lecture notes of abstract algebra iiruiw10/pdf/alg2.pdf · 2017-07-31 · 7.1. deﬁnition of...

LECTURE NOTES OF ABSTRACT ALGEBRA II

RUI WANG

CONTENTS

1. Review of Group Theory 22. Definition of Ring 32.1. Definition of the ring 32.2. Homomorphism 42.3. The Multiplication of a Ring 83. Integral Domains 104. Characteristic of a ring 125. Fermat’s and Euler’s Theorem 125.1. Preparation and key points 125.2. Fermat’s little theorem and application 145.3. Euler’s generalization and its application 156. The Fields of Fractions 176.1. The construction 176.2. Uniqueness 186.3. Examples 197. The ring of polynomials 197.1. Definition of polynomial ring 197.2. The evaluation homomorphism 247.3. Relation between solving polynomial equations and ev 268. Polynomials over a field 268.1. Division Algorithm for F [x] 268.2. Quotient homomorphism and ideal 288.3. Ideals and quotient rings 288.4. F [x] is a PID 298.5. Irreducible polynomials 328.6. Uniqueness of factorization in F [x] 339. Maximal Ideals and Prime Ideals 349.1. Terms for integral domains 349.2. PID is UFD 359.3. Maximal ideals and primes ideals 3710. Field extension 3810.1. Motivation 3810.2. Definition of field extension 39

Date: June, 1, 2016.1

2 RUI WANG

10.3. Algebraic and transcendental elements 4011. Vector Spaces 4211.1. Definition of vector spaces 4211.2. Linear independence and basis 4312. Algebraic Extension 4412.1. Finite extensions 4412.2. Algebraically closed field and algebraic closures 4712.3. Summary 4813. Final project 49

1. REVIEW OF GROUP THEORY

Definition 1.1 (Group). A group is a set G with a binary operation ⇤ satisfying the following proper-ties:

• (Associativity) For any a, b, c 2 G, (a ⇤ b) ⇤ c = a ⇤ (b ⇤ c);• (The unity) There exists (unique) e 2 G, such that for any a 2 G, e ⇤ a = a ⇤ e = a;• (Inverse) For every a 2 G, there exists (unique) a�1 2 G such that a ⇤ a�1

= e = a�1 ⇤ a.

For a group G, if any a, b 2 G have a ⇤ b = b ⇤ a, then G is called an abelian group (commutativegroup). Otherwise, G is called a nonabelian (noncommutative) group. If a group has only finitelymany elements, then it is called a finite group. Otherwise, it is called an infinite group.

Example 1.2 (Finite Groups). (1) Cyclic group Zn

. This is an abelian group.(2) Symmetric group S

n

.(3) Alternating group A

n

, i.e., all even elements in the symmetric group of Sn

. It is a subgroupof S

n

of order n!/2.(4) Dihedral group D

n

, i.e., group of symmetries of the regular n-gon. In particular, D3 = S3 isthe smallest nonabelian group.

(5) Klein 4-group V : {e, a, b, c|a2 = b2 = c2 = e, ab = ba = c, }. This is also abelian group andactually isomorphic to Z2 ⇥ Z2 by the Fundamental Theorem of Finitely Generated AbelianGroup.

Theorem 1.3 (Cayley). Every finite group G is isomorphic to a subgroup of Sn

, n = |G|.

Theorem 1.4. [Fundamental Theorem of Finitely Generated Abelian Group] Every finitely generatedabelian group G is uniquely isomorphic to a direct product of cyclic groups in the form

Z(p1)r1 ⇥ Z(p2)r2 · · ·⇥ Z(pn)rn ⇥ Z⇥ · · ·⇥ Z.

Here pi

are prime numbers. The rank (Betti number) of G is defined as the number of copies of Z.

Remark 1.5. |Z(p1)r1 ⇥ Z(p2)r2 · · ·⇥ Z(pn)rn | = (p1)r1 · · · (p

n

)

rn .

Remark 1.6. Z3 ⇥ Z4 ⇥ Z35⇠=

Z4 ⇥ Z3 ⇥ Z5 ⇥ Z7. It has only one generator (1, 1, 1) and henceisomorphic to Z420. (How to see it has only one generator? )

MATH 120B 3

On the other hand, the order 420(= 2

2 ⇥ 3⇥ 5⇥ 7) abelian group can only be isomorphic to either

Z4 ⇥ Z3 ⇥ Z5 ⇥ Z7 or Z2 ⇥ Z2 ⇥ Z3 ⇥ Z5 ⇥ Z7.

The latter hence is not isomorphic to Z420, i.e., has more than 1 generators.

Remark 1.7. The classification of finite nonabelian groups is much harder than the abelian ones.In fact, the classification of all simple finite groups has been done in some more than then thousandpages book. The classification of all finite groups is still kind of untouchable problem in group theory.

Example 1.8 (Infinite Groups). (1) (Z,+);(2) nZ;(3) (Q,+);(4) (Q⇤, ·).(5) (R,+);(6) (R⇤, ·).(7) (C,+);(8) (C⇤, ·).(9) S1

= R/Z.(10) (gl

n

(F),+); (GLn

(F), ·).

2. DEFINITION OF RING

2.1. Definition of the ring.

Definition 2.1 (Ring). A ring is a set R with two binary operation + and · satisfying the followingproperties:

• (Abelian group structure) (R,+) is an abelian group;• (Associativity of ·) For any a, b, c 2 G, (a · b) · c = a · (b · c);• (Compatibility of + and ·) For any a, b, c 2 G, a · (b+ c) = a · b+a · c, (a+ b) · c = a · c+ b · c.

Further more, if · is commutative, then call it an commutative ring.

Example 2.2. (1) (Z,+, ·), (Q,+, ·), (R,+, ·), (C,+, ·) are rings.(2) (M

n

(R),+, ·) is a ring if (R,+, ·) is a ring. Notice that, even R is commutative, when n � 2,M

n

(R) is not commutative.Lecture 1 stopped here.

(3) Assume X is a set and (R,+, ·) is a ring. Denote by C(X,R) the set of maps from X to R.Define• (f + g)(x) = f(x) + g(x);• (f · g)(x) = f(x) · g(x).

Then (C(X,R),+, ·) is a ring. If R is commutative, then (C(X,R),+, ·) is also a commutativering.

(4) Assume R is a commutative ring. Its polynomial ring is defined as

R[x] := {⌃i=0,··· ,naix

i|ai

2 R, n 2 N},

where x is the indeterminate. We are going to study intensively on this one in Section 22.

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(5) (nZ,+, ·) is a commutative ring.(6) (Z

n

,+, ·) is a commutative ring.(7) If R1 and R2 are rings, then R1 ⇥R2 is a ring. If both R1 and R2 commutative, then R1 ⇥R2

is also commutative.

Proposition 2.3. Assume R is a ring with 0 as the additive identity, i.e., the unity in the abelian group(R,+). Then

(1) 0 · a = a · 0 = 0;(2) a · (�b) = (�a) · b = �(a · b);(3) (�a) · (�b) = a · b.

Hence, we can use the following symbols, na := a + · · · + a if n is a non-negative integer, andna := (�a) + · · ·+ (�a) if n is a negative integer, and consider n as a multiplier.

Lecture 2 stopped here.

Definition 2.4. Assume (R,+, ·) is a ring. A subset S ⇢ R is called a subring of R, if

(1) it is closed under the two binary operations + and ·;(2) (S,+, ·) is a ring.

From the definition, if S is a subring of R, then (S,+) is a subgroup of (R,+). It follows that0 2 S, and any a 2 S, �a is also in S.

Actually, to see a subset S is a subring, it is enough to check the following.

Proposition 2.5. Assume (R,+, ·) is a ring and S ⇢ R is a subset. S is a subring if and only if

(1) (S,+) is a subgroup of (R,+);(2) S is closed under the multiplication ·.

Moreover, (1) is equivalent to check: Any a, b 2 S, a� b 2 S.

Example 2.6. nZ is a subring of Z.

2.2. Homomorphism. Recall the definition for group homomorphisms, similarly, we introduce theconcept of ring homomorphism.

Definition 2.7. Assume (R,+R

, ·R

) and (R0,+R

0 , ·R

0) are two rings. A map � : R ! R0 is called a

homomorphism if

• �(a) +R

0 �(b) = �(a+R

b);• �(a) ·

R

0 �(b) = �(a ·R

b).

Remark 2.8. The general principle for a homomorphism is “the map communicates with operations”.

Remark 2.9. A homomorphism of ring is first a homomorphism of the abelian groups (R,+R

) to(R0,+

R

0). Hence, it follows from the group homomorphism properties,

�(0R

) = 0

R

0 , �(�a) = ��(a).

However, a ring homomorphism does require more than a group homomorphism. For example, wehave the following example.

MATH 120B 5

Example 2.10. Rings (Z,+, ·) and (2Z,+, ·). Consider

� : Z ! 2Z, a 7! 2a.

Show that, it is a group homomorphism (Actually, a group isomorphism), however, it is not a ringhomomorphism. For example,

�(1) · �(1) = 2 · 2 = 4 6= �(1 · 1) = 2.

Definition 2.11. If a ring homomorphism � as a map is injective, then we say � is injective. If a ringhomomorphism � as a map is surjective, then we say � is surjective. If � as a map is bijective, i.e.both injective and surjective, we call � an isomorphism.

Of course, an injective/surjective/bijective ring homomorphism is a injective/surjective/bijectivegroup homomorphism with respective to the abelian group structures in the two rings.

Lemma 2.12. Assume R,R0, R00 are three rings and � : R ! R0, : R0 ! R00 are two ringhomomorphisms. Then the composition

� � : R ! R00

is also a ring homomorphism.

The following theorem is not hard to prove but fundamental.

Theorem 2.13. Isomorphism is an equivalence relation.

Proof. • (Symmetric)• (Reflective)• (Transitive)

⇤

Example 2.14. Assume R is a ring. Show that

� : R ! R, a 7! �a

is a ring isomorphism of R to itself.

In fact, the objectives of the group theory are equivalence classes of ring isomorphisms.Denote by

ker� = {a 2 R|�(a) = 0

R

0},

and call it the kernel of �.First, from the abelian group structure, ker� is a normal subgroup of R, and we can conclude from

this that

Lemma 2.15. The ring homomorphism map � is injective if and only if ker� = {0}.

Now we show that more than the abelian group structure from the addition, in fact we have

Proposition 2.16. The kernel ker� is a subring of R.

In fact, it is an ideal of R. We are going to discuss more about this in Section 26.

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Example 2.17. Consider the map

� : Z ! Zn

= Z/nZ, a 7! [a],

where Z/nZ is the quotient group. Prove it is a surjective ring homomorphism with kernel

ker� = nZ.

Lecture 3 stopped here.In fact, we can and will prove in general,

Theorem 2.18. [Fundamental Homomorphism Theorem 1] If � : R ! Q is a surjective ring homo-morphism. Then R/ ker� is also a ring, and

[�] : R/ ker�! Q

is an ring isomorphism.

Proof. (1) Show that R/ ker� has a natural ring structure induced from the ring structure of R.(2) Show [�] is ring isomorphism.

⇤

Example 2.19. As a conclusion of this theorem, as rings

Z/nZ ⇠=

Zn

.

If the ring homomorphism is not surjective, we can modify the fundamental homomorphism theo-rem using the following fact.

Proposition 2.20. If � : R ! R0 is a ring homomorphism, then the image �(R) is a subring of R0.

Hence assume � : R ! R0 is a ring homomorphism, then

� : R ! �(R)

is a surjective ring homomorphism, we can restate the fundamental homomorphism theorem as

Theorem 2.21 (Fundamental Homomorphism Theorem 2). If � : R ! R0 is a ring homomorphism.Then

[�] : R/ ker�! �(R)

is an ring isomorphism.

Here come more facts for ring homomorphisms. Now you can test the FHT for them.

Lemma 2.22. Assume R is a ring and S is a subring of R. Denote by i : S ! R the inclusion of theset S into R. Then the inclusion i is a ring homomorphism.

Usually, we also say a ring S is a subring of R, if there exists an injective ring homomorphism fromS to R. In fact, in this abuse of terminology, we identify the ring S with its image in R, since they areisomorphic.

MATH 120B 7

Lemma 2.23. Assume R1, R2, · · · , Rn

are rings. The projection

⇡k

: R1 ⇥R2 ⇥ · · ·⇥Rn

! Rk

is a surjective ring homomorphism for every k = 1, 2, · · · , n.

Example 2.24. Assume X is a set and R is a ring. Then we have seen that C(X,R) is a ring. Fixx 2 X , we can define a map

evx

: C(X,R) ! R, evx

(f) = f(x).

Notice that evx

is a ring homomorphism and it is surjective. It is called the evaluation homomorphismat x.

In general, the space C(X,R) is very big if we include all possible maps from X to R. To getgood theory, we usually add requirement for types of maps so that we shrink C(X,R). For example,consider the following case.

Example 2.25. Denote by C[x] the set of all polynomials over C. In our previous notations, C[x] ⇢C(C,C), moreover, it is a commutative subring of C(C,C). For each z 2 C, the evaluation mapcan be not restricted to C[x], for which we still use ev

z

to denote. It is still a surjective (why?) ringhomomorphism. The kernel

ker evz

= {f 2 P|f(z) = 0}.

