M3P8 Algebra III

July 13, 2011

Life gives you Lemons.

- Professor Kevin Buzzard

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Chapter 1

Basics Properties of Rings

Definition 1.1. A ring (with 1) is a set R along with elements 0, 1 R andmaps + : RR R, : RR R (write a+ b for addition and abbreviatea b by ab) such that

(1) (R,+) is an abelian group with 0 as identity

(2) (R,) is a semigroup with identity 1 (i.e. a, b R, (ab)c = a(bc))(3) a, b, c, a(b+ c) = ab+ ac and (a+ b)c = ac+ bc.

Example.

- Z,Q,R,C with the obvious choices for 0, 1,+, are rings- R[x] (the polynomials with real coefficients) is a ring

Observe that R is a group under +. So a R, inverse b R such thata + b = 0. We will call this inverse a: (a) + a = 0. Further, we definex y = x+ (y) as subtraction. In general we cannot do division in rings!Note. Some people do not demand 1 R and do not demand 1r = r = r1. Forthese people, I define a ring with 1. Other people demand 0 6= 1 as an extraaxiom. This barely makes any difference as 0 = 1 = R = {0}.Example. Mn(R) along with usual +, and 0 = 0n is a ring. The identityelement is 1 = In (Note that (AB)C = A(BC) but AB 6= BA in general.Definition 1.2. A ring R is commutative if ab = ba a, b R.Example. Residue Class Rings. Take m 1 to be an integer and define anequivalence relation on Z by

a b m divides a b a b mod m.

Let Z/mZ denote the set of equivalence classes:

Z/mZ = {0, 1, . . . ,m 1} = {[0]m , [1]m, . . . , [m 1]m}.

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It turns out that Z/mZ is a ring under the operations[a] + [b] = [a+ b] [a] [b] = [ab]

E.g. Z/5Z is a ring.

Note. Our intuition is bases on rings like Z,Q, . . . which are all well-behavedrings. In general rings are not so well-behaved!

Example.

- Let R = M2(R) and a = b =(

0 10 0

). Then a 6= 0, b 6= 0, but ab = 0.

- Let R = Z/6Z,= [2], b = [3]. Then a 6= 0, b 6= 0, but ab = [0].- Polynomials can have too many roots, e.g. consider x2 1 in the ringZ/8Z: x = [1], [3], [5], [7] are all roots of the polynomial.

- Cancellation can also fail, i.e. ra = rb ; a = b. E.g. let r = [2], a =[2], b = [0] in Z/4Z.

All this happens because we cannot divide.

Definition 1.3. A ring R is called a division ring if R/{0} is a group undermultiplication with 1 as the identity (1 6= 0). In other words, a 6= 0,b suchthat ab = ba = 1.

Example.

- Z is not a division ring (as 2 does not have an inverse under )- Z,Q,R are all division rings under the usual operations of addition andmultiplication.

A commutative division ring is called a field. For example, Z,Q,R are fields.

Fields are great: the whole theory of vector spaces works over a general field.

Example. The ring of polynomials in R, R[x], is not a field, as 1x is not apolynomial.

Definition 1.4. If R is a ring and S R is a subset, we say S is a subring if0, 1 S and t + s, st S s, t S and furthermore if S becomes a ring itselfwith this 0, 1,+,, i.e. S satisfies the axioms.Lemma 1.5. If R is a ring and S R is a subset of R such that 0, 1 S ands+ t, st, s t S s, t S then S is a subring of R.

Proof. We need to check the three axioms: (S,+) is a group, because s, t S = s t S. So inverses exist. Also it is obviously abelian and the otheraxioms are obvious, e.g. say r, s S. Need r(s + t) = rs + rt : but this is truein R, so it must be true in S.

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Example.

- Z is a subring of Q

- Q is a subring of R

- R is a subring of C.

- Let d Z be an integer that is not a square. Define Z[d] to be thefollowing subset of C :

Z[d] = {a+ b

d|a, b Z}.

Lemma 1.6. Z[d] is a ring.

Proof. It is a subset of C so lets use Lemma 1.5: We need to check that ifa + b

d = r Z[d] and a + bd = r Z[d], then r r, rr Z[d].

r r Z[d] is easy to check and

rr = (a+ bd)(a + b

d) = (aa + bbd

Z) + (ab + ba

Z)d.

A slightly less obvious fact about Z[d]: if a + d

d = a + d

d, then a =

a, b = b (where a, a, b, b Z). For a+ dd = a + dd,

a a =bd d

d

= (b b)d.

If b 6= b, then d = aabb Q, butd 6 Q. Hence b = b = a a = 0 =

a = a.

Example. Q[d] = {a + bd|a, b Q} (for d 6 Z) is a ring (by the same

proof as before: subring of C).

Lemma 1.7. Q[d] is a field.

Proof. Q[d] is clearly commutative, so all I need to do is to check that if

0 6= r Q[d], then 1r Q[d]. So assume 0 6= r = a + bd Q[d]. Then

a2 b2d 6= 0 for if a2 b2d = 0, then a2 = b2d and either b = 0 or d = (ab )2 .But d is not square by assumption, hence

b = 0 = a2 = 0 = a = 0 = r = 0

contradiction. So Q 3 t = a2 b2d 6= 0 and from above we see that 1r =at bt

d.

Example. The set of all functions f : [0, 1] R is naturally a ring. the roleof 0 is played by the function sending any x [0, 1] to 0 and 1 is the functionsending any x [0, 1] to 1. Define f + g by (f + g)(x) = f(x) + g(x) and fg by(fg)(x) = f(x)g(x).

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Exercise. This defines a ring.

Proposition 1.8. Let R be a ring and say r, s, ri, si R.

(a) r0 = 0r = 0 r, s R(b) if r denotes the inverse of r under addition, then

(r)s = r(s) = (rs) (r)(s) = rs r, s R

(c) (ni=1 ri)

(nj=1 sj

)=ni,j=1 risj

(d) if r R and rs = s s R, then r = 1.(e) if R is a ring and 0 = 1 in R then R = {0} has one element (conversely,{0} is a ring).

Proof.

(a) standard exercise in group theory: 0 + 0 = 0, hencer(0 + 0) = r(0) = r0 + r0 = r0 = r0 = 0. Similarly for 0r = 0.

(b) to check (r)s = (rs) is need to check that (r)s+ rs = 0. Then bydistributivity, it suffices to prove that(r + r)s = 0. But r + r = 0 and0s = 0 by (a). Hence r(s) = rs. Now

(r)(s) = (r(s)) = (rs) = rs

since R is an additive group.

(c) tedious induction on m+ n using distributivity.

(d) set s = 1.

(e) if r R, then r = r1 = r0 = 0 by (a). Conversely, check that {0} satisfiesall the axioms.

Convention: By definition 0, 1 R and define 2 R to be 1 + 1. Similarly for3, 4, ..., 73, .... Further, define 1 R to be the additive inverse of 1 such that1 + (1) = 0 and similarly n to be the additive inverse of n. We obtain a mapZ R which may or may not be an injection, e.g. 73 = 0 in Z/73Z and 73 = 1in Z/72Z.

Definition 1.9. If R is a ring and if 0 6= a R, then we say a is a left-divisorof zero if b 6= 0 in R such that ab = 0 (similarly for right-divisor of zero. Notethat if R is commutative, these notions coincide and we say that a is a zerodivisor). If a R and b R such that ab = ba = 1, then we say a is a unit(for R commutative, we only need ab = 1 for a to be a unit). Write R for theset of units in R.

Remark. R is a group, as associativity and identity are ring axioms and inversesexists by definition of a unit.

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Example.

- 2 is a zero divisor in Z/6Z as 2 3 = 0 in this ring but 2 6= 0, 3 6= 0.- 5 is a unit in Z/6Z since 5 5 = 1 in this ring.

- The matrix(

0 10 0

)is a left and ring zero divisor in M2(R) as

(0 10 0

)2= 0

(note that if a R and an = 0 for some n 1 then a is nilpotent).- the units in M2(R) are the invertible matrices (i.e. GL2(R)) .

- R = Z has no zero divisors as ab = 0 = a = 0 or b = 0.- The units in Z are Z = {1}.- If R is a field (or even a divison ring), then R = R \ {0}.

Definition 1.10. A ring R is an integral domain if

(1) R is commutative

(2) 0 6= 1(3) R has no zero divisors (i.e. if ab = 0, then a = 0 or b = 0).

Example.

- Z is an integral domain

- any field is an integral domain

- the zero ring {0} is not an integral domain (which is a wise convention).- any subring of a an integral domain is again an integral domain = anysubring of C, e.g. Z[

d],Q[

d] etc., is an integral domain.

Lemma 1.11. Let m be a positive integer and let R be the ring Z/mZ. ThenR is an integral domain iff m is prime.

Proof. Note first that if m = 1 is not prime, then Z/1Z = {0} is not an integraldomain. Ifm = p is prime, then I need to check that Z/pZ is an integral domain:clearly, we have that Z/pZ is commutative and 0 6= 1. Now say that a, b 6= 0 inZ/pZ : lift a to A Z and b to B Z. Since a, b 6= 0 we have that p - A, p - B,hence p - AB. So AB 6= 0 mod p = ab 6= 0. So Z/pZ is an integral domainif p is prime. To show the converse, assume that m is not prime, i.e. m = abwith 1 < a < b < m. Then a, b 6= 0 and ab = 0 in Z/mZ. So Z/mZ is not anintegral domain.

Lemma 1.12.

