rings lect 2

8/6/2019 Rings Lect 2

http://slidepdf.com/reader/full/rings-lect-2 1/8

2 Ideals, Quotients, Homomorphisms

This section is a brief review of some of the key ideas of any algebraic system: the manufactureof ‘quotient structures’ and the analysis of the ‘homomorphisms’.

2.1 Ideals

We begin with two definitions.

Definition 1. We say that a subset I of a ring R is an ideal and write I ¡R if

(i) I is a subgroup under +;

(ii) for all a ∈ R and all i ∈ I we have ai ∈ I .

Examples. {0} and R are always ideals. If K is a field then the only ideals of K are {0}and K. If n ∈ Z then the set of all multiples of n, nZ is an ideal of Z. Generally if R is aring and a ∈ R then aR = {ax : x ∈ R} is an ideal of R. The ideal aR is sometimes denoted

by < a > or (a) and is known as the principal ideal generated by a.Proposition 2.1.1 (Test for ideals). Let R be a ring. Then I ⊆ R is an ideal of R if

(i) 0 ∈ I ;

(ii) if a, b ∈ I then (a − b) ∈ I ;

(iii) if a ∈ I , r ∈ R then ar ∈ I .

Proof. Easy.

Definition 2. Suppose that R is a ring and I ¡R. For each a ∈ R we call

a := a + I := {a + i | i ∈ I }

the coset of a.

The notation a is neat, but needs care if there are different ideals around as it doesn’tidentify I in the way that the ‘a + I ’ notation does.

2.2 Quotients

Suppose that R is a ring and that I ¡R, I = R. Then the set of cosets, which we denote byR or by R/I can be made into a ring. We recap briefly from the first year work.

2.2.1 Operations

(i) as zero element, the class 0;

(ii) −a := −a;

(iii) a + b := (a + b);

(iv) as identity element, the class 1;

(v) a · b := (a · b).

At once we have a doubt: are these good definitions? Let us deal only with the last one.Suppose that a = a′ and b = b′; can we be sure that a · b = a′ · b′? Of course we can, but ittakes a moment to check.

6



2.2.2 Axioms

Now that we have the operations we need to see whether the axioms are satisfied. Again letus do only one, (R5) say. We need to prove that for all a,b,c ∈ R the following holds:

a · (b + c) = a · b + a · c.

Well

LHS = a · (b + c) assumption

= a · ((b + c) definition of + on R

= (a · (b + c) definition of · on R

= (a · b + a · c) Axiom (R5) in R

= (a · b + a · c) definition of + on R

= (a · b + a · c) definition of · on R (twice)

= RHS assumption

and we are done.

2.3 Applications

2.3.1 Modular Arithmetic

Clearly dZ := {dn | n ∈ Z} is an ideal. We can therefore carry out the construction, andmanufacture the quotient ring; in this case the coset notation Z = Z/dZ helps us keep trackof the modulus d. We call this the ring of integers modulo d and we denote it by Zd.

We can use these rings to illustrate other things we have done: for example, when is Z/dZan integral domain? Let 0 = a ∈ Z/dZ, and suppose that for some 0 = b ∈ Z/dZ we havethat ab = 0. Then ab is divisible by d. If d = d1d2 is composite this is always possible, justtake a := d1 and b := d2. So for an integral domain d must be prime (or 0). In both thesecases (for detailed proof see later, but most of us think it is obvious) we do get an integraldomain.

Perhaps more interestingly, what are the units of Z/dZ? Here we are seeking those u forwhich we can find a v such that uv − 1 is divisible by d. That is, given d and u we ask whenwe can find v and m such that vu + md = 1. If this is possible then the only common factorsof u and d are ±1; conversely if the only common factors are ±1 then by Euclid’s algorithm

we can find v and m. That is,

(Z/dZ)∗ = {u | (u, d) = 1}.

We denote the order of this group by φ(d); for example φ(12) = 4, since there are four units,±1 and ±5.

2.3.2 The Complex Numbers

Suppose we start with the real numbers. We can construct the polynomial ring R[X ]. In thisring the multiples of (X 2 + 1) form an ideal, which we write as X 2 + 1—trivial calculation.So we can form the quotient ring R[X ]/X 2 + 1.

7



What is it? It is, once again, the complex numbers C. Write i := X = X + X 2 + 1,then every element can be expressed as a + bi (use the Division Algorithm to see that everypolynomial can be written as a + bX + g(X )(X 2 + 1)). We have that

i2

=

X 2

= X 2

= X 2

+ 1 − 1 = X 2

+ 1 − 1 = 0 − 1 = −1.

We’ve now provided a theoretical underpinning for ideas like ‘adjoin a new number whosesquare is −1’.

