ bi)(c− di)

(c+ di)(c− di)

)= τ

((ac+ bd) + (bc− ad)i

c2 + d2

)=

(ac+ bd) + (ad− bc)i

c2 + d2

but we also have

τ(a+ bi)

τ(c+ di)=a− bi

c− di=

(a− bi)(c+ di)

(c− di)(c+ di)=

(ac+ bd) + (ad− bc)i

c2 + d2

so

τ

(a+ bi

c+ di

)=τ(a+ bi)

τ(c+ di).

Note that τ is its own inverse (τ−1 = τ because τ ◦ τ is the identityfunction on C), and hence is bijective. Therefore τ is an automorphismof C.

14.1.3 Determine the fixed field of complex conjugation on C.

Let τ be complex conjugation, as in the previous problem. Certainlyτ fixes R, for if a ∈ R then τ(a) = τ(a + 0i) = a − 0i = a. Supposea, b ∈ R and a + bi is fixed by τ . Then τ(a + bi) = a− bi = a + bi, so0 = 2bi, hence b = 0, hence a+ bi = a ∈ R. Thus the fixed field of τ isR.

14.1.4 Prove that Q(√

2) and Q(√

3) are not isomorphic.

Assume, to the contrary, that there is some σ which is an isomorphismfrom Q(

√2) onto Q(

√3). Then σ(1) = 1 and, more generally, σ(q) = q

for all q ∈ Q. Consider the element σ(√

2) ∈ Q(√

3). Since Q(√

3) is aquadratic extension of Q and has basis {1,

√3}, there are a, b ∈ Q such

that σ(√

2) = a + b√

3. Squaring both sides and using the hypothesisthat σ is an isomorphism, we get

(49) 2 = σ(2) = σ(√

2)2) = (σ(√

2))2 = (a+b√

3)2 = a2+2ab√

3+3b2.

If ab 6= 0 then, since a, b ∈ Q,

√3 =

2− a2 − 3b2

2ab∈ Q.

94

But this is a contradiction, so ab = 0, hence a = 0 or b = 0. If a = 0

then, by (49), 2 = 3b2, which implies that√

23 = b ∈ Q, a contradiction.

If b = 0, then (49) implies√

2 = a ∈ Q, again a contradiction. Thisshows no such isomorphism σ can exist, hence Q(

√2) 6∼= Q(

√3).

14.2.1 Determine the minimal polynomial over Q for the element√

2+√5.

Let p = 2 and q = 5 and α =√p+

√q. Then α2 = p+ q + 2

√p√q, so

4pq = (α2 − (p+ q))2 = α4 − 2α2(p+ q) + (p2 + 2pq + q2)

so0 = α4 − 2α2(p+ q) + (p− q)2

With p = 2 and q = 5, we get 0 = α4 − 14α2 + 9 and the minimalpolynomial of α is x4 − 14x2 + 9.

14.2.2 Determine the minimal polynomial over Q for the element 1 +3√

2 + 3√

4.

Let p = 2 and α = 1 + 3√p+ 3

√p2. Then α− 1 = 3

√p+ 3

√p2, so, cubing

both sides, we have

α3 − 3α2 + 3α− 1 = (α− 1)3 = p(1 + 3√p)3

= p(1 + 3 3√p+ 3 3

√p2 + p)

= p(1 + 3(α− 1) + p)

= p(p− 2 + 3α)

= p2 − 2p+ 3pα

It follows that

α3 − 3α2 + 3(1− p)α− (p− 1)2 = 0.

The minimal polynomial of α over Q is

x3 − 3x2 + 3(1− p)x− (p− 1)2.

With p = 2 we conclude that the minimal polynomial of α over Q is

x3 − 3x2 − 3x− 1