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(i) if R is a ring and a R with ar = as, then r = s(ii) if R is an integral domain and if a 6= 0 and ar = as, then r = s.

Proof. For the first part, choose b R such that ba = 1. Then

ar = as = bar = bas = 1r = 1s = r = s.

For the second part, let a 6= 0 and ar = as. Then a(r s) = 0. But R is anintegral domain and a 6= 0. So

r s = 0 = r = s.

Note. (ii) is not a special case of (i), for example 2 Z is non-zero but not aunit.

Abstract Polynomial Rings

Let R be any commutative ring. Define the polynomial ring R[x] of polyno-mials to be, formally, the set of all infinite sequences (c0, c1, . . . , cn, . . .) withci R i but all but finitely many ci equal to zero. Informally, we thinkof (c0, c1, . . . , cn, 0, 0 . . .) as being c0 + c1x + . . . + xn. Define 0 = (0, 0, . . .),1 = (1, 0, 0, . . .) and

(a0, a1, . . .) + (b0, b1, . . .) = (a0 + b0, a1 + b1, . . .)

and(a0, a1, . . .)(b0, b1, . . .) = (c0, c1, . . .)

with cn =ni=0 aibni.

Exercise. This defines a ring.

Notation. Call this ring R[x] and write f = (a0, a1, . . . , ad, 0, 0, . . .) = a0 +a1x+. . .+ adx

d. If ad 6= 0 then we say that d is the degree of f if an = 0 n > d.Proposition 1.13. If R is an integral domain then R[x] is also an integraldomain.

Proof. Say 0 6= a, b R[x] . Write

a = a0 + a1x+ . . .+ adxd

b = b0 + b1x+ . . .+ bexe

with ad, be 6= 0. Then ab = STUFF+adbexd+e. Now, as R is an integral domain,adbe 6= 0 = ab 6= 0.

If R is a commutative ring, define R[x1, x2] = (R[x1])[x2].

Corollary 1.14. If R is an integral domain, then so is R[x1, . . . , xn].

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Proof. Do induction on n.

We say that a subfield of a ring R is a subring S R that is a a field. Forexample, R is a subring of R[x] (the constant polynomials) but also a subfieldof R[x].Remark. If K is a subfield of the ring R, then R is naturally a vector space overK. For example, C is a vector space over R.

Lemma 1.15. A finite integral domain is a field.

Proof. Say that 0 6= a R with R being a finite integral domain. We need tofind an inverse for a, i.e. b such that ab = 1. Consider the map ma : R Rgiven by ma(r) = ar for r R. ma is injective, for if ma(r) = ma(s), thenar = as = r = s by 1.12(ii). Hence ma is injective. Also ma is surjectivesince R is finite. Hence it is a bijetion, so we can choose b such that ma(b) =ab = 1.

Trickier: a finite division ring is a field. This is known as the Artin-WedderburnTheorem.

Corollary 1.16. The ring Z/mZ is a field iff m is prime.

Proof. By 1.11 and 1.15 and the fact that a field is an integral domain.

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Chapter 2

Homomorphisms, Ideals &Quotient Rings

Definition 2.1. Let R and S be rings. A map : R S is a ring homomor-phism if

(i) (0) = 0, (1) = 1

(ii) (a+ b) = (a) + (b)

(ii) (a b) = (a) (b)

Remark. From Group Theory, we know (x) = (x) and so (x y) =(x) +(y)Definition 2.2. : R S is an isomorphism if a ring homomorphism : S R such that and are the identity map.In practice, a ring homomorphism : R S is an isomorphism iff is abijection.

Special case: R = S. A ring homomorphism : R R is called an endomor-phism and an isomorphism : R R is called an automorphism.Example.

1. Z ZnZ (where n is a positive integer) given by t t mod n is a ho-momorphism: [0] is the zero in ZnZ , [1] is the one in

ZnZ and if a, b Z,

then(ab) mod n = (amod n)(bmod n)

(this is the definition of the product in ZnZ ). And similarly

(a+ b) mod n = (amod n) + (bmod n)

(this is the defintion of addition in ZnZ ). So this is a ring homomorphism.

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2. R = C with f : R R given by f(z) = f(z) is a ring homomorphism:0 = 0, 1 = 1 and

f(a+ b) = a+ b = a+ b = f(a) + f(b)

f(ab) = ab = ab = f(a)f(b).

Notice that since f is bijective, this is in fact an isomorphism (and anautomorphism). f is its own inverse, i.e. f f = identity.

3. R = R[x], S = R. Choose some R. Define : R[x] R by (f) =f(), where f(x) R[x] (e.g. if = 2 and f = x2 +1, then (f) = f(2) =5). is called the evaluation homomorphism. Note that the polynomial 1is not the same as the polynomial x: (x) = , but (1) = 1. is easilychecked to be a ring homomorphism.

4. The Frobenius Homomorphism: Say R is a commutative ring and sayp = 0 in R (e.g. R = ZpZ or

ZpZ[x] ). Define : R R by

(x) = xp = (x x x . . . x)p times

(or (r) = rp). This is also a ring homomorphism as (0) = 0, (1) = 1and

(rs) = (rs)p = rpsp = (r)(s)

(r + s) = (a+ b)p = ap + pap1b+ +(pi

)apibi + + bp.

But p |(pi

)if 1 i p 1. Therefore

(pi

)= 0 in R and hence

(a+ b)p = ap + bp = (a) + (b).

5. R = Q(

2) = {a+ b2 : a, b Q}. Define : R R by

(a+ b

2) = a b

2.

is a ring homomorphism because:- (0) = 0

- (1) = 1

- ((a+b

2)(c+d

2)) = (ac+2bd+

2(bc+ad)) = ac+2bd2(bc+ad)

(a+b

2)(c+d

2) = (ab

2)(cd

2) = ac+2bd(bc+ad)

2 = (xy) = (x)(y).6. Inclusions: Q R,R C,M2(R) M2(C) are all ring homomor-

phisms.

Remark. Injective ring homomorphisms R S are the same as subrings of S.Lemma 2.3. Let : R S be a ring homomorphism and let T be the imageof , i.e. T = {(r)|r R}. Then T is a subring of S.

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Proof. By Lemma 1.5, we need to check that T contains 0,1 and that T is closedunder +,,. Clearly (0) = 0, (1) = 1, so 0, 1 T. Now say a, b T . Leta = (r) and b = (s). Then

a+ b = (r) + (s) = (r + s)

a b = (r) (s) = (r s)ab = (r)(s) = (rs).

Hence T is a subring of S.

In fact, any map : X Y between sets factors as X pi Z i Y with pi asurjection and i an injection (Z Y is image of ). The above Lemma 2.3.shows that the same is true for rings: any ring homomorphism : R S isR

pi T i S, pi a surjection, i an injection and pi, i are ring homomorphisms.We have already seen an example of a surjective ring homomorphism: : ZZ/nZ, n 1.Question: Are there any more surjections : Z R, where R is ring of acompletely different type to Z/nZ?Answer: We will answer this soon.Here is a problem that we need to solve first: Say : R S is a ring homomor-phism. We have seen that Im() is a subring of S. Is it also true that ker() ={r R : (r) = 0} is a subring of R (for example the kernel of Z Z/mZ isthe set of integers which are multiples of m = {. . . ,m, 0,m, . . . })? No, as thisis not a subring of Z in general (for example it is very likely that 1 / ker()).Definition 2.4. A subset I R (where R is a ring) is called a left ideal if

(1) I is a subgroup of R (under +)

(2) If r R and i I, then ri I.Similarily for right ideals. A subset I R (R a ring) is called a bi-ideal, or a2-sided ideal, if I is a left and right ideal.Remark. If R is a commutative ring, then all three of these notions coincide,and we will call I an ideal, i.e. if

(1) I is a subgroup of R under +

(2) ri I for r R, i I.Notation. If I is an ideal of R, we write I E R or I / R.Example.

1. If R is a ring, then {0} and R are both bi-ideals of R.2. Let R = R[x] be the set of all polynomials with real coefficients. Let I =xR[x] be the polynomials with no constant term. If f = a1x + +, g =b1x + + I, then f g = (a1 b1)x + . . . has no constant term andso I. Also, 0 I, therefore I is a subgroup of (R,+). Next, we needto check that if f I and g R, then fg I (i.e. R is a commutativering). f = a1x + a2x2 + . . . , g = b0 + b1x + . . . (b0 6= 0 is okay), thenfg = a1b0x+O(x

2). Therefore, fg I, so I E R, i.e. I is an ideal of R.

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3. Say m 1. Set I = mZ = {mt : t Z} R = Z (m is an integer), I isthe set of multiples of m.

Remark. I is the kernel of the map from Z Z/mZ.Claim. I is an ideal. For 0 = 0m I. If a, b I, then a = ms, b = mtfor s, t Z. Therefore a b = m(s t) I, hence I is a subgroup of(Z,+). Finally, if r R = Z, and i I, then i = mu for some u Z andso ri = rmu = m(ru) I. Therefore, I is an ideal of R = Z (note that Ris a commutative ring).

Remark. If m = 0, then {mr : r Z} = {0} is also an ideal, and every-thing in the above proof works, giving another proof that {0} E Z.