From FHT,

C[x]/ ker evz

⇠=

C.

Lemma 2.26. For a particular polynomial p, the equation p(z) = 0 has a solution, if and only if

p 2 [z2C ker ev

z

.

Can we answer the question that if p(z) = 0 has a solution by understanding the structure of[z2C ker ev

z

? In fact, the real difficulty is, what kind of structure on [z2C ker ev

z

we should look at.A lot of modern math can be more or less fit into this picture.

Proposition 2.27. The statement that every algebraic equation of C has a solution is equivalent toshow that

C[x] = [z2C ker ev

z

.

Example 2.28. The map

� : C ! M2(R), �(a+ ib) =

a b

�b a

!

is an injective ring homomophism.


8 RUI WANG

2.2.1. HW for the homomorphism part.

(1) Check the map

� : C ! M2(R), �(a+ ib) =

a b

�b a

!, a, b 2 R

is an injective ring homomorphism.(2) Assume R,R0, R00 are three rings, and

� : R ! R0, : R0 ! R00

are two ring homomorphisms. Prove that

� � : R ! R00

is also a ring homomorphism.(3) Assume H is a subring of R, does the quotient group R/H always carry the ring structure

induced from R? If yes, prove it. Otherwise, give a counter example. (You may refer 26.9.)(4) Assume H,R are two rings. People also say H a subring of R if there is some injective ring

homomorphism from H to R. In this sense, can Zn

be a subring of Z? Explain your answer.(5) Assume R is a ring and X is a set. Denote by C(X,R) the set of all maps from X to R. It has

a ring structure induced from R. For any x 2 X , we can define the evaluation map

evx

: C(X,R) ! R, evx

(f) = f(x).

Prove it is a surjective ring homomorphism.

2.3. The Multiplication of a Ring. Now we look more closely at the multiplication in a ring struc-ture. Assume (R,+, ·) is a ring.

2.3.1. The unity.

Definition 2.29. If there is an element 1 2 R such that

1 · a = a · 1 = a

for any a 2 R, then we call R is a ring with the unity 1. (The unity if exists, is unique. )

We don’t require that the multiplication · has unity in general.

Lemma 2.30. If R is a ring with unity 1 = 0, then R = {0}.

Example 2.31. (1) (Z,+, ·), (Q,+, ·), (R,+, ·), (C,+, ·) are commutative rings with unity.(2) If R is a ring with unity, then (M

n

(R),+, ·) is a ring with unity.(3) If R is a ring with unity. Then C(X,R) is a ring with unity.(4) Assume R is a commutative ring with unity. Then its polynomial ring R[x] is also a commu-

tative ring with unity.(5) (nZ,+, ·) is a commutative ring. However, if n 6= 1, it doesn’t have unity.(6) (Z

n

,+, ·) is a commutative ring with unity [1].(7) About direct product of rings, we have the following

MATH 120B 9

Lemma 2.32. The direct product R1⇥ · · ·Rn

of rings is commutative or has unity if and onlyif each R

i

is commutative or has unity, respectively. Moreover, if 1i

is the unity of Ri

, then(11, · · · , 1n) is the unity of R1 ⇥ · · ·R

n

.

Lemma 2.33. Assume R,R0 are two rings, and R has the unity 1. For any ring homomorphism� : R ! R0, the subring �(R) of R0 then has the unity �(1). In particular, if � is surjective, then R0

has unity �(1).

Example 2.34. If R is a ring with unity, while R0 is a ring without unity. It is impossible to exist asurjective ring homomorphism from R to R0. For example, no surjective homomorphism from Z tonZ for n � 2. Actually, there is no nontrivial ring homomorphism from Z to nZ since any nontrivialsubring of nZ has no unity.

Example 2.35. If both R,R0 are rings with unity, and � is some isomorphism from R to R0. Then�(1) = 1

0. In particular, if R is a cyclic group, this already determines the isomorphism.For example, construct an isomorphism between Z

ab

and Za

⇥Zb

, for a, b are positive integers andgdc(a, b) = 1. The unity of Z

ab

is (1) and the unity of Za

and Zb

is (1, 1). The isomorphism must be

�(1) = (1, 1).

Then any �(n) = �(n · 1) = n�(1) = (n, n).

2.3.2. Units. Now assume that R is a ring with unity 1 6= 0, we can further ask if every elementa 6= 0 can have an inverse. It is equal to ask, if R⇤

:= R� {0} has the structure of group.

Definition 2.36. Assume R is a ring with unity 1 6= 0. An element u 2 R is a unit of R if it has amultiplicative inverse (the unique element u�1 2 R such that u ·u�1

= u�1 ·u = 1). If every nonzeroelement in R is a unit, then call R a division ring (or skew field). If moreover R is a commutativering, then we call R a field. A noncommutative division ring is called a strictly skew field.

Example 2.37. (Q,+, ·), (R,+, ·), (C,+, ·) are fields, and (Z,+, ·) is not a field. The only units inZ are ±1.

Proposition 2.38. Assume R is a unital ring. Denote U ⇢ R the set of all unites in R. Then (U, ·) isa group

Proof. (1) Any x, y 2 U , we have

(xy)(y�1x�1) = 1 = (y�1x�1

)(xy).

This means that xy 2 U .(2) Since 1 2 U and x1 = 1x = 1, so 1 is the identity in (U, ·).(3) Any x 2 U , we have x�1 2 U .

Hence (U, ·) is a group. ⇤

Example 2.39. Find all units in Z14. In fact, all units in Zn

are m 2 Zn

with gcd(m,n) = 1. (SeeProposition 5.2. )

As a consequence, Zp

is a field if p is a prime number, and usually, people use Fp

to denote it.These are called finite fields.


10 RUI WANG

3. INTEGRAL DOMAINS

Definition 3.1. Assume R is a ring. Then any two nonzero elements a, b 2 R such that a · b = 0 arecalled 0-divisors or divisors.

Lemma 3.2. R doesn’t have any divisor if and only if the cancelation rule holds, i.e.

a · b = 0 if and only if a = 0 or b = 0.

Proposition 3.3. Ring isomorphisms map divisors to divisors.

Example 3.4. Find all divisors of Z6.

Solution: (1) [0] and [1] are obviously not divisors.(2) [2][3] = [6] = [0], so [2] and [3] are divisors.(3) [4][3] = [12] = [0], so [4] is also a divisor.(4) [5] is not a divisor since gcd(5, 6) = 1.

⇤

This example makes us guess the general statement

Proposition 3.5. In the ring Zn

, [k] is a divisor if and only if [k] 6= 0 and gcd(k, n) 6= 1.

Proof. (1) We first show that if [k] is a divisor, then [k] 6= 0 and gcd(k, n) 6= 1.If [k] is a divisor, then [k] 6= [0] and there exists some [l] 6= 0 such that [kl] = 0, i.e.

n|(kl). Then if gcd(k, n) = 1, we must have n|l, which contradicts with [l] 6= 0. Hence,gcd(k, n) 6= 1.

(2) Then lets show that, if gcd(k, n) 6= 1, then k is a divisor.Assume [k] 6= 0 and gcd(k, n) = d 6= 1, then

[k · nd] = [0], and [

n

d] 6= [0].

⇤

Example 3.6. Find all divisors of Z24.Using Proposition 3.5, they are

[2], [3], [4], [6], [8], [9], [10], [12], [14], [15], [16], [18], [20], [21], [22].

Example 3.7. For prime number p, Zp

has no divisor.

Example 3.8. Consider the ring R⇥ R. What are the divisors it? Show that any element in the set

{(x, 0), (0, y)|x 6= 0, y 6= 0}

is a divisor. If R is a ring without divisor, then this set consists of all divisors of R⇥R.This shows that R⇥ R is not isomorphism to C as rings since C has no divisors.

Definition 3.9. An integral domain D is a commutative ring with unity 1 6= 0 and containing nodivisors of 0.

Example 3.10. (1) Q, R, C are integral domains, which are also fields;(2) Z

p

, for p prime is integral domain, it is also a (finite) field;

MATH 120B 11

(3) Z is an integral domain, but not a field.

We are going to prove the following two statements:

(1) A field is an integral domain.(2) A finite integral domain is a field.


Proposition 3.11. A field is an integral domain.

Proof. Assume F is a field. Then for any

x · y = 0,

assume x 6= 0, then we multiple x�1 to both sides, and get

x�1 · (x · y) = x�1 · 0 = 0.

The l.h.s is y. Hence it shows that F has no divisor, i.e. an integral domain.⇤

Corollary 3.12. Q,R,C, Fp

are integral domains.

Proposition 3.13. Every finite integral domain is a field.

Proof. We only need to show that every 0 6= x 2 R has an inverse.For any x 6= 0, we consider a map

fx

: R ! R, y 7! x · y.

If x · y1 = x · y2, thenx · (y1 � y2) = 0.

Because R is an integrable domain, and x 6= 0, so y1 = y2. This shows that the map is injective.Since R is finite, then an injective map must be surjective too. So there must be some y 2 R such

that x · y = 1.Since R is also commutative, y · x = 1. This shows that x has inverse too. ⇤

Corollary 3.14. Assume p is a prime number, then Zp

is a field.

Proof. Because we have seen from Proposition 3.5 that Zp

is an integral domain. Since it is finite, itis field. ⇤

Example 3.15. Assume � : R ! R0 is a ring homomorphism between unital rings R and R0. If � isnot trivial, then �(1) 6= 0

0 (why?). Assume �(1) = x 6= 0

0. Then we have

x · x = x.

This is equivalent to say thatx(x� 1

0) = 0

where 1

0 is the unity of R0.Hence if R0 is a field, we can conclude that x must be 1

0.Otherwise, if we want x 6= 1

0, then x must be a divisor of R0.For example, we can consider

12 RUI WANG

(1) Z2 ! Z6, [1] 7! [3]. ([3] is the unity of the subring {[0], [3]} of Z6. )(2) R ! R⇥ R, 1 7! (1, 0).

These are all examples of ring homomorphisms which maps 1 to some element different from 1

0.

4. CHARACTERISTIC OF A RING

Assume R is a ring.

Definition 4.1. The minimal positive integer n such that n · a = 0 for all a 2 R, is called thecharacteristic of the ring R, and denote it by char. If There is no such n, call the characteristic is zero.

Example 4.2. (1) 0 = char(Z) = char(Q) = char(R) = char(C).(2) char(Z

n

) = n.

Proposition 4.3. If R is a ring with unity 1 6= 0, then

char(R) = min{n|n · 1 = 0}

if {n|n · 1 = 0} 6= ;. Otherwise, char(R) = 0.

Proof. Obviously, char(R) � min{n|n ·1 = 0}. We only need to show that char(R) min{n|n ·1 =

0}. For this, it is enough to check that, every n 2 Z+ which makes n · 1 = 0, we have n · a = 0 forevery a 2 R. In fact, it follows from

n · a = a+ a+ · · ·+ a

= 1 · a+ 1 · a+ · · ·+ 1 · a

= (1 + 1 + · · ·+ 1) · a

= (n · 1) · a

= 0 · a

= 0.

⇤

Example 4.4. (1) char(Zm

,Zn

) = lcm(m,n), for m,n � 2.(2) char(Z

m

,Z) = 0.For these two example, because they are all unital rings. The characteristics are the same as the

order of the unities (in the abelian group).

5. FERMAT’S AND EULER’S THEOREM

5.1. Preparation and key points. We first prove two facts. One is a general statement for unitalrings, which we mention when we introduce the concept of units. The other is in particular for theunital ring Z

n

. These two simple facts together will lead to some interesting application in under-standing integers, names as Fermat’s little theorem and Euler’s generalization.

Proposition 5.1. Assume R is a unital ring. Denote U ⇢ R the set of all unites in R. Then (U, ·) is agroup


MATH 120B 13

Proof. (1) Show U is closed under multiplication. Any x, y 2 U , we have

(x · y) · (y�1 · x�1) = 1 = (y�1 · x�1

) · (x · y).

This means that x · y 2 U .(2) Since 1 2 U .(3) Any x 2 U , we have x�1 2 U .

Hence (U, ·) is a group.⇤

For Zn

, all units can be characterized by the following statement.

Proposition 5.2. Assume m,n are two positive coprime integers. Then [m] 2 Zn

is a unit if and onlyif gcd(m,n) = 1.

Proof. (1) Let’s first prove that if [m] 2 Zn

is a unit then gcd(m,n) = 1.Assume [m] 2 Z

n

is a unit is equivalent to assume that there exists some k 2 Z such that

km ⌘ 1 mod n.

Assume d = gcd(m,n), then d|m and d|n. Hence d|1. This concludes that d = 1.(2) Let’s prove that if gcd(m,n) = 1, then [m] is a unit. To do this, we need to find its inverse.

(The way of finding its inverse is very similar to the method we used when prove any finiteintegral domain is a field. )

Construct the following map

fm

: Zn

! Zn

, fm

([a]) = [am].

(Check it is well-defined!) We show that, the assumption gcd(m,n) = 1 indicates that themap f

m

is injective.If [a1m] = [a2m] for some [a1], [a2] 2 Z

n

, then we have

[(a1 � a2)m] = 0.