2.3.3 The square root of 2

For a moment suppose we are Ancient Greeks. With much hard work we have constructed(in our own way) the rational field Q. Then we start drawing right-angled triangles and tryto find the length of the hypotenuse of the isoceles right-angled triangle of side 1. To ourhorror we find we need a number whose square is 2, and of course we have good proofs thatno such rational number exists. Our construction above would save the day: just look at the

ring Q[X ] and consider the ideal X 2 − 2 consisting of the multiples of X 2 − 2. The cosetα := X + X 2 − 2 in the quotient Q[X ]/X 2 − 2 is the number we are looking for.

(There is much more to be said here: see later in the course.)

2.4 More about ideals

Suppose that R is a ring, and that I, J ¡R. The following are easy to check.

Proposition 2.4.1. The set I ∩ J is an ideal of R, and whenever K ¡ R with K ⊆ I and K ⊆ J we have that K ⊆ I ∩ J .

Proposition 2.4.2. The set I + J := {i + j | i ∈ I, j ∈ J } is an ideal of R, and whenever K ¡R with I ⊆ K and J ⊆ K we have that I + J ⊆ K .

Proposition 2.4.3. The set I · J := {

r ir · jr | ir ∈ I, jr ∈ J } is an ideal of R, and I · J ⊆ I ∩ J .

Definition 3. An ideal I of a ring R is said to be maximal if I = R and I ⊆ J ¡R impliesthat I = J or J = R.

Theorem 2.4.4. Let R be a ring and I ¡R. Then R/I is a field if and only if I is maximal.

Proof. Suppose that I is maximal, and that x ∈ I . Then J := I + < x >= I . Clearly J ¡R,so J = R. As 1 ∈ R we can write 1 = i + a · x for some i ∈ I and a ∈ R. Then 1 = i + ax,or ax = 1, and we have found an inverse for x.

Suppose that R/I is a field, and that I ⊆ J ¡ R with I = J . We can therefore choosex ∈ J \ I . Then x = 0, and so has an inverse a. That is, ax − 1 = 0, so that ax − 1 ∈ I . Asax ∈ J and I ⊆ J we get that 1 ∈ J . For any t ∈ R then, we get t = t · 1 ∈ J : hence R = J .

Since R/I has at least two elements I = R.

Definition 4. An ideal I of a ring R is said to be prime if I = R and xy ∈ I implies that either x ∈ I or y ∈ I .

8



Theorem 2.4.5. Let R be a ring and I ¡R. Then R/I is an integral domain if and only if I is prime.

Proof. Suppose that I is prime, and that x y = 0. Then xy = 0, so xy ∈ I . It follows thateither x = 0 or y = 0. So R/I is an integral domain.

Suppose that R/I is an integral domain. Then R/I has at least two elements so I = R.Suppose now that xy ∈ I . Then xy = 0 so x y = 0. As R/I is an integral domain x = 0 ory = 0. So either x ∈ I or y ∈ I . Hence I is prime.

Corollary 2.4.6. Let R be a ring and I ¡R. If I is maximal then I is prime.

Proof. If I is maximal then R/I is a field. So R/I is an integral domain, hence I is prime.

2.5 Constructing new fields

One can use theorem 2.4.4 to construct examples of fields.Let K be a field. Recall that a polynomial of positive degree p(X ) ∈ K[X ] is called

irreducible if there are no polynomials of positive degree f (X ), g(X ) ∈ K[X ] such that p(X ) = f (X )g(X ).

In the first year course you defined the highest common factor of two polynomials f, gand you saw that there exist polynomials m, n such that

hcf (f, g) = mf + ng

The proofs were done only in the case of R[X ] but the same proofs apply for any field K.

Proposition 2.5.1. Let K be a field, p(X ) ∈ K[X ] and I =< p(X ) >. Then I is maximal if and only if p(X ) is irreducible. It follows that the quotient ring K[X ]/I is a field if and only if p(X ) is irreducible.

Proof. Assume that I is a maximal ideal. Let

p(X ) = a(X )b(X )

be a factorisation of p(X ). Clearly I ⊆< a(X ) >. Since I is maximal either < a(X ) >= I or < a(X ) >= R.

If < a(X ) >= I then a(X ) = p(X )q(X ) for some q(X ) ∈ K[X ] so

p(X ) = p(X )q(X )b(X )

and q(X )b(X ) = 1. It follows that deg b(X ) = 0.

If < a(X ) >= R then 1 = a(X )q(X ) for some q(X ) ∈ K[X ], so deg a(X ) = 0. Weconclude that either deg a(X ) = 0 or deg b(X ) = 0. So p(X ) is irreducible.