4. All ideals of Z are of the form mZ = {mr : r Z} for m = 0, 1, 2, . . . .For, say I Z is an ideal, then if I contains a negative number n < 0,then (I is a subgroup) and n > 0 and n I. So either I = {0} or Icontains some positive integer. Let m be the smallest positive integer inI. Easy check

I = {. . . ,2m,m, 0,m, 2m, . . . } = mZ.For, certainly mZ I as m I and I is a group, and if t I, with t nota multiple of m, then n such that

mn < t < m(n+ 1)

and t mn I, but 0 < t mn < m, contradicting the definition of mbeing the smallest positive integer in I. So I = mZ.

Ideals are hardly every subrings. In fact:

Lemma 2.5. If I is a left-ideal of R, and 1 I, then I = R.Proof. Say r R. Then

I I = r1 I = r I,so I = R.

Remark. This lemma (2.5) also holds for right ideals.

If R is a non-commutative ring, then we can define a new ring Ropp by Ropp = R(as a set) and 0, 1 as before, with the same rule of addition, but a b in Roppis defined to be b a in R. So the left ideals of Ropp are the right ideals of R.Definition 2.6. If : R S is a ring homomorphism, define the kernel of to be

1 ({0}) = {r R : (r) = 0}.Proposition 2.7. The kernel of a ring homomorphism is an ideal.

Proof. is by definition a group homomorphism (with the group law + andidentity 0), so the kernel of is a subgroup by M2PM2. Now say that r Rand i ker, i.e. (i) = 0. We want to show that ri and ir are in ker. But

(ri) = (r)(i) = (r) 0 = 0 = ri kerby Proposition 1.8a. Similarly, ir ker. Hence ker is a bi-ideal.

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Next, we will define the quotient ring R/I, where R is a ring and I is a bi-idealof R. This is well-defined, for if I E R is a bi-ideal, then I is a subgroup of(R,+), so we can define the quotient group R/I per group theory. Recall thatan element of R/I is a subset r + I of R,

r + I = {r + i : r I}.

We will now aim to put a ring structure on R/I such that a natural mapR R/I is a ring homomorphism with kernel I.

Question: Is every bi-ideal the kernel of a homomorphism?Set up: Let R be a ring and I E R a bi-ideal of R. Our goal is to form thequotient ring R/I. So far we know that (R,+) is a group and I R is a normalsubgroup. Hence the quotient group R/I exists and has well-defined addition.Recall that the elements of R/I are I-cosets in R, i.e. subsets of R of the formr + I = {r + i : i I}.We will now make R/I a ring.Define 0 of R/I to be 0 + I = I.Define 1 of R/I to be 1 + I.Define + on R/I to be

(r + I) + (s+ I) = (r + s+ I) .

By group theory, we know that this is well-defined.For multiplication, define

(r + I)(s+ I) = rs+ I.

We need to check that this is well-defined. More precisely, that r = r+ i, i Iand s = s + j, j I. Then r + I = r + I and s + I = s + I. So we need tocheck that

rs+ I = rs + I,

i.e. that is rs = rs+ k, for some k I. Well,

rs = (r + i)(s+ j) = rs+ is+ rj + ij.

Set k = is+ rj + ij. We want to show that k I. Once we have checked that,we are done. But i, j I and r, s R, so is, rj I since I is a bi-ideal. Also,i, j I = ij I since I is also a left-ideal. So k K (as (I,+) is a group).So we have a well-defined product on R/I induced from the product on R. Wenow claim that R/I is a ring.

1. R/I is a group under addition, by group theory.

2.(1 + I)(r + I) = 1r + I = r + I = (r + I)(1 + I),

so 1 + I works as the multiplicative identity. Moreover,

((r + I)(s+ I)) (t+I) = (rs+ I) (t+I) = rst+I = (r+I) ((s+ I)(t+ I)) .

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Finally,

(x+ I) ((y + I) + (z + I)) = (x+ I)(z + y + I) = x(y + z) + I= xy + xz + I by distributivity in R= (xy + I)(xz + I)= (x+ I)(y + I) + (x+ I)(z + I)

and similarly for the other distributivity law.

Therefore, R/I is a ring.

Definition 2.8. Let R be a ring and I be a bi-ideal of R. We say that R/I isthe quotient ring.

Now, it is easy to check that the natural map R R/I given by r 7 r + Iis a ring homomorphism: the image of 0 is 0, the image of 1 is 1 and if r 7r + I, s 7 s + I, then r + s 7 r + s + I and rs 7 rs + I. It is just as easy toshow that the kernel of R R/I is

= {r : r + I = I} = {r : r I} = I.

The First Isomorphism Theorem strengthens this. It says a surjective ringhomomorphsim is determined by its kernel.

Theorem 2.9. (First Isomorphism Theorem). Say that : R S is a ho-momorphism of rings. Say I = ker. This is an ideal of R by 2.7. Further letT = Im. This is a subring of S by 2.3. Then there is a natural isomorphismof rings

R/I = T(and indeed induces this natural isomorphism).

Proof. I is a bi-ideal and R/I is a well-defined ring. Our plan will be to definea map : R/I T . Say r + I R/I. Define

(r + I) = (r).

Is this well-defined? Say r = r + i. Then r + I = r + I (this is if and only if).Therefore we need to check (r) = (r). But

(r) = (r) + (i) = (r) + 0 = (r)

as I = ker. Hence is well-defined. Now, for injectivity of , say

(r + I) = (s+ I) r, s R.

Then by definition of ,

(r) = (s) = (r s) = 0

since is a ring homomorphism. Therefore r s ker = I. So set r s = i.Then

r = s+ i = r + I = s+ I.

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Hence is injective. Next, surjectivity: say t T. We need to find r R suchthat (r + I) = t. Well, T = Im(), and so r R such that (r) = t. Then

(r + I) = (r) = t,

so is surjective. Combining, is bijective. Now, set : T R/I to be theinverse of . We leave it as an exercise to show that is a ring homomorphism.Then + and are the identities. Therefore, is an isomorphism.

We saw already that the image of a ring is a ring. However, it is not true thatthe image of an ideal is an ideal. For example, consider the map Z C givenby x 7 x. Then 2Z is an ideal of Z, but not of C.However, the pre-image of an ideal is also an ideal. That is

Proposition 2.10. Say f : R S is a homomorphism of rings. Say I S isa left ideal (resp. right ideal, resp bi-ideal). Then

f1(I) = {r R : f(r) I}

is a left ideal (resp. right ideal, resp. bi-ideal) of R.

Proof. Set J = f1(I). If , J, then f() I, f() I. Therefore, f( ) I. Hence J is closed under and 0 J and f(0) = 0. Therefore J is asubgroup of R (under addition). Now say r R and j J. We need to showthat rj J (resp. jr J, resp rj.jr J). But

f(rj) = f(r)S

f(j)I

= f(r)f(j) I

(since I is a left ideal). Therefore f(rj) I = rj J (for the case of a rightideal or a bi-ideal, the working is just the same).

Proposition 2.11. If R is a commutative ring, then R is a field R hasexactly two ideals (namely, {0} and R).

Say R is a commutative ring. What are all the ideals of R? We have seen thatif R = Z, then the ideals are {0} and nZ, n 6= 0. Another class of examples isgiven by Proposition 2.11. above.

Remark. R = {0}, the zero ring, is not a field, by definition.

Proof. ( = ) Easy. Firstly note {0} 6= R for a field, so there are at least 2 ideals.Now say that R is a field and I R is an ideal and I 6= 0. We want toshow that I = R. Choose 0 6= x I. As R is a field, there exists y Rsuch that yx = 1. By definition of an ideal, 1 = xy I. Now say thatr R is arbitrary, then r = r1 I.

(=) We have two ideals, R 6= {0} (as otherwise, R = {0} has only one ideal).We need to show that if 0 6= r R, then r has an inverse. So choose0 6= r R. Set I = {ar : a R}. It is easily shown that I is an ideal.Furthermore, r = 1r I, so I 6= 0, hence I = R, by assumption. Therefore1 R = a such that ar = 1. Hence r has an inverse. So R is a field.

15

Therefore the only ideals of C are 0 and C (and similarly for R and Q).

Definition 2.12. Let R be a commutative ring. An ideal P of R is said tobe prime, or a prime ideal, if P 6= R and if a, b R with ab P, then eithera P or b P . An ideal M R is maximal if M 6= R and if J is an ideal withM J R, then either J = R or M = J .Proposition 2.13. R a commutative ring and I R an ideal. Then R/I is afield if and only if I is maximal.

Proof.(=) Say I is a maximal ideal. We want to show that R/I is a field. By

definition, I 6= R, therefore R/I 6= {0}. Now, need to show check that anon-zero element of R/I has an inverse. So choose x+ I R/I with x+ Inot the zero element, i.e. x+ I 6= I, i.e. x / I. We need to invert x+ I inR/I. Define a subset J R thus:

J = {ax+ i : a R, i I}.

We claim that J is an ideal. We have that 0 J, since 0 R and 0 I.Further, if ax+ i1 and bx+ i2 are in J, a, b R, i1, i2 I, then

(ax+ i1) (bx+ i2) = (a b)R

x+ (i1 + i2)I

J,

therefore J is a group under addition. Finally if r R, a R, i I, then

rR

(ax+ iJ

) = (ra)R

x+ riI J.

Therefore J is an ideal.Now, clearly I J (simply set a = 0). So I J R. But I is maximal.Therefore J = I or J = R. But J 6= I, as x / I, but x J (a = 1, i = 0).So J = R. Therefore 1 J, and so we can write 1 = ax + i for somea R, i I.We now claim that a+ I is an inverse to x+ I. For

(a+ I)(x+ I) = (ax+ I) = 1 i+ I = 1 + I = 1 of R/I.