It means that n|((a1 � a2)m). Since gcd(m,n) = 1, it follows n|(a1 � a2), i.e. [a1] = [a2].This shows that f

m

is injective.Because Z

n

is finite, a map to itself is injective is equivalent to it is surjective. Hence thereexists some element [a] 2 Z

n

such that fm

([a]) = [am] = [a][m] = 1, which is the inverse of[m] in Z

n

.⇤

Let’s combine the results of the two results above into the following

Theorem 5.3. The subset

Gn

:= {[m] 2 Zn

| gcd(m,n) = 1}

is a group of multiplication of order '(n).

14 RUI WANG

5.2. Fermat’s little theorem and application. Fermat’s little theorem is nothing but Theorem 5.3under the situation of Z

p

where p is a prime number. For this case,

Gp

:= {[m] 2 Zp

|[m] 6= [0]},

and '(n) = p� 1. (Theorem 5.3 for this case is equivalent to state that Zp

is a field.)

Theorem 5.4 (The little Theorem of Fermat’s). If p is a prime not dividing a, then

ap�1 ⌘ 1 mod p,

i.e. [ap�1] = [1] in Z

p

.

Proof. Because p doesn’t divide a, so [a] 6= [0] and hence in the group Gp

. Since Gp

has only p � 1

elements, we must have [a]p�1= [ap�1

] = [1] since p� 1 must be multiple of the order of [a] (why?),i.e. ap�1 ⌘ 1 mod p.

⇤

Corollary 5.5. If p is prime. Thenap ⌘ a mod p,

i.e. [ap] = [a] in Zp

.

Proof. Since we have shown that [a]p�1= [1], so [a]p = [a]. ⇤

Fermat’s little theorem has interesting application. It can help us handle very large integers.

Example 5.6. Calculate the remainder of 10203 when divided by 13.

Solution: Because 13 is a prime number, we have

[10]

12= [1].

Since 203 = 12 · 16 + 11, so

[10

203] = [10]

203= [10

12]

16[10]

11= [1]

16[10]

11= [10]

11.

Since [10]

11 · [10] = [1], and also [10]

�1= [4]. So by the uniqueness of inverse, we get

[10]

11= [4].

⇤

Example 5.7. Show that 21111 � 3 is not divided by 11.

Solution: We calculate the remainder of 21111 when divided by 11. Using Fermat’s little theorem, inZ11,

[2

1111] = [2

10]

111[2] = [2].

Hence [2

1111 � 3] = [2]� [3] = [10] 6= [0]. So it is not divided by 11. ⇤


Example 5.8. Show that 15|(n33 � n).

Proof. Show that 3|(n33 � n) and 5|(n33 � n).

MATH 120B 15

(1) [n33] = [n2

]

16[n] = [n].

(2) [n33] = [n4

]

8[n] = [n].

⇤

From this prove, you can construct a lot of examples like this. e.g. 77|(n31 � n), 14|(n13 � n), · · · .General rule is, you pick distinct k prime numbers, p1, · · · , pk. Take m = lcm(p1 � 1, · · · , p

k

� 1).Then

p1 · · · pk|(nlm+1 � n), any l.

Remark 5.9. When mentioned Fermat’s name, I can not help mentioning another theorem of his,which is referred as Fermat’s big theorem or Fermat’s last theorem. This statement is quite easy,which is the equation

xn

+ yn = zn

always has positive integer solutions when n � 3. This theorem was first stated by Fermat in 1637in the margin of a book where he claimed he had a proof that was too large to fit in the margin. Noone knows if Fermat really provided a correct proof, but it took 358 years by mathematicians untilAndrew Wiles in 1995 first published the correct proof.

The unsolved problem stimulated the development of algebraic number theory in the 19th centuryand the proof of the modularity theorem in the 20th century. The proof and the theories developed forit are much more important than the statement of Fermat’s last theorem itself.

Here is a short story. When being asked why he didn’t try to prove Fermat’s last theorem, Hilbert(one of the most influential and universal mathematicians of the 19th and early 20th centuries) re-marked: ”Why should I kill the goose that lays golden eggs?” It is true that the Fermat’s last theoremin mathematics is just like a goose that lays golden eggs.

5.3. Euler’s generalization and its application. Then Euler’s generalization for Fermat’s little the-orem is an easy corollary of Theorem 5.3.

Theorem 5.10 (Euler’s Theorem). For any a 2 Z with gcd(a, n) = 1, have

a'(n) ⌘ 1 mod n,

i.e. [a'(n)] = [1] in Zn

.If n = p is prime, then �(p) = p� 1. It reduces to Fermat’s theorem.

Euler’s theorem can be used to solve the equation

[a][x] = [b] in Zn

(1)

for n � 2.

Proposition 5.11. If gcd(a, n) = 1, then the equation (1) has unique solution.

Proof. We have seen that gcd(a, n) = 1 indicates that [a] has inverse in Zn

. Then [x] = [b][a]�1 is asolution.

Assume both [x1] and [x2] are solutions, then

[a][x1] = [a][x2].

Multiply [a]�1 to both sides, we have [x1] = [x2]. This shows that there is unique solution. ⇤

16 RUI WANG

Example 5.12. Solve the equation [3][x] = [5] in Z10.

Solution: Because gcd(3, 10) = 1, there is only one solution which is

[x] = [3]

�1[5] = [7][5] = [5].

⇤

Now let’s prove a more general result.

Theorem 5.13. If gcd(a, n) = d, then the equation (1) has solution if and only if d|b. If d|b, thenthere are d solutions.

Proof. (1) We first prove the equation (1) has solution if and only if d|b.If d|b, then we consider the integer equation in Zn

d

[

a

d][x] = [

b

d].

Notice now gcd(

a

d

, nd

) = 1, so using Proposition 5.11, there exists a unique solution, denote itby x0. It means that

a

d· x0 �

b

d=

n

d· l, for some l 2 Z.

Multiply d to both sides, we get

ax0 � b = n · l, i.e. [a][x0] = [b] in Zn

.

Conversely, if there exists some solution x such that

[a][x] = [b],

then [ax� b] = [0] in Zn

. Since d|(ax) and d|n, we must have d|b too.(2) Now we show that if d|b, there are exactly d solutions.

If x1 and x2 are both solutions, then

[a(x1 � x2)] = [0] in Zn

.

This shows that (x1 � x2) must be multiple of n

d

. Conversely, it also shows that if x is asolution, any x+ l · n

d

is also a solution for any l 2 Z.Now we fix one chosen solution x. We have shown that the solution set of (1) is the same

as the set

{[x], [x+

n

d], [x+ 2 · n

d], [x+ 3 · n

d] · · · }.

Notice that, this set has d elements.⇤

Example 5.14. (1) Solve 12x ⌘ 27 mod 18;(2) Solve 15x ⌘ 27 mod 18.

Solution: (1) Since gcd(12, 18) = 6 - 27, there is no solution.

MATH 120B 17

(2) It is equivalent to solve

15x ⌘ 9 mod 18.

Since gcd(15, 18) = 3|9, there are 3 solutions. Solve

5x ⌘ 3 mod 6.

The unique (in Z6!) solution is [x] = [3]. So all solutions are

[3], [3 + 6] = [9], [3 + 12] = [15] in Z18.

⇤


6. THE FIELDS OF FRACTIONS

We have shown before that every finite integral domain is a field. In this section, we show that foran arbitrary integral domain (infinite or finite) D, we can canonically construct a field Q(D) from it,which is named the field of fractions of this integral domain.

In particular, Q(Z) = Q.

6.1. The construction. Assume D is an integral domain, which means D is a unital commutativering without divisors. We now construct the field Q(D) from it.

6.1.1. The set Q(D). Denote by D⇤= D \ {0}. Take D ⇥D⇤ consider a relation on it defined as

(a, b) ⇠ (a0, b0) if and only if ab0 = a0b.

Lemma 6.1. ⇠ is an equivalence relation on D ⇥D⇤.

Example 6.2. [(0, 1)] = [(0, b)] for any b 2 D⇤.

Example 6.3. If D = Z, what is the relation ⇠?

Definition 6.4. Define the set Q(D) := (D ⇥D⇤)/ ⇠. We denote elements in Q(D) be of the form

[(a, b)] for (a, b) 2 D ⇥D⇤.

6.1.2. Abelian group structure on Q(D).

Definition 6.5 (Lemma). Define

[(a, b)] + [(a0, b0)] := [(ab0 + a0b, bb0)].

One can check, it is well-defined. Also, from definition, it is commutative.

Lemma 6.6. (Q(D),+) is an abelian group with identity [(0, 1)] and inverse as

�[(a, b)] = [(�a, b)].

18 RUI WANG

6.1.3. Group structure of multiplication on Q⇤(D).

Definition 6.7 (Lemma). Define

[(a, b)] · [(a0, b0)] := [(aa0, bb0)].

One can check,

(1) it is well-defined;(2) it is commutative;(3) it satisfies distribution law with +.

Hence we have a ring structure (Q(D),+, ·).

Denote by Q⇤(D) := Q(D) \ {[(0, 1)]}.

Lemma 6.8. (Q⇤(D),+) is an abelian group with identity [(1, 1)] and inverse as

[(a, b)]�1= [(b, a)].

Remark 6.9. The last point, [(a, b)] = [(b, a)] is the key point of introducing this fraction field.Though we can not find inverse of a and b, we can use [(b, a)] as its multiplication inverse in this newring.

So far, we have

Proposition 6.10. (Q(D),+, ·) is a field. (As a result, it is also an integral domain. )

6.1.4. Embed D to Q(D). Now we construct an injective ring homomorphism which maps D intoQ(D). Moreover, the image of D in Q(D) becomes an subdomain. (Recall that a subring in anintegral domain may not be an integral domain, e.g., it may not have unity.)

Define◆ : D ! Q(D), a 7! [(a, 1)].

Lemma 6.11. (1) ◆ is injective ring homomorphism.(2) the image ◆(D) is a subdomain of Q(D).

In this sense, the field Q(D) we just constructed actually contains D. (D is a global slice of D⇥D⇤.)

6.2. Uniqueness. In fact, there are a lot of fields that contains D as subdomain. We are going toshow, Q(D) is the minimal one, i.e. any such field F contains Q(D) as a subfield.

Example 6.12. The integral domain Z, it is contained in the fileds Q,R,C. What we are going toprove here is

Q ⇢ R, Q ⇢ C.

Theorem 6.13. If there exists some field F such that D ⇢ F and D is a subdomain of F . Then thereexists some injective ring homomorphism

i : Q(D) ! F

such that iD

= id.

MATH 120B 19

Proof. Define

i([a, b]) = ab�1

and check it is an injective ring homomorphism. Moreover,

i([a, 1]) = a, a 2 D.

⇤

Corollary 6.14. Every field F contains the integral domain D must contain Q(D).

Corollary 6.15. If D is already a field, then Q(D) = D.

6.3. Examples.

(1) Q(Z) ⇠=

Q;(2) D := {m+ ni|m,n 2 Z} is a subdomain in C. Q(D) = {p+ qi|p, q 2 Q}.


7. THE RING OF POLYNOMIALS

7.1. Definition of polynomial ring.

7.1.1. Abelian group structure. Assume R is a ring. Defined a polynomial as a sequence of infiniteelements in R

~a := (a0, a1, a2, · · · )

where a0, a1, a2, · · · 2 R and there are only finitely many ai

6= 0. Let’s call such ~a a finite sequenceof R.

Denote by R[x] as the set of all finite sequences of R (we will explain why we put x here), i.e.

R[x] := {~a = (a0, a1, a2, · · · )|a0, a1, a2, · · · 2 R, and only finitely many ai

6= 0}.

Definition 7.1. The degree of a finite sequence ~a = (a0, a1, a2, · · · ) is defined as the minimal k 2 Zsuch that any i � k, a

i

= 0. In particular, define the degree of (0, 0, · · · ) is �1.

For example, the degree of (1, 2, 0, 0, · · · ) is 1.Using the concept of degree, we can write

R[x] = {~a = (a0, a1, a2, · · · )|a0, a1, a2, · · · 2 R, deg(~a) < +1}.

We can define addition and R-multiplication for R[x] as follows:

(a0, a1, a2, · · · ) + (b0, b1, b2, · · · ) := (a0 + b0, a1 + b1, a2 + b2, · · · )

r · (a0, a1, a2, · · · ) := (r · a0, r · a1, r · a2, · · · ).

Check that the addition and R multiplication are closed in R[x].

Lemma 7.2. (R[x],+) is an abelian group with

(1) the identity as (0, 0, 0, · · · );(2) the inverse of (a0, a1, a2, · · · ) as (�a0,�a1,�a2, · · · ).

20 RUI WANG

Remark 7.3. In fact, with the R-multiplication introduced, R[x] is more than an abelian group but aR-module, which is a generalization of linear spaces over a field.

Assume R is a ring. A set M is called a module if it is an abelian group (M,+) and has R-multiplication defined, satisfying the following properties:

(1) 1 · x = x;(2) (ab)x = a(bx);(3) (a+ b)x = ax+ bx;(4) a(x+ y) = ax+ ay.