Assume now that p(X ) is irreducible. Let’s say that I ⊆ J where J is an ideal of R,J = I . Let f (X ) ∈ J \ I . Since p(X ) is irreducible hcf (f (X ), p(X )) = 1. So there area(X ), b(X ) ∈ K[X ] such that

a(X ) p(X ) + b(X )f (X ) = 1

Since p(X ), f (X ) ∈ J we have that 1 ∈ J . So J = R and I is maximal.Finally we remark that by theorem 2.4.4, K[X ]/I is a field if and only if I is maximal.

So K[X ]/I is a field if and only if p(X ) is irreducible.

9



Remark 2.5.2. A polynomial of degree 2 or 3 in K[X ] ( K a field) is irreducible if and only if it has no roots in K.

Example 2.5.3. The quotient ring

Z3[X ]/ < X 2 + 1 >

is a field.

Proof. Indeed using the previous proposition it suffices to show that X 2 + 1 is irreducible.However X 2 + 1 has no roots in Z3 since 02 + 1 = 1, 12 + 1 = 2, 22 + 1 = 2 in Z3.

2.6 Homomorphisms

When we study algebraic objects the appropriate maps to consider are the maps that pre-serve the structure of these objects. So for K-vector spaces the appropriate maps are lineartransformations, for groups it is the group homomorphisms and so on. So we make thefollowing definition.

Definition 5. Let R and S be commutative rings with identity 1. We will say that a mapf : R → S is a homomorphism if, for all x, y ∈ R,

(i)f (1) = 1; (ii)f (x + y) = f (x) + f (y); (iii)f (x · y) = f (x) · f (y).

For an example, take R := R[x] and S := C and let for any a ∈ C, σa be the ‘evaluation’map, σa :

N 0

ckxk →N

0ckak. This is a homomorphism.

For another example, take R := Z and for any d ∈ Z let S := Z/dZ. Then the map: R → S defined by : n → n (which maps each n to its equivalence class modulo d) is a

homomorphism. Our construction of the quotient Z/dZ achieved precisely this.For a non-example, consider the map p : Z ⊕ Z → Z ⊕ Z given by p(n1, n2) = (n1, 0).This satisfies the conditions (ii) and (iii), for being a homomorphism: but fails to map theidentity to the identity.

We also make the following definition.

Definition 6. Let f : R → S be a homomorphism of rings. We say that f is an isomor-

phism if f is 1–1 and onto. In this case we write R ∼= S .

2.6.1 The Kernel

Definition 7. Let f : R → S be a homomorphism of rings. We say that f −1(0) :=

{z | f (z) = 0} is the kernel of the homomorphism. Sometimes we denote it by ker f .

Suppose that we have a homomorphism f : R → S of rings. Which elements get mappedto the same place? Well, it is clear that ‘mapping to the same place’ is an equivalence relationon R. f (x) = f (a) if and only if f (x − a) = f (x) − f (a) = 0, and so using x = a + (x − a)we have

{x | f (x) = f (a)} = a + {z | f (z) = 0} = a + ker f.

Note that we have at once a good test for f to be one-to-one: this is equivalent toker f = {0}.

1It matters here, so we emphasise the with identity .

10



Lemma 2.6.1. Let f : R → S be a homomorphism of rings, with kernel K . Then K is an ideal of R.

Proof. Clearly f (0) = 0 so 0 ∈ K . If a, b ∈ K then

f (a − b) = f (a) − f (b) = 0so a − b ∈ K . We conclude that K is a subgroup under +. Also if a ∈ K and r ∈ R then

f (ra) = f (r)f (a) = f (r)0 = 0

so ra ∈ K . We conclude that K is an ideal.

2.7 The Image

Suppose that we have a homomorphism f : R → S of rings.For example, we might be looking at σ0 : Q[x] → C. In this case there is a lot of S = C

which is quite irrelevant to the homomorphism. All that really matters is the part of S consisting of elements mapped from R.

We make the definition.

Definition 8. Let f : R → S be a homomorphism of rings. We say that

f (R) := {y ∈ S | for some x ∈ R, f (x) = y}

is the image of the homomorphism. Sometimes we denote it by im f .

Note that at once we have a silly test for f to be onto: this is equivalent to im f = S .

Lemma 2.7.1. Let f : R → S be a homomorphism of rings, with image f (R). Then f (R)is a subring of S .

Proof. We use the subring test. 1 = f (1) ∈ im f . If x, y ∈ im f then there are x1, y1 ∈ R

such that x = f (x1), y = f (y1). Sox − y = f (x1) − f (y1) = f (x1 − y1) ∈ im f , xy = f (x1)f (y1) = f (x1y1) ∈ im f

It follows that im f is a subring of S .

2.8 The Isomorphism Theorem

We can now give a complete description of any homomorphism.

Theorem 2.8.1 (The Isomorphism Theorem for Commutative Rings with Identity).Let R and S be commutative rings with identity, and let f : R → S be a homomorphism.Then ker f is an ideal of R and im f is a subring of S .