( = ) Want R/I a field = I is maximal. Firstly, R/I a field = I 6= R.Now say I J R and say J 6= I. We want J = R, then we are done. Solet us choose j J such that j / I. Then j+I 6= I in R/I (i.e. j+I 6= 0).But R/I is a field. Hence j + I has an inverse, say k + I. Therefore

(j + I)(k + I) = 1 + I = jk I + i = jk + i = 1

for some i I. Finally, i I = i J and j J = jk J. Thereforejk + i = 1 J. Hence J = R.

Proposition 2.14. R is a commutative ring, I is an ideal. Then I is prime iffR/I is an integral domain.

Corollary 2.15. Maximal ideals are prime in a commutative ring.

16

Proof. M maximal 2.13= R/M is a field obvious= R/M is an integral domain =M is prime.

Proof. (of 2.14)Case 1. I = R. Then I is not prime and R/I is not an integral domain. Nowsay

Case 2. I 6= R. Then I is not prime iff a, b R such that ab I, a / I, b / I.

a, b R st. ab+ I = 0 R/I

with a+ I 6= 0, b+ I 6= 0 in R/I. But

a+ I, b+ I R/I s.t. a+ I 6= 0, b+ I 6= 0 and (a+ I)(b+ I) = 0.

This is iff R/I is not an integral domain.

Remark. It is not hard to prove 2.15 directly (i.e. no quotients). We leave thisan an exercise.

Corollary 2.16. {0} is a prime ideal of a commutative ring R R is anintegral domain and {0} is a maximal ideal of R iff R is a field.

Corollary 2.17. Prime ideals are not always maximal.

Proof. There exists an integral domain that is not a field. for instance, Z, and{0} is prime, but not maximal.Corollary 2.18. Maximal ideals of Z are those of the form pZ, where p isprime. Prime ideals of Z are of the form pZ, where p is prime, and {0}.

Proof. Z is an integral domain, but not a field. Hence {0} is prime, but notmaximal. All other ideals of Z are nZ, n > 0, where n is the smallest positiveelement of the ideal after definition 2.4, and Z/nZ is a field 1.11 it is an integraldomain 1.16 n is prime.Remark. We have seen that it is not true that prime = maximal. Is it true,however, that prime and non-zero = maximal? This would be consistent witheverything we have seen so far. Yet, the answer is no. For example, consider theset R = C[x, y] with I = {rx : r R}. Clearly I 6= 0 as x I. To check that Iis prime but not maximal, we need to check that R/I is an integral domain butnot a field. Well, consider the map C[x, y] C[y] given by f(x, y) 7 f(0, y).It is easy to check that f is a surjective ring homomorphism with kernel beingthe multiples of x. By the First Isomorphism Theorem, RI = C[y], which is anintegral domain, but not a field (by Proposition 1.13).

Generators of Ideals

17

Definition. Let R be a commutative ring (out of sheer laziness) and X R asubset. I want to talk about the ideal generated by X. We say that the idealgenerated by X in R is

IR an ideal,XII.

However, we will ignore this definition implicitly use

Lemma. If is a set and , I is an ideal of R, then

I =

I

is an ideal.

Proof. 0 I = 0 I. Also,i, j I = i j I = i j I

i I, r R = i I = ri I = ri I.

Therefore there is a better definition: In the case that X is finite,

Definition 2.19. Let R be a commutative ring and X R be a finite subsetof R. Say X = {x1, . . . , xn}. The ideal generated by X is the set

I = {r1x1 + + rnxn : ri R}.Notation. We write I = (x1, . . . , xn).

Lemma 2.20. I as defined above is an ideal and it is the smallest ideal of Rcontaining X.

Proof. 0 I (set ri = 0 i). Also I is closed under :(r1x1+ +rnxn)(s1x1+ +snxn) = (r1+s1)x1(r2+s2)x2 (rn+sn)xn.Finally, if r1x1 + + rnxn I and a R, then

a(r1x1 + + rnxn) = (ar1)x1 + + (arn)xn I.Furthermore, if J is any ideal of R with X J,x1, . . . , xn J = r1x1, . . . , rnxn J = r1x1 + + rnxn J = I J.

Remark. We just showed that the ideal (x1, . . . , xn) is the smallest ideal of Rcontaining {x1, . . . , xn}. Therefore

(x1, . . . , xn) =

XIRI.

Hence both definitions (hard and easy one) are the same!

18

If X is infinite, the ideal generated by X is

{r1x1 + + rnxn : xi X}

where the sum is finite, and n is as big as you like. Check that this is an ideal.

Special case: n = 1 andX1 = {x1} = {x}, x R. Then (x) = {rx : r R} = Rxis called a principle ideal.

Not every ideal is principal, for example, consider I C[x, y] defined by I =(x, y), i.e.

I = {fx+ gy : f, g C[x, y]}.Check that I is the set of polynomials in x and y with no constant term. Further,I is the kernel of the map C[x, y] C given by

f(x, y) f(0, 0).

We claim that I cannot be principal, as if I = (f), then x I, then f dividesx and

= f = , 6= 0 or f = x, 6= 0and y I = f 6= x. So f = 6= 0, but / I. Therefore (x, y) cannot beprincipal.

However, if R = Z, and

I = (6, 8) = {6m+ 8n : n,m Z},

then 2 I as 8 6 = 2, and therefore 2t I t Z, as I is an ideal. On theother hand 6m+ 8n is even for all n,m, and so I = 2Z = (2). Therefore, (6, 8)is a principal ideal.

If fact, we have seen that every ideal of Z is of the form nZ = (n) for somen Z. Hence every ideal is principal.

Question: Let a, b Z, not both 0. Let I = (a, b) be an ideal of Z. I mustbe (d) for some d. What is d?

Definition 2.21. Let R be a commutative ring. We say that an ideal I isfinitely generated if

I = (x1, . . . , xn), xi R.We say that I is principal if I = (x) for some x R. Further, we say that R isNoetherian if all ideals of R are finitely generated. Finally, we say that R is aprincipal ideal domain (PID) if

1. R is an integral domain

2. all ideals of R are principal.

We think of these definitions in the following way:

19

- R noetherian R is finite dimensional. Therefore R not Noetherian= R pathological (much too big).

- R a PID = R is one-dimensional

Here are some things we cannot prove yet:

- C[x] is a PID

- C[x1, . . . , xn] is not a PID if n > 1. But it is Noetherian

- C[x1, . . . ] is not Noetherian.

We have proved that Z is a PID, as it is clearly an ID and all ideals are of theform nZ, and hence principal.

20

Chapter 3

Factorisation in IntegralDomains

The purpose of this chapter is to axiomatise and generalise the proof that anyn Z 2 is uniquely a product of primes. It will turn out that an analogoustheorem is true in Z[i] and Z[

2], but not Z[

5].

Throughout this chapter, R will denote an integral domain (so R is commuta-tive).

Definition 3.1. Say x R is a unit if y R such that xy = 1. Write R =the set of units in R (note that R is a group under multiplication with identity1). We say x divides y (denoted x | y) if q R such that y = qx. We call xand y associatives if y = ux for some unit u R.Exercise. Show that x R is a unit (x) = R. Show also that x | y y (x) (y) (x).Note. The notion of divides is the usual one in, for example, Z or R[x] etc.

Lemma 3.2. x and y are associatives (x) = (y).

Proof. First note that if y = ux and u R, then v such that uv = 1 andvy = vux = x. Note that the notion of being associatives is symmetric. If x = 0,then ux = 0, and the only associative of 0 is 0 = 0 1. On the other hand, if(0) = (y), then y (0) = {0}, and so y = 0. Hence the lemma is true for x = 0(or y = 0, by symmetry). Now say x 6= 0. Then y = ux, u a unit. = x = vy(where v = u1) and then (x) = Rx = Rvy = Ry = (y) as Rv = R. Conversely,if (x) = (y), then x (y) = x = ry and y (x) = y = sx. Hence

x = rsx = x(rs 1) = 0

and as x 6= 0 and R is an integral domain, = rs 1 = 0 = r Rand xand y are associatives.

Corollary. Being associatives is an equivalence relation.

21

Example. R = Z. The units of Z are {r Z : r | 1} = {1}. Hence theassociatives of n are n.Definition 3.3. We say r R is irreducible if r 6= 0, r not a unit, and if r = ab,then either a or b is a unit (as an example, R = Z : irreducible = usual notionof prime, up to sign: r Z irreducible r = p, p prime). We say r Ris prime if r 6= 0, r 6= an unit, and if r | ab = r | a or r | b.Exercise. Assume that the integers factor uniquely into primes. Check thatthe primes of Z are exactly p, for p a prime number.Example. _

1. Let R = Z. 2 is irreducible (as is any prime number) and 3 is too.

2. R = Z[i] = {a + bi : a.b Z}. The units of R are found as follows:Define N : R Z0 by N(a + ib) = a2 + b2, i.e. N(z) = zz = |z|2 .It is easy to see that N(rs) = N(r)N(s). Say r R, i.e. s suchthat rs = 1. Then N(r)N(s) = 1 = N(r) = 1 as N(r) Z0.So if r = a + ib and r R, then a2 + b2 = 1 = r = 1,i.Conversely, 1 and i are units. But 2 Z[i] is no longer irreducible,because 2 = (1 + i)(1 i), which is a product of two non-units (henceirreducibility of 2 depends on R). But 3 is still irreducible in Z[i], as if3 = rs, r, s Z[i], then 9 = N(3) = N(r)N(s). Let r = a+ bi, s = c+ di.Then (a2 + b2)(c2 + d2) = 9 = a2 + b2 {1, 3, 9}. But a2 + b2 = 3 hasno solutions in Z. So either N(r) = 1 or N(s) = 1. So r or s is a unit.