Using this definition, you can check that (R[x],+) is a R-module. In fact, together with the multi-plication we are going to introduce, R[x] is an associative algebra over the ring R (usually, peoplerequire the concept for algebra only when R is a commutative unital ring).

7.1.2. Multiplication. Now let’s try to introduce multiplication for it.

Remark 7.4. A naive way is to define

(a0, a1, a2, · · · ) · (b0, b1, b2, · · · ) := (a0 · b0, a1 · b1, a2 · b2, · · · ).

It is a multiplication and does make (R[x],+, ·) into a ring. However, it is not unital even when R isunital! (Why?)

Enlightened by the polynomials over R, we define the polynomial multiplication as

Definition 7.5.(a0, a1, a2, · · · ) · (b0, b1, b2, · · · ) := (c0, c1, c2, · · · )

withci

= ⌃

i

k=0(ak · bi�k

), i = 0, 1, 2, · · · .

From definition, we see that

deg(~a ·~b) deg(~a) + deg(

~b),

and hence R[x] is closed under the multiplication.

Remark 7.6. It is possible that

deg(~a ·~b) < deg(~a) + deg(

~b).

For example, consider Z6[x]. deg((0, 0, [2], · · · )) = 2 and deg((1, 0, [3], · · · )) = 2. However,

deg((0, 0, [2], · · · ) · (1, 0, [3], · · · )) = deg((0, 0, [2], 0, · · · )) = 2 < 2 + 2 = 4.

However, if R is an integral domain, we always have

deg(~a ·~b) = deg(~a) + deg(

~b).

(Prove it. )

Lemma 7.7. The multiplication is associative.

Proof. You may refer book P200 the proof of Thm 22.2 for this proof. Since we are going to provideanother proof later, I skip the steps. You may compare these two proofs. ⇤

MATH 120B 21

Remark 7.8. If we don’t require the finiteness, we can also get a similar ring, which is called the ringof formal power series and denoted by R[[x]]. We are not going to use it in our class.

7.1.3. The inderterminate [x]. Now let’s introduce a useful notation for R[x]. It will simplify thecalculations sometimes, among with many other advantages.

For any sequence ~a = (a0, a1, a2, · · · ) 2 R[x], we can write it as a(x) = a0x0+a1x

1+a2x

2+ · · · ,

where x is called the inderterminate. There are only finitely many nonzero ai

terms. The degree of ~ais the highest oder a(x). We use deg(a) to denote the degree of a polynomial. (The point here is thepowers of x are increasing. ) Let’s call this a finite polynomial with increasing order. Conversely, foreach finite polynomial with increasing order, we can write it into a sequence in R[x]. These two setscan be identified with each other using this way.

When we do multiplication for finite polynomials with increasing order, alternatively, we can usethe following way:

First calculate out all multiplications of monomials without repeating. Each term is of the form(a

i

xi

)(bj

xj

). Then put them together using additions, and rearrange them following the followingprocedures.

(1) Write terms (axi

) · (bxj

) into (a · b)xi+j;(2) Sum all terms with the same order using axi

+ bxi

= (a+ b)xi,(3) Change orders using axi

+ bxj

= bxj

+ axi and make the orders x increase.

Then we will get the multiplication as defined before.We notice that

(1) The orders of the monomials in summation won’t change the outcome of rearrangements;(2) Even we don’t start with a finite polynomial with increasing order, we will still get the same

outcome by this rearrangement procedures;(3) The zero terms can be omitted.

This way makes the calculation of multiplication become easier in some sense. For example, forR[x], when we do

(1x0+ 1x1

+ 0x2+ 1x3

+ 0x4+ · · · ) · (1x0

+ 0x1+ 0x2

+ 2x3+ 0x4 · · · ),

we don’t need to use that formula but can first write it as

(1x0+ 1x1

+ 1x3) · (1x0

+ 2x3)

and then calculate out every monomial, combine common order terms, and change orders into in-creasing as

1x0 · 1x0+ 1x0 · 2x3

+ 1x1 · 1x0+ 1x1 · 2x3

+ 1x3 · 1x0+ 1x3 · 2x3

= 1x0+ 2x3

+ 1x1+ 2x4

+ 1x3+ 2x6

= 1x0+ 1x1

+ 3x3+ 2x4

+ 2x6.

(Check by yourself this is the same as by directly using the multiplication definition. )Using this way, the proof of associativity gets easier as follows. (Compare them!)

22 RUI WANG

Proof. Since the order of summation for a polynomial is not important when doing multiplication, wewrite

a(x) = ⌃

1i=0aix

i, b(x) = ⌃

1j=0bjx

j, c(x) = ⌃

1k=0ckx

k 2 R[x].

As we have show, we can do formal calculations. Here the stuff inside [·] are formal calculations, andthe bracket [·] means rearrangement.

Using the formal calculation,

(a(x) · b(x)) · c(x) = [(⌃

1i=0aix

i · ⌃1j=0bjx

j

) · ⌃1k=0ckx

k

]

= [⌃

1i,j=0ai · bjxi+j · ⌃1

k=0ckxk

]

= [⌃

1i,j,k=0(ai · bj) · ckxi+j+k

].

a(x) · (b(x) · c(x)) = [⌃

1i=0aix

i · (⌃1j=0bjx

j · ⌃1k=0ckx

k

)]

= [⌃

1i=0aix

i · (⌃1i,j=0bjckx

j+k

)]

= [⌃

1i,j,k=0ai · (bj · ck)xi+j+k

].

For each triple i, j, k, since(a

i

· bj

) · ck

= ai

· (bj

· ck

),

we have⌃

1i,j,k=0(ai · bj) · ckxi+j+k

= ⌃

1i,j,k=0ai · (bj · ck)xi+j+k.

After rearrange terms, they are still the same and this proves that

(a(x) · b(x)) · c(x) = a(x) · (b(x) · c(x)).

⇤

From the addition and multiplication above, we have seen that it is safe to write a0x0 as a0 and

omit all 0xk terms. We are going to simply our notations in this way.

7.1.4. Polynomial ring. Now we are ready to state the the following theorem. The details are left toyou.

Theorem 7.9. Assume R is a ring, then (R[x],+, ·) is a ring which contains R as a subring. Moreover,

(1) If R is commutative, R[x] is also commutative;(2) If R is unital, R[x] is also unital;(3) If R is an integral domain, R[x] is also an integral domain.

Remark 7.10. F is a field doesn’t imply F [x] is a field. However, we can use the fractional field toconstruct a field from the integral domain F [x], which is denoted as

F (x) := Q(F [x]).

It is called the field of rational functions of one variable.

Example 7.11. (1) Z[x], Q[x], R[x], C[x];(2) Z

n

[x], Fp

[x].

MATH 120B 23

Proposition 7.12.

char(R) = char(R[x]).

This provides examples of infinite rings with finite characteristics.

Proof. Since R is a subring of R[x], by definition,

char(R) char(R[x]),

and if char(R) = 0, char(R[x]) is also 0.Now we assume char(R) is a positive number n, then for any

a(x) = a0 + a1x1+ · · ·+ a

k

xk,

n · (a(x)) = n · a0 + n · a1x1+ · · ·+ n · a

k

xk

= 0 + 0 + · · ·+ 0 = 0.

This shows

char(R) � char(R[x]).

Above all, char(R) = char(R[x]). ⇤


Proposition 7.13. If � : R ! R0 is a ring homomorphism, then it induces a ring homomorphism

¯� : R[x] ! R0[x], ¯�(⌃n

i=0aixi

) = ⌃

n

i=0�(ai)xi.

In particular, if � is injective, ¯� is also injective, i.e., if R is a subring of R0, then R[x] is also asubring of R0

[x].

7.1.5. Polynomial rings of multivariables. Assume R is a ring, we have seen that R[x] is a ring. Sowe can add another indeterminate [y] and denote it as (R[x])[y]. Similarly, we can also construct(R[y])[x].

Lemma 7.14. The ring (R[x])[y] is isomorphic to the ring (R[y])[x]. Denote this isomorphism classas R[x, y], and name it the ring of polynomials in two indeterminates x and y with coefficients in R.

Similarly, R[x1, x2, · · · , xk

] is called the ring of polynomials in indeterminates xi

and y with coef-ficients in R.

In particular, if D is an integral domain, D[x1, x2, · · · , xk

] is also an integral domain.If F is a field, we can construction the fraction field from the integral domain F [x1, x2, · · · , xk

]

and denote it by

F (x1, x2, · · · , xk

) := Q(F [x1, x2, · · · , xk

]).

This is called the field of rational functions in k indeterminates over F .

24 RUI WANG

7.2. The evaluation homomorphism. Now we relate polynomials with the ring of maps C(X,R)

we considered before. Recall that assume R is a ring and X is any set, we can define a ring structurefor the set of all maps from X to R, which we denote as C(X,R).

Now we prove that

Proposition 7.15. Assume R0 is a commutative ring and R is a subring, then the map

� : R[x] ! C(R0, R0)

define as

(�(a0 + a1x+ · · ·+ ak

xk

))(↵) = a0 + a1 · ↵ + · · ·+ ak

· ↵k

=: ⌃

k

i=0ai↵i

(In this notation, we omit the ↵0 term.)is a ring homomorphism.

After we prove this proposition, recall that we have proved that for any ↵ 2 R0, the evaluation map

ev↵

: C(R0, R0) ! R0

is a surjective ring homomorphism. Now, together with the homomorphism � : R[x] ! C(R0, R0),

we get the composition homomorphism

ev↵

� � : R[x] ! R0.

Let’s abuse notation and also denoteev

↵

: R[x] ! R0.


Corollary 7.16. For any commutative ring R0 and its subring R,

ev↵

: R[x] ! R0

is a ring homomorphism for any ↵ 2 R0.

Example 7.17. (1) If F = F 0, then ev0, ev1 are both onto.(2) Take F = F 0

= R.

ev2(2� x) = 0, ev2(�6 + x+ x2) = 0, ev2(4� 4x+ x2

) = 0, · · ·

Can we claim that ker ev2 = {(2� x)f(x)|f(x) 2 R[x]}?(3) Take F = F 0. For any 0 6= r 2 R, is the map ev

r

has nontrivial kernel? The answer is yes,because r � x is in the kernel.

(4) Take F = Q and F 0= R. For any 0 6= r 2 R, is the map ev

r

has nontrivial kernel? Forexample, r =

p2,

evp2(2� x2

) = 0.

However, we can show thatev

⇡

: Q[x] ! Ris one-to-one.

(5) Similarly,ev

i

: R[x] ! C, (1 + x2) 2 ker ev

i

.

MATH 120B 25

Now we prove the proposition.

Proof. (1) First show that

�(a(x) + b(x)) = �(a(x)) + �(b(x)) for any a(x), b(x) 2 R[x].

Assume

a(x) = ⌃

10 a

i

xi, b(x) = ⌃

10 b

i

xi.

Then

a(x) + b(x) = ⌃

1i=0(ai + b

i

)xi

For any ↵ 2 R0,

�(a(x) + b(x))(↵) = �(⌃1i=0(ai + b

i

)xi

)(↵)

= ⌃

1i=0(ai + b

i

)↵i.

On the other hand,

(�(a(x)) + �(b(x)))(↵) = �(a(x))(↵) + �(b(x))(↵)

= ⌃

10 a

i

↵i

+ ⌃

10 b

i

↵i

= ⌃

10 (a

i

+ bi

)↵i.

The last equality comes from the distribution law. We don’t use commutative condition here.(2) Next we show that

�(a(x) · b(x)) = �(a(x)) · �(b(x)) for any a(x), b(x) 2 R[x].

For any ↵ 2 R0,

�(a(x) · b(x))(↵) = �(⌃1k=0(⌃

k

i=0ai · bk�i

)xk

)(↵)

= ⌃

1k=0(⌃

k

i=0ai · bk�i

)↵k.

On the other hand,

(�(a(x)) · i(b(x)))(↵) = �(a(x))(↵) · �(b(x))(↵)

= ⌃

10 a

i

↵i · ⌃10 b

i

↵i

= ⌃

1k=0(⌃

k

i=0ai · bk�i

)↵k.

This last equality essentially uses R0 is commutative.So we finish the proof. ⇤

From now on, let’s restrict to the case that R0= F 0 is a field and R = F is a subfield of F .

Remark 7.18. This homomorphism � is NOT injective if char(F ) 6= 0. For example, for Fp

, considerthe polynomial

(1� xp�1)x = x� xp.

Using Fermat’s little theorem, for any x 6= 0, 1� xp�1= 0. Hence, (x� xp

) 2 ker�.

26 RUI WANG

7.3. Relation between solving polynomial equations and ev.

Definition 7.19. Take fields F ⇢ F 0 and f(x) 2 F [x]. If ↵ 2 F 0 such that

ev↵

(f) = 0,

then we say ↵ is a zero of f . Denote by Z(f) ⇢ F 0 for all zeros of f .

To solve an equation f(x) = 0 is equal to say to find zeros of it.

Example 7.20. Show that for Q ⇢ R,

Z(2� x2) \Q = ;.

Lecture 13 stopped here. Lecture 14 is midterm.