Moreover, f : R/ ker f → im f where f : x → f (x) is a well-defined isomorphism.Proof. We have already shown the first part in lemmas 2.6.1, 2.7.1. Clearly if x = y thenf (x) = f (y). So f is well defined. It is obvious that f is onto. Finally f (x) = 0 if and onlyif f (x) = 0 if and only if x ∈ ker f if and only if x = 0. So f is also one-to-one. Hence anisomorphism.

Just as for groups, we usually write the Isomorphism Theorem much more briefly: some-thing like ‘Let f : R → S be a homomorphism; then R/ ker f ∼= im f ’.

We remark that there is an Isomorphism Theorem for vector spaces too, but at least inthe finite dimensional case it doesn’t say more (or less) than the Rank–Nullity Theorem.

11



2.8.1 A Key Example — Evaluation

Let R := R[X ] and let S := C; consider the evaluation homomorphism σa : R → S given byσa :

N 0

ckX k →N

0ckak.

What is the kernel of σa? By definition, ker σa := {φ(X ) | φ(a) = 0}. So we must

decide which polynomials vanish at a. If a ∈ R, then the Remainder Theorem tells us theanswer: it is those polynomials which are exactly divisible by (X − a); that is, ker σa ={ψ(X )(X − a) | ψ ∈ R[X ]}. If a ∈ R then things are a bit more complicated. If φ(a) = 0and φ has real coefficients, then we also have that φ(a) = 0—the complex conjugate is also aroot. Now, by the Remainder Theorem in C[X ] we see that both (X − a) and (X − a) divideφ; whence the product (X − a)(X − a) = (X 2 − 2ℜaX + |a|2) divides φ. These conditionsare also clearly sufficient, so that we have

ker σa = (X − a) := {ψ(X )(X − a) | ψ ∈ R[X ]} if a is real,

and

ker σa = (X 2 − 2ℜaX + |a|2) := {ψ(X )(X 2 − 2ℜaX + |a|2) | ψ ∈ R[X ]} if a is not real.

What is the image of σa? If a ∈ R then surely φ(a) ∈ R. Moreover, given c ∈ R theconstant polynomial c evaluates to c. So we get that im σa = R in this case. When a ∈ R

then we get more than R, for instance X evaluates to a, and so (X − ℜa) evaluates to(ℑa)i = 0. Now it’s clear we get every complex number ζ = ξ + iη; it is the evaluation of thereal polynomial ξ + η

ℑa(X − ℜa).

What does the Isomorphism Theorem tell us? Well,(i) if a ∈ R then R[X ]/(X − a) ∼= R;(ii)if a ∈ R then R[X ]/(X 2 − 2ℜaX + |a|2) ∼= C.

(Note that any real monic quadratic polynomial with non-real roots can be expressed,for some a, as (X 2 − 2ℜaX + |a|2).)

2.8.2 Another—Modular Arithmetic

Let R := Z and S := Zd. The mapping n → n of n to its coset modulo d is a homomorphism;the image is all of S and the kernel is (unsurprisingly) dZ. The Isomorphism Theorem tellsus that Zd ∼= Z/dZ which is not really very surprising given its definition!

12



Which of the following are true?

1. If I, J are ideals then I ∪ J is an ideal.

2. If F, K are fields and φ : F → K is an onto homomorphism then φ is an isomorphism.

3. If R is an integral domain and I is an ideal of R then R/I is an integral domain.

4. There is a ring homomorphism f : Zn → Z.

5. There is a ring homomorphism f : Z → Zn.

6. There is a ring homomorphism f : Q → Z.

7. There is a ring homomorphism f : Q → Z p ,where p is a prime.

8. The rings Z[i] and Z[X ] are isomorphic.

9. C and R are isomorphic.

10. If an ideal I ⊂ R contains a unit of the ring R then I = R.

11. If p ∈ Z is a prime number and xy ∈ ( p) then either x ∈ ( p) or y ∈ ( p).

12. Q is an ideal of R.

13. If R is a ring and a1,...,an ∈ R then the set {a1r1 + ... + anrn : r1,...,rn ∈ R} is anideal of R.

14. < X > is a prime ideal of Z[X ].

15. Every prime ideal of Z[X ] is a maximal ideal.

16. If f : R → S is an isomorphism of the rings R, S then I ⊂ R is an ideal of R if andonly if f (I ) is an ideal of S .

17. If I ⊂ R is an ideal and f : R → S is a ring homomorphism then f (I ) is an ideal of S .

18. If I ⊂ S is an ideal of S and f : R → S is a ring homomorphism then f −1(I ) is anideal of R.

19. a + I ∈ R/I is a unit of R/I if and only if < a > +I = R.

13

rings lect 2

Documents