3. Consider R = Z[5] = {a + b5 : a, b Z}. Define N : R Z0

by N(a + b5) = a2 + 5b2 (= zz). If r is a unit, then rs = 1 for some

s R. Then N(rs) = N(r)N(s) = 1 = N(r) = 1. r = a + b5 anda2 + 5b2 = 1. So b = 0, a = 1. Hence r = 1 and these are both units.Here, there is no solution to a2 + 5b2 = 2 or a2 + 5b2 = 3 with a, b Z.Hence 2, 3 are irreducible in Z[

5]. For example,

2 = rs = 4 = N(2) = N(r)N(s) = N(r) {1, 2, 4} N(r) = 1 or N(s) = 1

and 1 +5 is irreducible as N(1 +5) = 6 and factors of 6 in Z1 are

1,2,3,6 and 2,3 are not possible. Also 15 is irreducible (norm is also6). Now 6 = 2 3 = (1 +5)(15), i.e. there are two factorisationsof 6 into irreducibles. On the other hand, 6 has no factorisations intoprimes, because none of 2,3,1 +

5, 15 are prime. For example, 2divides

6 = (1 +5)a

(15)b

but2 - 1 +5, 15 (as 1

52 / R) etc. Hence we see that Z[

5]has irreducibles that are not prime and some elements factor into irre-ducibles in more that one way.

On the other hand,

Lemma 3.4. All primes are irreducible in an integral domain.

22

Proof. Say r is prime. Then r 6= 0 and r is not a unit. Say r = ab. We willshow that one of a, b must be a unit. Now, r = ab = r | ab. But r is prime,hence, wlog, r | a (could be r | b as well). So

a = sr, s R = r = ab = srb = r(bs 1) = 0.As r 6= 0, we must have that bs = 1, since we are in an integral domain, andhence b is a unit.

Lemma 3.5. If 0 6= r R, then r is prime (r) is a prime ideal.

Proof. If r is a unit, then r is not prime and (r) = R is not a prime ideal. Sosay that r is not a unit. Then (r) is a prime ideal ab (r) = a (r) or b (r) r | ab = r | a or r | b. This is equivalent with saying thatr is prime.

Definition 3.6. An integral domain is called an Euclidean domain (ED) if thereis some function : R\{0} Z0 such that

(1) (ab) (a) if a, b 6= 0(2) If a, b R and b 6= 0, then one can write a = qb+ r with q, r R (we call q

the quotient and r the remainder) such that either r = 0 or (r) < (b).

Example.

1. R = Z and (r) = |r| .2. For F a field, R = F [x], (r) = deg(f).

Exercise. Verify whether the last two examples are indeed Euclidean domains.

Theorem 3.7. R is an Euclidean Domain = R is a prime integral domain.Remark. Compare this with the proof that all ideals of Z are principal.

Proof. Say that R is an Euclidean domain and I R is an ideal. If I = {0} =(0) than this is clear. Now, assume that I 6= {0}. Choose n I\{0} with (n)minimal. We claim that I = (n). Certainly (n) I. Now say i I. We want toshow that i (n). For (i = a, n = b), we can write i = qn+ r with either r = 0or (r) < (n). But r = i qn I. Hence (r) < (n) cannot be true by thedefinition of n. Hence r = 0 and so i = qn (n). Hence I = (n).

Our next goal is to show that things factor uniquely into primes if we are in anprime integral domain.

Corollary 3.8. F a field = F [x] is a PID.

Proof. Obvious.

At this stage, recall that in Z, prime = irreducible, but in Z[5], prime 6=

irreducible. In an general integral domain R, prime = irreducible.Lemma 3.9. In a PID, all irreducibles are prime.

23

Proof. Say R is a PID and r is irreducible. Then r 6= 0 and r is not a unit. Sayr | ab, a, b R and assume r - a. We want to show that r | b. Define I = (r, a).As R is a PID, we must have that I = (x) for some x R. So r, a (x), and sor = sx and a = tx. But r is irreducible, therefore either s or x is a unit. But scannot be a unit. For if s is a unit, su = 1 for u R and

r = sx = ur = x = a = bx = tur = r | acontradiction. So xmust be a unit. Hence I = (x) = R = i I, and therefore, R such that r + a = 1 = b = rb + ab and r | rb, r | ab (asr | ab). Hence r | b.Definition 3.10. An integral domain R is a unique factorisation domain (UFD)if

UF1 (factorisation) Any non-zero r R can be written r = ur1 . . . rn for somen 0 with u a unit and ri irreducible

UF2 (uniqueness) If r = ur1 . . . rn = vs1 . . . sm with m,n 0 with and u, vunits and ri, si irreducibles, then m = n and after reordering the si, ifnecessary, ri and si are associates i.

Remark. UF2 is necessary to deal with, e.g, 15 = 3 5 = 5 3 = 3 5 =1 35 etc.Example.

1. Z is a UFD.

2. F [x] is a UFD.

3. Any ED is a UFD.

Remark. We have seen that in any PID, prime = irreducible. This is, moregenerally, true in a UFD:For prime = irreducible in an ID (shown before). For the converse, say ris irreducible. Then r 6= 0 and r is not a unit. Hence we only need to checkr | ab = r | a or r | b. So say r | ab. If a = 0 = r | 0 = done. Saya, b | 0.Say rs = ab. Factor s, a, b :

- s = us1 . . . sm

- a = va1 . . . an

- b = wb1 . . . bp

where u, v, w are units and si, rj , bk irreducible. Now, get two factorisations ofrs = ab :

rs = us1 . . . smr = vwa1 . . . anb1 . . . bp.

By UF2, these two factorisations are the same up to order and associates. Hencer is an associate of some ai or some bj . Wlog, say ai = ur. Then r | ai | a =r | a. Hence prime = irreducible in a UFD.

24

Remark. We have seen that ED = PID and we will see that PID = UFD.The converses, however, are both false. In fact it is a theorem that if R is aUFD, then so is R[x]. In particular, we see that C[x, y] is a UFD: C is a field C is a PID (the only ideals are (0) and (1)) Hence C[x] is a UFD, and so C[x, y]is a UFD.But the ideal (x, y) is not principal. In fact, we have that PID = dim 1,and C[x, y] has dim 2.It is much harder to find a PID that is not an ED.

Example. Z[ = 1+

192

]= {a+ b : a, b Z}. Note,

2 =

(1 +192

)2=18 + 219

4= 9 +

192

= 5.

= Z[] is a ring. By M3P15, this is also a PID, and, by a messy calculation,is it not a ED.

Example. Z[i] is a UFD. It suffices to prove that Z[i] is an ED. Define (a+ib) = a2 + b2. We need to check that if x, y Z[i], then x = qy + r with(r) < (y) or r = 0.How to find q : consider xy C. So Z[i] is an ED Z[i] is a UFD.Exercise. _

1. Show that Z[2] is a UFD.

2. Why does the above procedure fail to work for Z[5]?

Theorem 3.11. A PID is a UFD.

Proof. Say R is a PID. Assume UF1 fails. Choose r R such that r 6= 0and r 6= ur1 . . . rn, ri irreducible for i = 1 n. ||||| Clearly r is not a unit(n = 0) and r is irreducible (n = 1). Hence r = r1s1 for some r1, s1, not unitsin R. If r1 = ut1 . . . tn and s1 = vw1 . . . wm, ti, wj irreducible and u, v units,then r = uvt1t2 . . . w1 . . . wm is a factorsation of r. This is a contradiction tothe definition of r. Hence one of r1, s1 does not factor into irreducibles either.Wlog, take r1. By the same trick, r1 = r2s2 where s2 is not a unit and r2 isnot the product of irreducibles. Similarly, r2 = r3s3, s3 not a unit and r3 notthe product of irreducibles. By repeating this procedure, we obtain an infinitesequence

r = r0r1 . . .

where ri = ri+1si+1, si+1 not a unit. Now, r = r1s1 ideal (r1) contains r andhence (r). Furthermore, (r) (r1) (for if (r1) = (r), then

r1 = rt, t R, = r = r1s1 = rts1 = r(1ts1) = 0 and r 6= 0 = s1 is a unit

contradiction). Similarly, (r1) (r2) with (r1) 6= (r2), and so we get an increas-ing chain of ideals

(r0) (r1) . . .

25

where all containments are strict. Now, let I =n0(rn). I is an ideal, since if

i, j I, then N 0 such thati, j (rN ) = i j (rN ) I.