8. POLYNOMIALS OVER A FIELD

We have known that, if F is a field, then F [x] is an integral domain. We are going to see thisintegral domain behaves similarly as the integral domain Z in many senses. (The reason is F [x]

naturally carries the structure of a graded ring.)

8.1. Division Algorithm for F [x]. The division algorithm for polynomials can be considered as aanalogue for division of integers, i.e. given n 2 Z+, any integer a can be uniquely written into

a = qn+ r,

with 0 r < n.Before we start, let’s recall we define deg(0) = �1. You will see this convention makes sense from

several points.

Theorem 8.1. [Divison Algorithm] Assume F is a field. For any g 2 F [x] with deg g � 0, one canwrite f 2 F [x] uniquely into the form f = qg + r with some q, r 2 F [x] and deg r < deg g.

Moreover, if deg(f) � deg(g), then deg q = deg f � deg g. If deg(f) < deg(g), then q = 0.

Before we prove this theorem, recall

Lemma 8.2. If D is an integral domain, then

deg(fg) = deg f + deg g.

Theorem 8.1 essentially depends on the field F is an integral domain.

Proof. We first show the existence. We fix g 2 F [x] with deg(g) = n � 0. If deg(f) < deg(g), thenwe can write

f = 0 · g + f,

which satisfies the requirement. For deg(f) � deg(g), let’s do induction.

(1) For deg(f) = deg(g) = n, write f(x) = ⌃

n

i=0aixi and g(x) = ⌃

n

i=0bixi. Then

f(x) = b�1n

an

g(x) + (f(x)� b�1n

an

g(x)).

Notice f(x)� b�1n

an

g(x) has at most degree n� 1 term. So this is the expression we want.

MATH 120B 27

(2) Now we assume that for any f 2 F [x] with deg(f) = k � n is settled down, let’s prove forthe case deg f = k + 1.

(3) Suppose we can write

f(x) = a0 + a1x+ · · · ak+1x

k+1, g(x) = b0 + b1x+ · · · bn

xn,

where ak+1 6= 0 and b

n

6= 0.Then we can write

f(x) = (ak+1b

�1n

xk+1�n

)g(x) + r1(x),

where

deg(r1) k.

• If deg(r1) < deg(g) = n, we are done.• Otherwise, deg(r1) � deg g. Since deg(r1) k now, we can apply the assumption above

to r1 and write it as

r1 = qg + r

with deg r < deg g. Plug it into the f , we get

f = (ak+1b

�1n

xk+1�n

+ q)g + r.

Notice that deg q = deg(r1 � deg g) < k + 1� n, hence

deg(ak+1b

�1n

xk+1�n

+ q) = k + 1� n.

This shows it satisfies the requirements for degrees.

Now let’s prove the uniqueness. Assume we can write

f = q1g + r1 = q2g + r2,

with deg(r1) < deg(g) and deg(r2) < deg(g). Then deg(r1 � r2) < deg(g) and

r1 � r2 = (q1 � q2)g.

If q1 � q2 6= 0, then using Lemma 8.2,

deg(r1 � r2) = deg((q1 � q2)g) = deg(q1 � q2) + deg g � 0 + deg(g) = deg(g).

We get contradiction. As a result, q1 = q2, and it follows r1 = r2. ⇤

For application, we can use long division to obtain q and r in practice.

Example 8.3. Consider in F5[x],

f(x) = 1 + 2x+ 3x2+ 4x3

+ x4, g(x) = 3 + x2+ x3.

Find q and r such that

deg(r) < 3, f = qg + r.

28 RUI WANG

Denote by F k

[x] the polynomials of degree k, by F<k

[x] the polynomials of degree less than k (i.e.at most k � 1 ) and by F>k

[x] the polynomials of degree at least k + 1. Theorem 8.1 tells us that,given any g 2 F k

[x] for k � 0, there is the remaider map

Rg

: F [x] ! Fk�1[x], f 7! r.

Recall for Z, we have the surjective ring homomorphism

�n

: Z ! Zn

, a 7! [a].

Next, we do the analogue to polynomials.

Remark 8.4. In fact, the degree n polynomial g should be made analogue to integer n+ 1. In partic-ular, the degree �1 polynomial corresponds to integer 0, and the degree zero polynomial correspondsto integer 1. In this sense, Z1 is not Z but {0}.

Lecture 15 stopped here

8.2. Quotient homomorphism and ideal. Given a polynomial g 2 F [x] of degree n � 0. Notice asa set,

ImRg

= F<n

[x].

Proposition 8.5. Define r1 ·g r2 = Rg

(r1r2). Then

(1) (ImRg

,+, ·g

) is a unital commutative ring.(2) The map R

g

: F [x] ! ImRg

is a surjective ring homomorphism.(3) kerR

g

:= (g) := {qg|g 2 F [x]}.(4) F [x]/(g) ⇠

=

ImRg

.

Proof. (1)(2)(3) follow from direct checkings and (4) are consequences from Theorem 2.18. Detailsare left to you to check. ⇤

This result is an analogue of Z/nZ ⇠=

Zn

.

Definition 8.6. For any g 2 F [x], the set

(g) := {fg|f 2 F [x]}

is called the principal ideal generated by g.

8.3. Ideals and quotient rings. Now we introduce the concept of ideal.

Definition 8.7. Assume R is a ring. An abelian additive subgroup I ⇢ R is called an ideal, if

RI ⇢ I, IR ⇢ I.

Obviously, if I is an ideal, it is also a subring by definition.If R is commutative, it is enough to check RI ⇢ I only. If R is unital, then if follows RI = I = IR.

By definition, any ring has two (if nontrivial) ideals. One is {0} and the other is R itself.The following lemma is important but is easy to check.

Lemma 8.8. The intersection of ideals is an ideal.

In fact, we have checked the following statement before but didn’t introduce the concept of ideal.

MATH 120B 29

Proposition 8.9. Assume � : R ! R0 is a ring homomorphism, then ker� is an ideal of R.

Conversely, any ideal in a ring can always be considered as a kernel of some ring homomorphismfrom this ring.

Proposition 8.10. (1) If I ⇢ R is an ideal, then R/I is a ring with multiplication defined as

[r1] · [r2] = [r1r2].

This is called the quotient ring of R by I .(2) The quotient map

�I

: R ! R/I

defined by taking cosets is a ring homomorphism with kernel

ker�I

= I.

Proof. Exercise. ⇤

It follows that in F [x], every principal ideal (g) is an ideal in the ring F [x]. This explains why wecall (g) principal ‘ideal’. More general, we one define the principal ideal in a commutative ring in thesame way.

Definition 8.11. Assume R is commutative ring. Any a 2 R, the set

(a) := {ga|g 2 R}

is an ideal and called the principal ideal generated by a.

Obviously from definition, for any ideal I ⇢ R,

(a) ⇢ I, for any a 2 I.


8.4. F [x] is a PID. The following theorem is an important property for F [x].

Theorem 8.12. F [x] is a principal ideal domain (PID), i.e. any ideal in F [x] is principal ideal.

Proof. Obviously, the zero ideal can be written as (0) and hence a principal ideal. Now consider anyideal I which is not (0), there must exist a nonzero polynomial of lowest nonnegative degree. Let’schoose and fix it, denote by g. (We can do this thanks for the choice axiom. )

Now any f 2 F [x],f 2 (g) +R

g

(f).

Notice deg(Rg

(f)) < deg(g), by our assumption to g, Rg

(f) must be zero. This shows that f 2 (g),and further actually I ⇢ (g). On this other hand, (g) ⇢ I , and we are done. ⇤

Now we can use this property to understand more for polynomials. The following fact tells uswhy the problem of solving an equation of polynomial can be related to understanding ideals in thepolynomial ring.

Recall we have introduced before. Assume F is a subfield of F 0. We have constructed the evalua-tion homomorphism

ev↵

: F [x] ! F 0, for any ↵ 2 F 0.

30 RUI WANG

Denote by I↵

:= ker ev↵

. Using Proposition 8.9, we know I↵

is an ideal of F [x]. Moreover, Theorem8.12 tells us, there must be some g 2 F [x] such that

I↵

= (g).

In particular, if ↵ 2 F , then we know I↵

= (↵� x).Let’s reinterpret the above statement purely in the language of polynomials.

Corollary 8.13. Assume F is a field. Any ↵ 2 F ,

f 2 I↵

if and only if f(x) = (↵� x)g(x) for some g 2 F deg(f)�1[x],

where I↵

:= {f 2 F [x]|f(↵) = 0}.

We can keep working in this way and use induction to show that

Corollary 8.14. For any k (k � 1) distinct elements

↵1,↵2, · · · ,↵k

2 F,

a polynomial f 2 I↵1 \ I

↵2 \ · · · \ I↵k

if and only if

f(x) = (↵1 � x)(↵2 � x) · · · (↵k

� x)g(x) for some g 2 F deg(f)�k

[x].

As a consequence, a nonzero polynomial f 2 F [x] of degree k has at most k distinct zeros in F .

Proof. The ‘if’ part is obvious, and we now prove the ‘only if’ part.We have seen that this statement is true for k = 1. Assume that k = l is true. We prove for the case

k = l + 1. For any f 2 \i=1,··· ,l+1I↵i ⇢ \

i=1,··· ,lI↵i , by our assumption,

f(x) = ⇧

i=1,··· ,l(↵i

� x)g(x), for some g 2 F deg(f)�l

[x].

Now since f(↵l+1) = 0, it follows

0 = ⇧

i=1,··· ,l(↵i

� ↵l+1)g(↵l+1).

Because each ↵i

� ↵l+1 6= 0 and F is a field, have g(↵

l+1) = 0, i.e.,

g 2 I↵l+1

.

Using Corollary 8.13, g(x) = (↵l+1)h(x) with deg(h) = deg(g)� 1 = deg(f)� (l+1). As a result,

f(x) = ⇧

i=1,··· ,l+1(↵i

� x)h(x)

and we are done.To see the statement that a nonzero polynomial f 2 F [x] of degree k has at most k distinct zeros in

F , we assume otherwise, if f has more than k distinct zeros, we can use the k distinct zeros to write

f(x) = ⇧

i=1,··· ,k(↵i

� x)g(x)

but now deg(g) = 0 and hence a constant a in F . Then we take an extra zero ↵, which makes

0 = ⇧

i=1,··· ,l(↵i

� ↵)a.

It follows a = 0, which contradicts with the assumption that f is nonzero. ⇤

MATH 120B 31

For a polynomial f 2 F [x], denote by Z(f) the set of all zeros of f in F . We have shown thatthe number of elements in Z(f) is at most deg(f) if f is nonzero (deg(f) � 0). However, this setactually can be empty.

Example 8.15. (1) Take F = R, #Z(1 + x2) = 0;

(2) Take F = C, #Z(1 + x2) = 2.

Given f 2 F [x], take each elements in Z(f) and use this way to write

f(x) = ⇧

i=1,··· ,#Z(f)(↵i

� x)g(x).

The polynomial g may or may not have zero, but if it has, the zeros must be in Z(f), i.e.

Z(g) ⇢ Z(f).

If it doesn’t have zeros, we are done. Otherwise, keep using zeros in Z(g) to write

g(x) = ⇧

i=1,··· ,#Z(g)(↵ji � x)h(x).

This procedure must stop for finite times, since degrees will exhaust.We conclude as in the following theorem

Proposition 8.16. Any f 2 F [x] can be written into the form

f(x) = ⇧

i=1,··· ,#Z(f)(↵i

� x)pig(x),

with Z(g) = ; and deg(g) = deg(f)� ⌃

i=1,··· ,#Z(f)pi � 0.As a consequence, a nonzero polynomial f 2 F [x] of degree k has at most k zeros (maybe not

distinct) in F .

We call the number ⌃i=1,··· ,#Z(f)pi the number of zeros with multiplicity for f 2 F [x] in the field

F , and use the notation |Z(f)| to denote it.

Example 8.17. (1) Take F = C, |Z(x2)| = 2;

(2) Take F = C, |Z(�1 + 2ix+ x2)| = 2;

(3) Take F = R, |Z(�1 + 2ix+ x2)| = 0.

Lecture 17 stopped here.The following statement is an interesting application of the above theorem.

Theorem 8.18. Assume F is a field. Then any finite subgroup of the multiplication group (F ⇤, ·) iscyclic. In particular, for a finite field F , F ⇤ itself is cyclic.

Proof. First we know that any finite subgroup G of (F ⇤, ·) must be a finite abelian group. It followsfrom the classification theorem of finitely generated abelian group Theorem 1.4,

G ⇠=

Zd1 ⇥ · · ·Z

dk,

with di

= pdii

, i = 1, · · · , k. From group theory we know that, if we can prove d1, d2, · · · , dk arecoprime, then

Zd1 ⇥ · · ·Z

dk⇠=

Zd1d2···dk ,

i.e., it is a cyclic group.

32 RUI WANG

To show d1, d2, · · · , dk are coprime, it is enough to show the least common multiple m = lcm(pr11 , · · · , prkk

)

is the same as their product d1d2 · · · dk.Obviously, m d1d2 · · · dk. To see the other direction, notice any (0, · · · , 0, x, 0, · · · , 0) with x

on the i-th position with x 2 Zdi satisfies the polynomial equation

xm � 1 = 0.