As R is a PID, I is pricipal. Hence I = (d) for some d R and d I = d (rN ) for some N 0. Therefore

(d) (rN ) ( (rN+1) ( I = (d)contradiction. Hence PID = UF1.For PID = UF2, consider lemma 3.9: irreducibles are prime in a PID. So,as r 6= 0 and r = ur1r2 . . . rn = vs1 . . . sm. We will prove that n = m and afterre-ordering ri and si are accociates by induction on n. If n = 0, then r = u isa unit and if m > 0, then s1 | u = s1 = 1. But s1 is irreducible, hence s1is not a unit, contradiction. Hence m = 0, and so the base case of inductionholds. Now the inductive step: Let n 1 and assume the statement is true forn < n. Then r = ur1 . . . rn, n 1 and r = vs1 . . . sm. So r1 | r = vs1s2 . . . smand r1 is irreducible, hence r1 is prime by 3.9. Therefore r1 | v or r1 | si forsome i. As v is a unit v | 1, so r1 | v = r1 | 1, contradiction. Hence r1 | sifor some i, 1 i m (and in particular m 1). After re-ordering the, wlog,si, r1 | s1. Say s1 = r1t for some t. s1 is irreducible, hence either r1 or t mustbe a unit. But r1 is not a unit (see above), so t must be a unit. Hence r1 ands1 are associates. Now, cancel r1 (which is fine, as R is an ID). So

ur1r2 . . . rn = vs1s2 . . . sm = vr1ts2 . . . sm

= ur2 . . . rn = vtunit

s2 . . . sm

and by our inductive hypothesis, we must have n 1 = m 1 = n = m andri and si are associatives for all i 2 after re-ordering, if necessary.Remark. As a consequence, if n 1 and t Z such that t2 1 mod n, thenn = a2 + b2, a, b Z.

Remark. If p is prime and p 1 mod 4, then t such that t2 1 mod p,for example t = p12 . To show this we could, alternatively, use the fact that(Z/pZ) is cyclic elements of order 4 (namely t).

Note. Z[3] = Z

[1+32

]= algebraic integers inQ(

3) = a+b(

1+32

), a, b

Z.

Questions:

(Q1) What are the algebraic integers in Q(d), d Z not being a multiple of

a square number?

(A1) This question will be answered in M3P15: Z[d] if d 6 1 mod 4 or

Z[

1+d

2

]if d 6 1 mod 4.

26

(Q2) When is the set of algebraic integers of Q(d) an ED, PID, UFD?

More generally, what about algebraic integers in a general number fieldlike Q( 3

2) etc.?

(A2) This question is fully answered for d < 0 : There are only 9 values ofd < 0 which give rise to rings which are UFDs (i.e. generally, this is notan UFD (and therefore, not a PID or ED). The largest value of |d| whichgives rise to a UFD is d = 163 and the algebraic integers in Q(163)is Z

[1+163

2

]. 1+

1632 is a root of

x2 + x+ 41 = 0

which is prime for 0 x 39. It was known ages ago that there exists a dwith d < 163 such that the algebraic integers of Q(d) is a UFD, but itwas not proved until the 1960s that no further such ds exist. For d > 0,it was conjectured by Gauss that Z[

d] was a UFD infinitely often. This

conjecture is still open.

(Q3) Another question is whether the integers of Q(d) form a UFD infinitely

often. The natural norm on Z[

d], d > 0, given by (a + b

d) =a2 db2 is a good canditate for the Euclidean norm.

(A3) It can be shown that this is not an Euclidean norm for d bein sufficientlylarge. An interesting example is Z[

14]. This is a UFD and a PID, but is

it an ED? It was known for a while that the generalised Riemann hypoth-esis would imply that Z[

14] is an ED. However, in 2004 it was proved

unconditionally that Z[

14] is an ED (note that (a+b

14) =a2 14b2

does not yield an ED).

Remark. Some final facts about UFDs: We have seen an example of a ring thatfails to be a UFD, because uniqueness of factorisation fails (Z[

5]): 2 3 =(1+5)(15). But clearly (as it can be shown using induction on a2+5b2)

every non-zero, non-unit element of Z[5] is a product of irreducibles: only

uniqueness happens to fail.

We will end this chapter with an example of a ring which has elements whichdo not even factorise into irreducibles (i.e, UF1 fails rather than UF2).

Definition 3.12. Let R be an integral domain. We say that a properly dividesb if

(1) a | b(2) a, b are not associates (i.e. b = sa, s not a unit).

Example. 3 divides 3 in Z, so 3 properly divides 6 in Z.

Lemma 3.13. In a UFD R, you cannot have a sequence r1, r2, . . . of non-zeroelements of R such that ri+1 properly divides ri i (c.f. proof of UF1 for aPID).

27

Proof. Clearly r1 6= 0 = r1 = u1 . . . n where i is irreducible and u is aunit. Hence r2 has at most n 1 irreducible factors, and continuing, we findthat rn must be irreducible and rn+1 must be a unit. But then there cannot bean rn+2 properly dividing rn+1.

Example. Now let R be the subring of Q[x] consisting of all polynomials f =a0 + a1x + + anxn with a0 Z. Note that R is an integral domain (asR Q[x], which is an ID) and 2 / R (as 12 / R). Now look at the sequencex, x2 ,

x4 , . . . . It is easy to see that each element properly divides the previous one.

Hence R cannot be a UFD. Indeed, we cannot factor x R into irreducibles:x = N xN for all N 1. This is a problem. In fact, reversing the implicationPID = UF1, we can prove that R is not Noetherian. Indeed, the idealI R of polynomials with zero constant term is not finitely generated: Iff1, f2, . . . , fn I and J = (f1, . . . , fn) I is the ideal generated by all the fiand if fi = aix+ . . . , then ai Q, and so N 0 in Z such that Nai Z i.Check that f J, f = ax+O(x2) and Na Z. Hence J 6= I, as 12N x I, butnot in J.

28

Chapter 4

Localisation

Let us start with a special case: the field of fractions of an Integral Domain.We have seen the construction of R/I, where R is a ring and I a bi-ideal. Also,R/I is smaller than R. The only way to make bigger rings, that we have seenso far, is by adding a polynomial variable R R[x] R. In this chapter wewill see another way of creating bigger fields out of smaller ones, e.g. Z Q.

As an example, consider a transcendental number in C, i.e. some x C suchthat x is no non-zero root of any polynomial in Q[x], e.g. x = pi. Consider thesubring Q[pi] = {a0 +a1pi+ +anpin : ai Q} of C. As abtract rings, we haveQ[pi] = Q[x]. But this is not a field. We want to ask the question: What is thesmallest subfield of C that contains pi (i.e.

FC,piF,F field F )? It is clear that

it most contain 0 and 1 and so it must contain Z, and as it is a field, it mustalso contain all elements of 1Z etc. As it contains Q and pi, it must also containQ[pi]. Since 1pi and in fact

1p(pi) Q[pi] for any p Q[x], p 6= 0 ( = p(pi) 6= 0).

Hence it must also contain a(pi)b(pi) for any a, b Q[x], b 6= 0. This works through:{z C : z = a(pi)b(pi) for a, b Q[x], b 6= 0} is indeed a field. This worked becauseQ C and pi C and C is a field. Hence a(pi)b(pi) makes sense in C.But what happens if you do not have a field?

We will try to answer the following question: If R is an integral domain, how dowe build a field F such that R F and F is (somehow) generated by R? Forexample, Z Q and Q[pi] Q(pi) = {a(pi)b(pi) : a, b Q[x], b 6= 0} are all within C.the construction is called the field of fractions.

Theorem 4.1. (Field of fractions of an integral domain). Let R be an integraldomain. Then a field F and an injective ring homomorphism : R F suchthat for any f F,a, b R, b 6= 0, such that f = (a)(b) . Furthermore, the pair(F, ) is unique up to unique isomorphism, i.e. if (F, ) and (F ) both work,then there is a unique isomorphism of fields i : F F such that

29

Fi

R

>>}}}}}}}

AAA

AAAA

F

and the maps commute, i.e. i = .

Proof. Let X = R (R\{0}) (think of this cross-product as a pair of numeratorand denominator). Define a relation on X by (a, b) (c, d) ad = bc.Claim. is an equivalence relation.

1. Reflexive: Obvious

2. Symmetric: Obvious (as R is an integral domain, hence commutative)

3. Transitive: Say (a, b) (c, d) (e, f). We want to show that (a, b) (e, f). Now ad = bc and cf = de, and so adf = bcf = bde and d 6= 0 byassumption. Hence af = be by cancellation in an integral domain, and so(a, b) (e, f).

Define F to be the set of equivalence classes and denote the equivalence class of(a, b) by [a : b]. Next, we make F a ring:Define 0 = [0 : 1] = [0 : t]t 6= 0 and 1 = [1 : 1] = [t : t] t 6= 0. Further, let[a : b] + [c : d] = [ad + bc : bd]. Is this well-defined? Say [a : b}] = [A : B] (i.e.(a, b) (A,B), is [a : b] + [c : d] = [A : B] + [c : d]?). Well, need to check

aB = bA = [ad+ bc : bd] = [Ad+Bc : Bd],

i.e. that (ad+ bc)Bd = bd(Ad+Bc). Therefore, we need to check

aB = bA = aBd2 + bcBd = Abd2 +Bcbd

and this is clear. Define [a : b] [c : d] = [ac : bd]. This is also well-defined(exercise!). Now, we only need to check the axioms:

30

Additive Inverse: Note that [a : b] + [a : b] = [ab ba : b2] = [0 : b2] = 0Associativity:

([a : b] + [c : d]) + [e : f ] = [ad+ bc : bd] + [e : f ] = [adf + bcf + bde : bdf ]

and

[a : b] + ([c : d] + [e : f ]) = [a : b] + [cf + de : df ] = adf + bcf + bde+ bdf ]

+ is associative (and clearly also commutative). Hence (F, 0,+) is agroup. Also is associative and 1 is the identity.

Distributivity We need [a : b]([c : d] + [e : f ]) = [a : b][c : d] + [a : b] [e : f ].But the left hand side is

[a : b][cf + de : df ] = [acf + ade : bdf ]

and the right hand side is

[ac : bd] + [ae : bf ] = [acbf + bdea : b2df ]

and we can cancel b because it is a denominator.We will show that these equivalence classes form a ring F (in fact, F is even afield): [a, b] = [0 : 1] a = 0). Therefore, if [a : b] 6= [0, 1] then a 6= 0 andinverse is [b : a]. It is easy to see that [a : b][b : a] = [1 : 1]. : R F definedby (a) = [a : 1] is a ring homomorphism (easily checked). Also is injective:the kernel is an ideal and (a) = 0 = [a : 1] = [0 : 1] = a = 0. So it isinjective.So we see that if R is an integral domain, then F is a field and : R R. Now,given f F, f = [a : b], b 6= 0, we have f = (a)(b) because

(a)

(b)= (a)(b)1 = [a : 1][b : 1]1 = [a : 1][1 : b] = [a : b].

All that is left is to show that F is unique in the following strong sense: if F is any other field, and R F is an injective ring homomorphism such thatf F , f = (a)(b) , for some a, b 6= 0, then a unique : F F such that

F

R

>>}}}}}}}

AAA

AAAA

F

commutes.For uniqueness of , say f F. Then f = (a)(b) for some a, b 6= 0 a, b R. Weknow ((a)) = (a), ((b)) = (b) and

(f) (b) = (f)((b)) = (f (b)) = ((a)) = (a).

31

So if exists it must be defined by (f) = (a)(b) for f =

(a)(b) . Finally, we claim

that we can define by if f F, write f = (a)(b) , and set (f) = (a)(b) . Need to

check that this is well-defined, i.e. if f = (a)(b) =(c)(d) , b, d 6= 0, then

(a)(b) =

(c)(d) .

Well,

(a)

(b)=(c)

(d)= (a)(d) = (c)(b) = (ad) = (cb) = ad = bc

as is injective. So (a)(d) = (b)(c) = (a)(b) = (c)(d) (as

is injective,we have that (b), (d) 6= 0).Now is a ring homomorphism (easy to check). Is an injection? Yes, F isa field = ker() is an ideal, but the only ideals of a field are 0 and F . But(1) = 1 (easy) = ker() 6= F = ker() = 0.Last thing to check: Is a surjection? Yes, because if f F , then by assump-tion f =

(a)(b) for some a, b, b 6= 0. Therefore f = ([a : b]).

Notation. F is called the field of fractions of R and it is denoted by Frac(R).

Corollary 4.2. Any integral domain is a subring of a field.

One can do a slightly more general construction here and go from Z to a smallerring than Q, i.e. Z a ring Q (there are lots of rings that meet thisrequirement).

Definition 4.3. Let R be a commutative ring. A multiplicative subset of R isS R if

(1) 1 S(2) s, t S = st S.

Example. _

1. Let R be a commutative ring, a R. S = {1, a., a2, . . . }.2. Let R be a commutative ring, P R a prime ideal. S = R\P (also, could

have S = {I}, I any ideal).3. A special case of 2.: R is an integral domain, P = (0), S = R\{0}.

Definition 4.4. Let R be an integral domain and let S be a multiplicativesubset and assume 0 / S (e.g. S = R\{0}) . We define the localisation of R atS to be the equivalence classes [r : s] R S such that

(r1, s1) (r2, s2) r1s2 = r2s1.

In other words, the localisation of R at S is either

(1) the equivalence classes of R S under (r1, s1) (r2, s2) r1s2 = r2s1where +,0, 1 are defined as in Frac(R) (special case: S = R = {0})

32

(2) subset of Frac(R) consisting of elements of the form r/s for r R, s S.Notation. We denote the localisation of R at S by RS or S1R.

Example.

1. S = R\{0}, we get Frac(R).2. R = Z, p prime = S = {1, p, p2, . . . }. Then the localisation is { apn :a Z, n 0} Q.

3. R = Z, P = (p), S = Z\P = {n Z : p|n}. Then the localisation is{ab : p - b}.

Note. There is an injection R RS given by r r/1 (that is why we demanded1 S).Lemma 4.5. Let R an integral domain and 0 / S a multiplicative set. If A isany commutative ring and if : R A is any ring homomorphism such that(s) is a unit in A s S, then : RS A extending (i.e. such that

R //

A

RS

>>

and is unique.

Proof. Uniqueness of : if r/s = [r : s] RS then (r/s) (s) must equal(r). This forces (r/s) =

(r)(s) =

(r)(s) . Conversely, this does work. Also

(r/s) = (r)(s) makes sense in A, as (s) is a unit by assumption. Now, ifrs =

rs , then s

r = rs and so

(s)(r) = (r)(s) = (r)(s)

=(r)(s)

.

Hence is well-defined. It is easy to show that is a ring homomorphism.

Remark. If R is an integral domain and P R is a prime ideal and S = R\P ,then people often write RP for RS , S1R.

Example.

(1) S = R\{0} (null subset as R is an integral domain), then S1R = Frac(R)(2) S = {1}, then S1R = R(3) p prime and R = Z, S = {1, p, p2, . . . } (write this as S1Z = Z[ 1p ]) then S1R ={ apn : a Z, n 0} which is a ring, a subring of Q.

Exercise. If q is a prime number, q 6= p, then (q) S1R is still a primeideal as Z ZqZ given by p something non-zero. But Z/qZ is a field. Hencep a unit. So Z Z/qZ extends to a map S1Z Z/qZ and one cancheck that the kernel is (q). Hence (q) is maximal ((q) = qS1Z = { apn : a Z multiple of q, n 0}). But p is a unit in S1Z. Hence (p) is no longer prime(Spec(S1Z) Spec(Z) is an open set. The complement is (p)).

33

(4) R = Z, P = (p), p prime, P a prime ideal generated by p. S = Z/P = {n Z :p - n}. Now S1R = {ab : a Z, b Z and p - b}. S1R is a ring called Z(p), Zlocalised at p. If q is prime, q 6= p then q S 1q Z(p) = q is a unit inZ(p). But 1p / Z(p) and indeed pZ(p)is a prime ideal and a maximal ideal: amap Z(p) Z/pZ extending Z Z/pZ as any s S is a unit mod p, and thekernel of this is (p).In fact, Z(p) has exactly two prime ideals, namely (0) and pZ(p), one containedin the other.

Remark. Spec(Z(p)) = 2 points, one closed and one open.

Exercise. Show that Q has uncountably many subrings.

Definition 4.6. A commutative ring is local if it has a unique maximal ideal.

Example. A field has only two ideals, 0 and itself. But by definition a fieldcannot be maximal in itself, hence 0 is the maximal ideal.

Let us construct some more interesting local rings:

Zorns Lemma

This is an axiom of mathematics (it is completely equivalent to the axiom ofchoice). A poset or partially ordered set is a pair (S,) where S is a set and is a binary relation such that

(1) s S, s s(2) if s t and t s, then s = t(3) if s t and t u, then s u.Remark. partial because we do not assume s, t either s t or t s.Example. A good example to bear in mind: if S is a subset of X, then s t s t.

A chain in a poset is a subset T S if t1, t2 T either t1 t2 or t2 t1.Example.

- The empty set is a chain.- {s} is a chain- If S = set of subsets of X and s1 s2 s3 . . . , then {sn} is a chain.

Definition. An upper bound for a subset W S is s S such that w Wwe have w s. A maximal element of a poset S is x S such that if y S andx y, then x = y.Remark. I did not just say that x is maximal if z, z x. It is fine to havex S maximal and z S such that z x and x z (maximal 6= bigger thaneverything but maximal = nothing bigger).

34

Example. s1 s2 . . . subsets of X, then s =n1 sn is an upper bound

and X is too.

Lemma 4.7. (Zorns Lemma) Say S is a poset with the property that everychain has an upper bound. Then S has a maximal element (possibly manymaximal elements).

Here is why we want Zorn:

Theorem 4.8. If R is a ring and I R is an ideal with I 6= R, then maximalideal m such that I m R.

Proof. S = {ideals J R such that I J, J 6= R}. Define J1 J2 J1 J2. Note that I S S 6= 0 (which is good, as the empty chain has an upperbound, namely I). If T S is a non-empty chain (T is a big collection of idealstotally ordered by inclusion), then J =

xT x. J is an upper bound for T as

x J x T and I J and furthermore J 6= R, as if J = R, then 1 J . Then1 x for some x T = 1 x for some x S. But 1 x = x = R andR / S. Hence chains have an upper bound by Zorns Lemma, S has maximalelements and it is easy to check that these are maximal ideals.

Corollary 4.9. If R 6= {0}, then R has a maximal ideal.

Proof. I = {0} and then use 4.8.Proposition 4.10. Say R is a commutative ring and I R is an ideal. ThenI is the unique maximal ideal of R iff R is the disjoint union of I and R.

Proof. Say I is the unique maximal ideal of R. Then I 6= R, so I = isnot a unit (as a unit = 1 I = I = R). Conversely, if / I, then letJ = (). We want J = R, as then 1 J = R such that = 1 and is a unit. But if J 6= R, then by 4.8, J m, the maximal ideal and so I is theunique maximal ideal. Hence m = I = J I = J I, contradiction.Conversely, say R is the disjoint union of R and I. Say J is any ideal, J 6= R.Then J = is not a unit = I. Hence J I and so I is theunique maximal ideal.

An Example of a Local Ring

Let R be an integral domain, P R be a prime ideal. Let S = R\P.Claim. S1R is a local ring and the unique maximal ideal is {as : a P, s S} = S1P.