By this way, we have d1d2 · · · dk different zeros. Using Proposition 8.16, there must be d1d2 · · · dk m too. Hence m = d1d2 · · · dk and we are done. ⇤

Using this example, for example, we know that any finite field, the multiplication group F ⇤ can notbe isomorphism to Z2 ⇥ Z2.

8.5. Irreducible polynomials.

Definition 8.19. We call a polynomial f 2 F [x] irreducible, if

(1) deg(f) � 1;(2) whenever f = gh for some g, h 2 F [x], then there must be either deg(g) = 0 or deg(h) = 0.

We first notice

Proposition 8.20. (1) Any f 2 F [x] of degree 1 is irreducible.(2) A f 2 F [x] of degree 2 or 3 is reducible if and only if Z(f) 6= ;.(3) For a field F , any polynomial of degree at least 2 is reducible if any only if any nonconstant

polynomial has zeros over F .

Proof. (1) Immediately from definition;(2) Using Proposition 8.16;(3) • For any f 2 F [x], if deg(f) = 1, then f(x) = a1x + a0 with a1 6= 0. Obviously, it has

zero x = �a�11 a0.

If deg(f) � 2, by our assumption, it is reducible, we can write

f = gh, 0 < deg(g) < deg(f), 0 < deg(h) < deg(f).

If deg(g) = 1 or deg(h) = 1, then we are done.Otherwise, deg(g) � 2 and deg(h) � 2. We can use assumption again and keep factoringg and h until we obtain factor polynomial of degree 1.This shows Z(f) 6= ;.

• For any f 2 F [x] with deg(f) � 2, take ↵ 2 Z(f), then using Proposition 8.16,

f(x) = (x� ↵)g(x)

with deg(g) = deg(f)� 1. This shows f is reducible.⇤

We are going to show later, any f 2 C[x] with deg(f) � 2 is reducible. Then using Proposition8.16, we know f(x) 2 C[x] always has zeros.

Today, let’s focus on f 2 Z[x] and f 2 Q[x] and introduce some methods in testing if f isirreducible or not. Basically, we are working on high school math in a somehow more systematicalway.

MATH 120B 33

The following theorem is not very hard to prove but the proof is a little messy, and so we skip theproof.

Theorem 8.21. A polynomial f 2 Z[x] factors into a product of two polynomials of lower degrees rand s in Q[x] if and only if it has such a factorization with polynomials of the same degrees r and s

in Z[x].

Using this statement, when we try to see if a polynomial f 2 Z[x] ⇢ Q[x] is reducible in Q[x], itis enough to check if it is reducible in Z[x]. The latter usually is easier to check.

Example 8.22. Show f(x) = x4 � 2x2+ 8x+ 1 is irreducible over Q[x].

Corollary 8.23. If f(x) = xn

+ an�1x

n�1+ · · · + a0 2 Z[x] with a0 6= 0, and if f(x) has a zero in

Q, then it has a zero in Z. Moreover, this zero must divide a0.

The following theorem is another useful tool in judging if a polynomial is reducible in Q[x].

Theorem 8.24 (Eisenstein Criterion). For f(x) = an

xn

+ an�1x

n�1+ · · ·+ a0 2 Z[x], if there exists

some prime number p such that

(1) an

6= 0 in Zp

;(2) for any i < n, a

i

= 0 in Zp

;(3) a0 6= 0 in Z

p

2 ,

then f(x) is irreducible over Q.

Example 8.25. Show f(x) = 25x5 � 9x4 � 3x2 � 12 is irreducible over Q[x].

Corollary 8.26. The polynomial

�

p

(x) = xp�1+ xp�2

+ · · ·+ x+ 1

is irreducible over Q for any prime p.


8.6. Uniqueness of factorization in F [x]. Given any f 2 F [x], if f is irreducible, then f = f , itis already a (trivial) multiplication of irreducible polynomials. Otherwise, we can write f = gh forsome g, h 2 F [x] with deg(g) < deg(f) and deg(h) < deg(f). If both g and h are irreducible, f is amultiplication of irreducible polynomials. Otherwise, we can keep working in this way. It must stopat finite steps since deg(f) is finite and such decomposition always lower degrees. We conclude,

Lemma 8.27. Every f 2 F [x] can be written into a multiplication of several irreducible polynomials.

However, we don’t know if it is unique. To be more concrete, if

f = p1p2 · · · ps = q1q2 · · · qt, pi

, qj

are all irreducible,

can we know s = t and pi

= qi

for every i = 1, · · · s?Actually, we are going to prove s = t and up to orders, p

i

differs from qi

by a nonzero constantelements in F . Such statement is usually referred as uniqueness of factorization. Then the statementcan be rephrased as

34 RUI WANG

Theorem 8.28. For any field F , the polynomial ring F [x] is a unique factorization domain (UFD).

The key point in the proof is, every irreducible polynomial is prime in the sense we are going tointroduce next.

9. MAXIMAL IDEALS AND PRIME IDEALS

Instead of directly proving the UFD theorem for polynomial rings, let’s make the context moregeneral. You will see, as PIDs, the integral domain Z and F [x] shares a lot of similar properties.

9.1. Terms for integral domains. Assume D is an integral domain, we introduce the followingconcepts for elements in D.

Definition 9.1. (1) An element u is called a unit, if u has inverse;(2) Call a divides b, if b = aq for some q 2 D. In this case, we also say a is a divisor of b;(3) Call a is a proper divisor of b, if a is a divisor of b and b = aq with neither a nor q is unit;(4) Call a and b are associates, if a = bu for some unit u. (It is equivalent to say b = au for some

unit u.)(5) Call a is irreducible, if a is not a unit and a has no proper divisor;(6) Call p is prime, if p is not a unit and p divides ab implies p divides a or p divides b.

Next, we give the interpretation of these definitions using the concept of principal ideals.

Lemma 9.2. Assume D is an integral domain.

(1) u is a unit, if and only if (u) = (1);(2) a divides b, if and only if (b) ⇢ (a);(3) a is a proper divisor of b, if and only if (b) ( (a) ( (1);(4) a and b are associates, if and only if (a) = (b);(5) a is irreducible, if and only if (a) 6= (1) and there is no c such that (a) ( (c) ( (1);(6) p is prime, if and only if (p) 6= (1) and ab 2 (p) implies a 2 (p) or b 2 (p).

Next we generalize the concept of greatest common divisor to integral domains.

Definition 9.3. Assume D is an integral domain. An element d is called the greatest common divisorof two elements a, b 2 D which are not both zero, if

(1) d is a divisor of both a and b;(2) d is greatest, i.e., any divisor c of both a and b is also a divisor of d.

Be careful that, in general for an integral domain, the greatest common divisor of two elementsmay not exist. However, we have the following results for PIDs.

Proposition 9.4. If D is a PID, then any two elements a, b 2 D which are not both zero, greatestcommon divisors exist. In fact, any d with

(d) = (a) + (b)

is a greatest common divisor, and it is unique up to a unit in D.


MATH 120B 35

Proof. It is straightforward to check that I := (a) + (b) is an ideal, and it is the smallest ideal whichcontains a and b.

Because D is a PID by our assumption, there exists some d 2 D such that I = (d).First, since a, b 2 I = (d), it follows d divides a and b.Any c divides a and b, have

(a) ⇢ (c), (b) ⇢ (c).

It follows (d) = I = (a) + (b) ⇢ (c) and hence c divides d.If (d) = (d0), we know d and d0 are associates. ⇤

Definition 9.5. In a PID, if (a) + (b) = (1), we call a and b are coprime or relatively prime.

Proposition 9.6. In a PID, the followings are equivalent:

(1) p is prime;(2) p is irreducible.

Proof. From (1) to (2) actually doesn’t need PID but true for any integral domains. See the followingLemma 9.7. So the only thing to need to prove here is from (2) to (1), i.e. an irreducible element mustbe prime.

Assume p is irreducible and p divides ab. Further assume p doesn’t divide a and let’s show then p

divides b.Since a PID, take d a greatest common divisor of p and b. If we assume p doesn’t divide b, since p

is irreducible, d must be unit. (Otherwise, d is associate of p, it follows d divides b and contradiction.) It follows 1 = rp+ sb for some r, s 2 D. Multiply a, we get

a = rap+ sab.

It follows p divides a which contradicts with out assumption.⇤

Lemma 9.7. Assume D is an integral domain. If p is prime, then p is irreducible.

Proof. We only need to show there is no c 2 D such that

(p) ( (c) ( (1),

if p is prime.Assume there is such c. Then we can write p = cd. It follows cd 2 (p). Since p is prime, it means

that c 2 (p) or d 2 (p).For the first case, it follows that (c) = (p), which contradicts to our assumption. For the second

case, write d = pe, then using p = cd = cpe, it follows ce = 1 and so c is unit, which contradictswith assumption too.

⇤

9.2. PID is UFD. Now we ready to prove the following promised statement that every PID is a UFD.

Definition 9.8. We call an integral domain D is a unique factorization domain, if

(1) Any element a 2 D which is neither 0 nor unit can be factored into products of irreducibleelements;

36 RUI WANG

(2) Such factorization is unique in the following sense: For any two factorizations

a = p1 · · · ps = q1 · · · qt,

we have s = t and after changing of orders, one can make pi

and qi

are associates.

Now, let’s prove a PID must be UFD.

Theorem 9.9 (PID implies UFD). Every PID is a UFD.

Proof. (1) Assume D is a PID. Any element a 2 D which is neither 0 nor unit, we first show itcan be factored into products of irreducible elements.

If a is irreducible, we are done. Otherwise, a can be factored into

a = a1b1 with neither a1, b1 a unit.

It follows (a) ( (a1) and (a) ( (b1).It both a1 and b1 are irreducible, then we are done. Otherwise, keep decomposing by this

way, and we get(a1) ( (a2).

At last, we get an ascending chain

(a) ( (a1) ( (a2) · · · ( (1).

Take union of this chainI = [1

i=1(ai).

Direct checking shows it is an ideal. Since PID, it is the same as some (c). Assume c 2 (ak

),then we get

(c) ⇢ (ak

) ⇢ (c).

It means the chain actually stops at (ak

).(2) Next, we show the factorization is unique. Assume

a = p1 · · · ps = q1 · · · qt.

First, p1 divides a and it follows there must exist an element qi

such that p1 divides qi

. We canchange orders and WLOG assume p1 divides q1. So we can write q1 = p1u1. However, sinceq1 is irreducible, u1 must be unit.

Plug them into a, we get

p1p2 · · · ps = p1u1q2 · · · qt.

We can cancel p1 since it is an integral domain, and then have

p2 · · · ps = u1q2 · · · qt.

Repeat the above arguments, we finally get

qi

= pi

ui

, i = 1, · · · s = t, 1 = u1 · · · us

,

after change of orders if necessary.⇤

MATH 120B 37

Corollary 9.10. (1) Z is UFD;(2) Assume F is a field, then F [x] is UFD.

Remark 9.11. In general, any Euclidean domain is a PID and so a UFD.

Remark 9.12. F [x1, · · · xn

] for n � 2 is not PID. However, we can still prove it is a UFD.

We postpone the proof of the following theorem to the end of this quarter.

Theorem 9.13. If D is a UFD, then D[x] is a UFD.

Corollary 9.14. Assume F is a field, then F [x1, · · · xn

] is UFD.

Remark 9.15. The final bonus problem asks you to prove the Novikov ring is PID, so it followsNov(�) and Nov(�)[x1, · · · , xn

] are both UFD.


9.3. Maximal ideals and primes ideals. In general for a ring R,

Definition 9.16. (1) An ideal M is called a maximal ideal, if M 6= R and there is no ideal I suchthat

M ( I ( R.

(2) An ideal P is called a prime ideal, if P 6= R and any ab 2 P implies a 2 P or b 2 P .

With these definitions, we immediately have

Lemma 9.17. (1) In an integral domain, an element p is prime, if and only if (p) is prime.(2) In a PID, a is irreducible if and only if (a) is maximal.

In general, we have

Proposition 9.18. Assume R is a commutative unital ring. Then every maximal ideal is prime ideal.

Proof. Assume M is an maximal ideal, we now prove it is a prime ideal.Take ab 2 M and assume that a /2 M , we now show b 2 M . Notice that M +(b) is an ideal, which

includes M . IfM = M + (b),

then b 2 M and we are done. Otherwise, since M is maximal ideal, we must have

M + (b) = (1).

As a consequence, we can write

1 = m+ bq, m 2 M, q 2 R.

Then a = am+ abq 2 M , we get contradiction.⇤

Theorem 9.19. Assume R is a commutative unital ring. Then

(1) An ideal I in R is a prime ideal, if and only if R/I is an integral domain;(2) An ideal I in R is a maximal ideal, if and only if R/I is a field.

38 RUI WANG

Proof. Recall for any ideal I ⇢ R, R/I is a commutative unital ring.

(1) We first show I is a prime ideal if and only if R/I is an integral domain.Take two elements [a], [b] 2 R/I , have [a] · [b] = [0] if and only if ab 2 I .If we assume a /2 I , then, b 2 I if and only if [b] = 0, i.e. I is prime ideal if and only if

R/I is an integral domain.(2) Since we have shown that a maximal ideal is a prime ideal, we know R/I is an integral

domain. We only need to show I is maximal if and only if every [0] 6= [a] 2 R/I has inverse.Notice a /2 I if and only if I + (a) 6= I . If I is maximal, I + (a) = (1), and

1 = i+ aq, i 2 I, q 2 R.

Then [a][q] = [1], and so [a] has inverse.Conversely, consider any ideal I ( I 0, take some element a 2 I 0, if it has inverse, then

1 = aa�1 2 I 0. It follows I 0 = (1) = R, and this shows I is maximal ideal.

⇤

In a PID, maximal ideal is the same as prime ideals. We now take an example which is not PID tosee their difference.

Example 9.20. Consider Z[x], the evaluation homomorphism

ev0 : Z[x] ! Z, f(x) 7! f(0)

is surjective. Using Theorem 2.18, we know ker ev0 is an ideal of Z[x] and

Z[x]/ ker ev0⇠=

Z.

Since Z is an integral domain but not a field, ker ev0 is a prime ideal but not a maximal ideal. Thisideal is

ker ev0 = {⌃1i=1aix

i 2 Z[x]},

i.e. polynomials without constant terms.

Remark 9.21. In fact, to see Z[x] is not a PID, you can for example, look at the ideal I := (x) + (3)

and argue it is not a principal ideal.

Example 9.22. Similarly, we can show F [x, y] is not a PID.


10. FIELD EXTENSION

10.1. Motivation. Given a field F , we have shown that any polynomial f 2 F [x] with degree k atmost has k zeros (with multiplicity).

Lemma 10.1. For a field F , the follows are equivalent:

(1) any nonconstant polynomial has at least one zero;(2) any degree k � 0 polynomial f 2 F [x] has k zeros with multiplicity.

MATH 120B 39

Proof. Form (2) to (1) is obvious. We only need to show from (1) to (2).In fact, if we assume any nonconstant polynomial has at least one zero, then the decomposition in

Proposition 8.16 must have deg(g) = 0, and then (2) follows immediately. ⇤

Definition 10.2. If a field satisfies (1) in Lemma 10.1, then call it an algebraically closed field.

Example 10.3. (1) Q is not algebraically closed;(2) R is not algebraically closed;(3) C is algebraically closed;

Lemma 10.4. Every algebraically closed field is infinite field.

Proof. Assume F is algebraically closed and has only k elements, and

F = {↵1, · · · ,↵k

}.

Consider the polynomial

f(x) = 1 + (x� ↵1)(x� ↵2) · · · (x� ↵k

).

Let’s show it does have zeros over F . Otherwise, say ↵i

. Then

f(↵i

) = 1 6= 0.

We get contradiction.⇤

The main result we want to show is, every field can be extended to an algebraically closed field.

10.2. Definition of field extension.

Definition 10.5. Given a field F , a field E has F as a subfield is called a field extension of F .

Example 10.6. Both R and C are field extensions of Q; C is a field extension of R.

Theorem 10.7 (Kronecker’s Theorem). Assume F is a field. Any f 2 F [x] with positive degree, thereis a field extension E of F such that being considered as a polynomial f 2 E[x], f has zeros in E.

Proof. We explicitly construct E.For any f 2 F [x], we decompose f into irreducible factors as

f = p1 · · · pr.

In fact, use each pi

, we can construct a such field. Take p1 as example.By Lemma 9.17 and Theorem 9.19, F [x]/(p1(x)) is a field and we denote it by E. Next we show,

this is the field we want.

(1) We show F can is a subfield of E. Consider the map

� : F ! E, a 7! [a].

Since � = � i, where i is the inclusion and is the quotient map

i : F ! F [x], : F [x] ! F [x]/(p(x)) = E,

and both are ring homomorphisms, we know � is also ring homomorphism.

40 RUI WANG

It is one-to-one, because for any [a] = [b], we have a � b 2 (p(x)). Since both a and b areof degree 0, we must have a = b.

(2) Denote by ¯f for considering f as a polynomial in E[x] as introduced in Proposition 7.13. Inparticular, if f(x) = a0 + a1x+ · · · a

n

xn, then

¯f(x) = [a0] + [a1]x+ · · · [an

]xn.

There is an obvious zero of f 2 E[x] which is [x]. We now see the reason. By plugging [x]

into ¯f , we get

¯f([x]) = [a0] + [a1][x] + · · · [an

][x]n

= [a0 + a1x+ · · · an

xn

]

= [f(x)]

= [p1(x)][p2(x) · · · pr(x)]

= [0][p2(x) · · · pr(x)]

= [0].

⇤

Example 10.8. Consider f(x) = x2+ 1 2 R[x]. It is irreducible in R[x]. Extend R to E :=

R[x]/(x2+ 1).

Construct map� : C ! R, a+ bi 7! [a+ bx], a, b 2 R.

Not hard to see, it is a field isomorphism. (Give the proof.) This map explains that E is the algebraicway of constructing C from R.

Example 10.9. Consider f(x) = x4 � 5x2+ 6 2 Q[x]. It can be factored into

f(x) = (x2 � 2)(x2 � 3),

such that both x2 � 2 and x2 � 3 are irreducible over Q.


10.3. Algebraic and transcendental elements. Assume E is a field extension of F . We now useelements in E to do field extensions for F . For this purpose, we need to distinguish two types ofelements by using the following definitions.

Definition 10.10. Assume E is a field extension of F . An element ↵ 2 E is called algebraic over F ,if ↵ 2 Z(f) for some nonzero f 2 F [x]. An element ↵ 2 E is called transcendental, if ↵ /2 Z(f) forany nonzero f 2 F [x].

Example 10.11. Consider C as extension of R, then i is algebraic over R since i is zero of the equationx2

+ 1 = 0 in R[x].In fact, we are going to prove that, every number ↵ 2 C is algebraic over R.

Example 10.12. Consider R as extension of Q, then ⇡ and e are transcendental over Q. (This is infact not easy to prove. You could refer wikipedia the page on Transcendental and some referencestherein for some ideas of proofs. )

MATH 120B 41

Theorem 10.13. Assume E is a field extension of F . Denote by

ev↵

: F [x] ! E

the evaluation homomorphism at some ↵ 2 E. Then ↵ 2 E is transcendental over F if and only ifev

↵

is one-to-one.

This proof is an immediate translation of definition and left to yourselves.As a result, the image ev

↵

(F [x]) is a domain of E. We use F (↵) to denote its fraction field.

Now if we take ↵ 2 E which is algebraic over F , consider the ring homomorphism

ev↵

: F [x] ! E.

It kernel can be identified with

I(↵) := {f 2 F [x]|↵ 2 Z(f)},

and hence an ideal of F [x] by Proposition 8.9. Since F [x] is a PID, it is a principal ideal, i.e. thereexists some p 2 F [x] such that I(↵) = (p). In fact, from the proof of Theorem 8.12, we know thatp is the polynomial in I(↵) of the lowest degree. It follows p must be irreducible. (Why?) Then itfollows F [x]/I(↵) is a field, which is isomorphism to the image of the evaluation map.

Definition 10.14. Assume E is a field extension of F and ↵ 2 E is an algebraic element over F . Callsuch p

↵

the irreducible polynomial for ↵ over F . Its degree is called the degree of ↵ over F . Denoteit by deg(↵;F ).

Obvious, deg(↵, F ) is well-defined. (Why?)

Example 10.15. (1) Consider C as a field extension of R. Then i 2 C is an algebraic elementover R. We have deg(i;R) = 2.p

2 2 C is another algebraic element over R. We have deg(

p2;R) = 1.

(2) Consider C as a field extension of Q. Thenp2 2 C is an algebraic element over Q. We have

deg(

p2;Q) = 2.

Notice, if ↵ 2 F , then I(↵) = (p↵

) = (x� ↵), and so deg(↵;F ) = 1.Let’s summarize and state the above discussions into the above theorem.

Theorem 10.16. Assume E is a field extension of F . Then for any ↵ 2 E which is an algebraicelement over F ,

(1) There is an irreducible p↵

2 F [x] such that I(↵) = (p↵

);(2) F (↵) := F [x]/(p

↵

) is a subfield of E; Moreover, elements in F (↵) can be identified withelements in F<deg(↵;F )

[x] as in Section 8.2;(3) ↵ 2 F , if and only if deg(↵;F ) = 1, if and only if F (↵) = F .

Definition 10.17. In general, assume E is a field extension of F . Call it is a simple extension if thereis some ↵ 2 E such that E = F (↵).

We now can reinterpret Kronecker’s theorem as

42 RUI WANG

Theorem 10.18 (Kronecker’s theorem). Assume F is a filed. Any irreducible polynomial p 2 F [x]

leads to a simple extension of F , which is

E = F [x]/(p).

To be more concrete, E = F [x]/(p) = F ([x]).

Example 10.19. Notice that x2+ x+1 is irreducible over Z2[x]. (Why?) Use it, we can create a new

field as a field extension of Z2, which is Z2[x]/(x2+x+1). It has a zero ↵ := [x] 2 Z2[x]/(x

2+x+1).

Hence it is a simple extension by ↵, i.e.

Z2(↵) = Z2[x]/(x2+ x+ 1).

We also Z2(↵) is the same as the set Z<2[x], which contains four elements

0, 1, x, 1 + x.

By this way, we obtain a field of four elements.

Lecture 23 stopped here.We end this section by stating the following theorem. We will come back to its proof after we

explain linear spaces in a systematically way.

Theorem 10.20. Assume E is a field extension of F and ↵ 2 E is an algebraic element over F .Assume deg(↵;F ) = n. Then F (↵) is an n-dimensional vector space over F with basis

{[1], [x], [x2], · · · [xn�1

]}.

Moreover, every element � in F (↵) is algebraic over R with degree deg(�;F ) deg(↵;F ).

11. VECTOR SPACES

Since we are familiar with vector spaces over R and C, here what we do is only to generalize thefields to general F . However, we will see, a lot of interesting application will come out.

11.1. Definition of vector spaces.

Definition 11.1. Assume F is a field. A vector space over F (or a F -vector space), is an abeliangroup (V,+) with scalar multiplication

F ⇥ V ! V, (a, v) 7! av,

satisfying, for a, b 2 F , u, v 2 V

(1) 1v = v;(2) a(bv) = (ab)v;(3) (a+ b)v = av + bv;(4) a(u+ v) = au+ av.

Here 1 is the unity of the field F .Elements in V are called vectors and elements in F are called scalers.

This definition implies

Lemma 11.2. If V is a vector space over F , then

MATH 120B 43

(1) 0v = 0 for any v 2 V ;(2) a0 = 0 for any a 2 F ;(3) (�a)v = �(av) = a(�v) for any a 2 F , v 2 V ;

Example 11.3. (1) Assume F is a field, then V := F n, i.e. direct product of n-copies of F is avector space over F . In particular, for F = R or C, these are the examples we are familiarwith.

(2) Assume F is a field, then F [x] is a vector space over F . (However, this is essentially differentfrom F n since it is infinitely dimensional. )

(3) If E is a field extension of F , then it is a vector space over F . (We are interested in a case offinitely dimensional, which is the simple extension. )

11.2. Linear independence and basis.

Definition 11.4. Assume V is a vector space over F . Call a subset S = {vi

|i 2 I} of V spans V , ifany v 2 V , can find finitely many v

i

2 S such that

v = a1v1 + · · ·+ an

vn

,

for some ai

2 F .A vector v 2 V of the form

v = a1v1 + · · ·+ an

vn

,

is called a linear combination of {v1, · · · , vn}.Denote by span

F

S the set of all linear combinations of elements in S.

Lemma 11.5. For any S ⇢ V , spanF

S is a vector space F .

Of course, a set which can span V always exists (e.g. V itself).

Definition 11.6. The minimal number of numbers of such sets is called the dimension of V , i.e.,

dim

F

V := min

spanFS=V

#S.

Call V a finitely dimensional vector space over F , if dim

F

V < 1. Otherwise, V is called aninfinitely dimensional vector space over F .

Lemma 11.7. V is of finite dimension over F if and only if there exists a finite set S ⇢ V such thatV = span

F

S.

Example 11.8. (1) F n is finitely dimensional since {ei

|i = 1, · · · , n} where

ei

= (0, · · · , 0, 1, 0, · · · , 0), (the i-th component is 1).

(2) F [x] is infinitely dimensional. (Prove it.)(3) Assume E is a field extension of F and ↵ 2 E is an algebraic element over F . F (↵) then is

of finitely dimensional. (See the last theorem in this section for its proof.)

Definition 11.9. Finite vectors are called linearly independent, if 0 has a unique way as a linearlycombination of these vectors.

44 RUI WANG

Proposition 11.10. Any finite set S ⇢ V has a linearly independent subset B ⇢ S such that

spanF

B = spanF

S.

Moreover, the number of elements in B is uniquely determined by S.

Corollary 11.11. Any finitely dimensional vector space V over F has a finite set of linearly indepen-dent vectors B such that V = span

F

B. Such B is called a basis of V . Moreover, #B = dim

F

V .

The most important example we want to use later is

Theorem 11.12. Assume E is a field extension of F and ↵ 2 E is an algebraic element over F . Then

(1) F (↵) is a vector space over F of dimension n = deg(↵;F ), with a basis

B = {1,↵, · · · ,↵n�1}.

(2) Every element � 2 F (↵), we have deg(�;F ) deg(↵;F ).

Proof. (1) Straightforward checking shows that B is a basis.(2) Consider the set

{1, �, · · · �n}.Since we have seen dim

F

F (↵) = n, these n + 1 elements must be linearly dependent. Inanother word, we can find b0, · · · , bn not all zero such that

b0 + b1� + · · · bn

�n

= 0.

Considerf = b0 + b1x+ · · · b

n

xn 2 F [x],

whose deg(f) = n. Notice f 6= 0 and � 2 Z(f), so � is algebraic and deg(�;F ) n.⇤


12. ALGEBRAIC EXTENSION

12.1. Finite extensions.

Definition 12.1. A field extension E of field F is called an algebraic extension, if every element in E

is algebraic over F .

Definition 12.2. If E is a field extension of F and dim

F

E = n, then we call E is a finite extensionof degree n over F . Denote by [E : F ] := n.

Example 12.3. [C : R] = 2.

Lemma 12.4. A field extension E of F is the same as F if and only if [E : F ] = 1.

Proof. (1) It is obvious that E = F indicates [E : F ] = 1.(2) Conversely, if [E : F ] = 1, then we can take 0 6= ↵ 2 E such that E = span

F

{↵}. Since1 2 E, can write 1 = k↵ for some k 2 F . Then ↵ = k�1

1, and so

E = spanF

{↵} ⇢ spanF

{1} = F.

⇤

MATH 120B 45

Theorem 12.5. Every finite extension is an algebraic extension.

Proof. Assume E is a finite extension of F and take n = [E : F ]. Let’s show that E is an algebraicextension by showing that any ↵ 2 E is algebraic over F .

Consider (n+ 1)-elements1,↵, · · · ,↵n,

then there exist c0, · · · , cn 2 F which are not all zero such that

c0 + c1↵ + · · ·+ cn

↵n

= 0,

i.e., ↵ 2 Z(f) where0 6= f(x) = c0 + c1x+ · · ·+ c

n

xn 2 F [x].

⇤

Theorem 12.6. Assume E is a finite extension of F and K is a finite extension of E. Then K is afinite extension of F and

[K : F ] = [K : E][E : F ].

Proof. Assume basis of E over F is{e1, · · · en}

and basis of K over E is{f1, · · · fm}.

We show B := {ei

fj

|i = 1, · · · , n; j = 1, · · · ,m} are basis of K over F .

(1) Show spanF

B = K;(2) Show B is linearly independent.

⇤

Corollary 12.7. Assume a sequence of finite extensions

F1 ⇢ F2 · · · ⇢ Fk

.

Then [Fk

: F1] = [Fk

: Fk�1][Fk�1 : Fk�2] · · · [F2 : F1].

Corollary 12.8. Assume E is a field extension of F and ↵ 2 E is algebraic over F . Then for any� 2 F (↵), we have F (↵) is a finite extension of F (�), and deg(�;F ) divides deg(↵;F ).

Example 12.9. There is no element in Q(

p2) that is a zero of x3 � 2. A reason is, if there is such

an element, say ↵ 2 Q(

p2), notice x3 � 2 is irreducible, this element has degree deg(↵;Q) = 3.

However, we know deg(↵;Q) must divide 2 = deg(

p2;Q). This is not possible.

Now we extend the notation of simple extension F (↵) to more general one, F (↵1, · · · ,↵k

).

Definition 12.10. Assume E is a field extension of F and ↵1, · · · ,↵k

2 E. Denote by F (↵1, · · · ,↵k

)

the minimal field that containing ↵1, · · · ,↵k

and call it the field from F by adjoining to F elements↵1, · · · ,↵k

2 E.

This definition is consistent with our previous definition in the following sense.

Lemma 12.11. (1) F (↵) defined in this way coincides with the simple extension defined above.

46 RUI WANG

(2) As fields, (F (↵1))(↵2)⇠=

(F (↵2))(↵1)⇠=

F (↵1,↵2).

Example 12.12. [Q(

p2,p3) : Q] = 4 and a basis is {1,

p2,p3,p6}.

Proof. Using Theorem 12.6, we know

[Q(

p2,p3) : Q] = [Q(

p2,p3) : Q(

p2)][Q(

p2) : Q].

We know [Q(

p2) : Q] = 2 since x2 � 2 2 Q[x] is irreducible. Now we calculate [Q(

p2,p3) :

Q(

p2)].

For this, using Lemma 12.11, consider Q(

p2,p3)

⇠=

(Q(

p2))(

p3). Notice

p3 /2 Q(

p2), (why?

– we prove this at the end of this proof.) so

[Q(

p2,p3) : Q(

p2)] � 2.

On the other hand, since Q ⇢ Q(

p2), we have deg(

p3;Q(

p2)) deg(

p3;Q) = 2.

Hence we get [Q(

p2,p3) : Q(

p2)] = 2 and so [Q(

p2,p3) : Q] = 4.

A basis can be got by multiplying the basis of Q(

p2) over Q as {1,

p2} and the basis of (Q(

p2))(

p3)

over Q(

p2) as {1,

p3}, and we get

1,p2,p3,p6.

(The proof of Theorem 12.6 ensures that this way always gives basis. )

Now we provep3 /2 Q(

p2). Assume it is not true, then together with

p2 2 Q(

p2), we have

p3 +

p2 2 Q(

p2).

Notice deg(

p3 +

p2;Q) = 4, which is not possible since otherwise we will have deg(

p3 +

p2;Q)

divides deg(p2;Q) = 2. So this leads to contradiction.

⇤

Example 12.13. [Q(2

12 , 2

13) : Q] = 6 and moreover, Q(2

12 , 2

13)

⇠=

Q(2

16).

Proof. This problem can be done in the similar way as previous example. First, use Theorem 12.6,we can write

[Q(2

12 , 2

13) : Q] = [Q(2

12 , 2

13) : Q(2

13)][Q(2

13) : Q].

Notice x3 � 2 2 Q[x] is irreducible, so [Q(2

13) : Q] = 3.

Next, we calculate [Q(2

12 , 2

13) : Q(2

13)]. Using Lemma 12.11, consider

Q(2

12 , 2

13)

⇠=

(Q(2

13))(2

12).

First, notice that 212 /2 Q(2

13)) for dimension reason. So it follows

[Q(2

12 , 2

13) : Q(2

13)] � 2.

On the other hand, since Q ⇢ 2

13), we have deg(2

12;Q(2

13)) deg(2

12;Q) = 2. So, [Q(2

12 , 2

13) :

Q(2

13)] = 2 and it follows [Q(2

12 , 2

13) : Q] = 6.

A basis can be got by multiplying the basis of Q(2

13) over Q as {1, 2 1

3 , 223} and the basis of

(Q(2

12 , 2

13) over Q(2

13) as {1, 2 1

2}, and we get

1, 213 , 2

12 , 2

23 , 2

56 , 2

76 .

MATH 120B 47

Notice we can find another basis by replacing 2

76 by 2

16 and this new basis is exactly the same as

basis of Q(2

16). As a result, Q(2

12 , 2

13)

⇠=

Q(2

16). ⇤

Theorem 12.14. Assume E is an algebraic extension of F . Then the followings are equivalent:

(1) E is a finite extension;(2) There are finite elements ↵1, · · · ,↵k

2 E such that E = F (↵1, · · · ,↵k

).

Proof. (1) Assume E is a finite extension, which means [E : F ] = n is a finite number. We haveshown that, if n = 1, then E = F = F (1) and we are done. Assume we can prove every casefor n = k. Now let’s prove for n = k + 1.

Assume F ⇢ F (↵1, · · · ,↵k

) ⇢ E.(2) Assume E = F (↵1, · · · ,↵k

) is an algebraic extension, then every ↵i

is algebraic over F , so[F (↵1) : F ] is finite. It follows ↵2 is algebraic over F (↵1), and so [F (↵1,↵2) : F (↵1)] isfinite.

Keep doing this way, we get F (↵1, · · · ,↵k

) is finite extension over F (↵1, · · · ,↵k�1).Using Corollary 12.7, [F (↵1, · · · ,↵k

) : F ] is finite.⇤


12.2. Algebraically closed field and algebraic closures.

Theorem 12.15. Assume E is a field extension of F . The the set

¯FE

:= {↵ 2 E|↵ is algebraic over F}

is a subfield of E. This is called the algebraic closure of F in E.

Proof. Take ↵, � 2 ¯FE

, then Theorem 12.14 tells us that F (↵, �) is a finite extension of F . ByTheorem 12.5 , F (↵, �) is algebra. It follows then

F (↵, �) ⇢ ¯FE

.

As a result, ↵± �,↵� 2 ¯FE

and ↵��1 2 ¯FE

if � 6= 0. It shows that ¯FE

is a field. ⇤

Example 12.16. Consider C as a field extension of Q. The subfield ¯QC is called the field of algebraicnumbers.

Definition 12.17. A field F is called algebraically closed, if every nonconstant polynomial f 2 F [x]

has a zero in F .

Theorem 12.18. Assume F is algebraically closed, then any algebraic extension E of F is the sameas F .

Proof. Assume E is an algebraic extension of F . Take any ↵ 2 E, since F is algebraically closed,using Proposition 8.20, deg(↵;F ) = 1. From Lemma 12.4, this means F (↵) = F and so ↵ 2 F . ⇤

Theorem 12.19 (Fundamental theorem of algebra). The field C is algebraically closed.

Proof. Assume f 2 C[z] and deg(f) � 1. We show it has a zero in C. Otherwise, for any z 2 C,f(z) 6= 0. There are several different ways to derive contradiction, and we here give two proofs.

48 RUI WANG

• (A proof using complex analysis). Consider g(z) := 1/f(z) which is a globally define ana-lytic function now. Moreover,

|g(z)| = 1

|f(z)|

bounded. By Liouville theorem, such analytic function must be constant and we get contra-diction.

• (A proof using topology). Construct g(z) := f(z)|f(z)| , then we get a continuous map from unit

disk to its boundary. Now we show it is not possible unless f is constant.The idea is, if f is not constant, then g(z) must turn around origin at least once. (This

statement needs some proof but it is not hard to see from intuition.) Then such way must tearoff the disk, so can not be continuous and we get contradiction. The precise language shouldbe, it must map the trivial fundamental group to some nontrivial one, however, since such mapis a group homomorphism, so not possible.

⇤

We state the following important theorem without proof.

Theorem 12.20. Every field F has an algebraic closed extension.


12.3. Summary. Assume E is a field extension of F . We have constructed an algebraic extension¯FE

of F such that

F ⇢ ¯FE

⇢ E.

Any finite elements ↵1, · · · ,↵k

2 ¯FE

, we have a finite extension F (↵1, · · · ,↵k

) and

F ⇢ F (↵1, · · · ,↵k

) ⇢ ¯FE

⇢ E.

In particular, if F is algebraically closed, then

F =

¯FE

⇢ E.

In fact, we can prove:

Proposition 12.21. F is algebraically closed, if and only if every extension E of F , we have F =

¯FE

.

Proof. We only need to prove if every extension E of F , we have F =

¯FE

, then F is algebraic closed.Otherwise, there exist an irreducible p 2 F [x] of degree � 2. Consider E := F [x]/(p). This is a

field as extension of F .Notice ¯F

E

6= F since [1], [x] 2 E are linearly independent over F , (The reason is, deg(p) � 2.– why?) and this indicates dim

F

¯FE

� 2. Hence it follows from Lemma 12.4 that ¯EE

6= F . Thiscontradicts with our assumption.

⇤

MATH 120B 49

13. FINAL PROJECT

• Due 6/8 10:30 before the final.• For extra 35 points to total (350 points).• Provide proofs with details to gain credits.• References, group discussions are Ok and recommended.

Assume � ⇢ R is an abelian subgroup of R. Denote by Nov(�) the set of all formal series withformal parameter t in the form

x(t) = ⌃

�2�a�t�, a

�

2 Zsatisfying the following condition

{� 2 �|a�

6= 0, � > C} is finite, for any C 2 R.

On Nov(�), define the addition +

N

and multiplication ·N

as

(⌃

�2�a�t�

) +

N

(⌃

�2�b�t�

) := ⌃

�2�(a� + b�

)t�

(⌃

�2�a�t�

) ·N

(⌃

�2�b�t�

) := ⌃

�2�(⌃�1+�2=�

a�1b�2)t

�.

(1) Check Nov(�) is closed under +N

and ·N

.(2) Prove (Nov(�),+

N

, ·N

) is an integral domain. This ring is called the Novikov ring.(3) Describe units in Nov(�).(4) Prove Nov(�) is a PID.

Remark 13.1. This ring is called the Novikov ring which was introduced by S. P. Novikov around1980s in his work on Morse theory.

On one side, it is an important ring in modern mathematics, in particular in the field of geometryand mathematical physics. On the other side, it is a good exercise for you to really understand basicconcepts in ring theory you have learnt from this course.

lecture notes of abstract algebra iiruiw10/pdf/alg2.pdf · 2017-07-31 · 7.1. deﬁnition of...

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