Proof. First check that S1R is an ideal:

a

s+b

t=at+ bs

st, st S, a, b P = at+ bs P

and as S1P , 0 = 01 S1P . Also as S1P and bt S1R = abst S1P as ab P, st S. Finally, if S1R and / S1P, then I claim that is a unit. For = as , s S, R and / P = a S = sa S1R.Hence as is a unit.

35

Example.

1. Z(p) is a local ring with unique maximal ideal {ab : p | a, p - b}, the quotientfield Z\pZ.

2. R = C[x, y] (polynomials in two variables). P = (x, y) = kernel ofevaluation map R C, F F (0, 0). Rp = { fg : f, g C[x, y], g /(x, y), i.e. g(0, 0) 6= 0}. Therefore RP = rational fuctions (ratio of twopolynomials) on C2 which are well-defined in a neighbourhood of (0,0).The evaluation map RP C is given by fg f(0,0)g(0,0) .

36

Chapter 5

More on Polynomial Rings

Throughout this chapter, R will be a commutative ring. Recall that R[x] ={a0 + + anxn : ai R,n 0 arbitrary}. The purpose of this chapter is todetermine which, good, properties of R are inherited by R[x]. For example, ifR is a field, then this does not imply that R[x] is a field (as x has no inverse).But R[x] is, at least, a PID (as it is a ED). Also, if R is an ID, then R[x] willalso be an ID (as proved in 1.13). We will now prove something deeper: wewill show that if R is noetherian, then R[x] is also noetherian. Recall that R isnoetherian if all its ideals are finitely generated, for example a field or a PID.An example for a non-noetherian ring is {f Q[x] : f(0) Z} (this was provedin lectures).

Theorem 5.1. If R is noetherian, then so is R[x].

Proof. Let I be an ideal of R[x]. Our goal will be to generate I with a finitenumber of elements. Let Xn be the set of polynomials in I of degree exactly n.Then I is a disjoint union: {0} X0 . Let Jn be the following set:

Jn = {0} {r R s.t.f Xn with f = rxn + . . . }.

So we have Jn R, i.e. Jn = 0+leading terms of elements of Xn.

37

Claim. Jn is an ideal of R.

Proof. Jn is closed under since if r, s Jn r, s 6= 0, then f, g I such thatf + rxn . . . , g = sxn + . . . , and so

f g I (r s)xn + Xn if r s 6= 0

r s Jn. The cases where one of r, sr s = 0 can be verified by hand. Forexample, if r 6= 0, r Jn, then to show thatr Jn,

r Jn = f = rxn + I = f I = r Jn.

Now say that r Jn, s R. We want to show that rs Jn. For rs = 0, this isclear. If rs 6= 0, then r 6= 0, and so

f = rxn + Xn I sf I

withsf = srxn + . . . = sr Jn.

Claim. J0 J1 J2 . . . .Proof. If r Jn, then either r = 0 r Jn+1 or r 6= 0 and

f = rxn + I.

x R[x] = xf I and xf = rxn+1 + Xn+1 = xf Xn+1 andr Jn+1.Claim. N such thatJN = JN+1 = = JN for all n N (c.f. proof that aPID is a UFD).

Proof. If J =n1 Jn, then J is an ideal (seen this before) and R is Noetherian.

Hence J is finitely-generated and j1, . . . jm J such that J = (j1, . . . jm). Now,each ji is in some J for some (and hence all) sufficiently large . Let N = maxof the . Then N is such that j1, . . . , jm JN . So

J = (j1, . . . jm) JN JN+1 J

and hence all inclusions are equalities.Now, for 0 n N, Jn is an ideal of R, and hence Jn is finitely-generated, sayby rn1, rn2, . . . , rnn , where n 6= the number of generators. Choose fnk Xnsuch that

fnk = rnkxn + . . . , 1 k n.

{fnk} is a finite set, and fnk Xn I for all n and for all k.Claim. The finite set {fnk} 0 n N

0 k ngenerates I.

Proof. We will check that f I, f (f01, f02, . . . , f0k, . . . ) = I I byinduction on deg f . Clear for deg f = 0, i.e.f R, f 6= 0. Then

f X0 f J0 = X0 {0} = f (f01, f02, . . . , f00)

38

as an ideal of R. Therefore

f

rif0i ri R f I.

Next case: 1 deg f N, f I. We can assume any polynomial in I of degreestricly less than deg f in I . Want f I . Set d = deg f . Then

f = rxd + . . . , r J.

By definition, Jd = (rd0, rd1, . . . ) and so i R such that r =i ifdi (think

of these as leading terms). Now consider the polynomial g = f i ifdi. fand all the fdi have degree d. Hence deg g d. But

g =

(r

i

irdi

)xd + = 0xd + . . .

deg g < d.

Furthermore, f and the fdi I g I. The inductive hypothesis implies thatg I . Hence

f = g +

iIfdi I .

Final case: deg f = d > N. Again, can assume (inductive hypothesis).

g I,deg(g) < d = g I .

Again, write f = rxd + . . . where r Jd = JN = (rN0, rN1, . . . ). Hence we canwrite r =

i irNi i R. Consider

g = f i

ixdN fNi

degreeN

degree d, leading term is irNi. Now

g = rxd i

(irNi)xd + l.o.t.

therefore deg g < d g I (by inductive hypothesis) and so f I .Corollary 5.2. If R is a field or a PID, then R[x1, . . . , xn] is noetherian.

Proof. R is noetherian. Now we use induction on n.

Lemma 5.3. If : A B is a surjective ring homomorphism, and if A isnoetherian, then B is noetherian.

Remark. If B A is a subring, then this is NOT true.

A noetherian = B noetherian.

For example A = Q[x], B =?????????

39

Proof. Say I B is an ideal of B. Set J = {a A|(a) I} = 1(I). ThenJ is an ideal of A by either checking hte axioms or by observing that J = kerwhere

A //

;;B// B/I.

Hence J = (a1, a2, . . . an) for some finite set {ai}. Set bi = (ai). I claim thatI = (b1, . . . , bn). For if b I, is surjective, we have that a A such that

(a) = b, (a) I = a J.Hence a =

ni=1 iai i A. By applying , we get

b =

ni=1

(i) ai (b1, . . . , bn).

Corollary 5.4. If R is a PID and I R[x1, . . . , xn] is an ideal, then R[x1, . . . , xn]/Iis noetherian.

Example. Z[5] is not a UFD (and in fact not a PID: the ideal (2, 1 +5)

turns out not to be principal). But is is noetherian. there is a surjectionZ[x] Z[5] given by x 7 5

x2 + 2x 7 7 (5)2 + 25 7 etc.In particular a+ bx 7 a+ b5. Hence this map is surjective (notice that itskernel is (x2 + 5)). As Z is a PID, it follows that Z is noetherian. Hence Z[x] isnoetherian and so Z[

5] is notherian by 5.1 and 5.3 respectively.Remark. In a P15 you will study the integer ring of a number field.

Our next goal will be to show that if R is a UFD, then R[x] is also a UFD.Say R is a UFD. Note that if we define a relation on a set of irreducibleelements by

a b a = ub, u a uniti.e. a b = a, b are associates, then is an equivalence relation. Henceit divides the set of irreducibles into equivalence classes. Now, choose a fixedrepresentative in each class and let X be the set of representatives. For exampleif R = Z, p p and p p,p p for p prime. These are all, hence theequivalence class of p is {p,p} and X = {2, 3, 5, 7, . . . } is a possible set ofrepresentatives. Now, each non-zero element r of R can be written uniquely (upto reordering) as r = uxe11 . . . x

enn for some n 0, u a unit, xj X, ej Z 1

. Hence if F = Frac(R), then any f 6= 0 F can be written as f = rs (factorsr and s) and hence f = uxe11 . . . x

enn with now ej Z. For example, in Q for

X = {2, 3, 5, . . . }, we have 12 = 223, 9 = 32 and 6217 = 21311171.Definition 5.5. Fix an irreducible element pi X R, where R is a UFD.The pi adic valuation on F = Frac(R) is the map v = valpi : F Z definedas follows: if 0 6= f F, then write

f = upie

xiX,xi 6=pixpii

40

with e Z (e = 0 is possible) and define valpi(f) = e. It is easily checked thatval(fg) = val(f)+val(g) (hence v is a group homomorphism (F ,) (Z,+)).Remark. Sometimes it is a useful convention to set valpi(0) = + so thatvalpi : F Z{+}, for example R = Z, X = {2, 3, . . . } and pi = 2. valpi(8) =3, valpi(70) = 1, since 70 = 21 5 7. Also valpi( 34 ) = 2 and valpi(x odd) = 0.Definition 5.6. Say 0 6= q F [x], q = f0 +f1x+ +fnxn. Say pi X. Define

vpi(q) = mini:fi 6=0

{vpi(fi)}.

Example.

- Let R = C[z], and f = z322 . The irreducibles in R are X = {z| C}.val2(f) = 2 and valz3(f) = 1, so valz(f) = 0 6= 0, 3.

- Let R = Z, f = 79 =732 . Hence val3(f) = 2, val7(f) = 1, valp(f) =

0 p 6= 3, 7.Lemma 5.7.

(i) valpi(xy) = valpi(x)valpi(y)

(ii) if valpi(x) = 0 and valpi(y) > 0, then valpi(x+ y) = 0.

Proof. (i) x = upieni=1 x

eii (it is okay if some ei = 0) and y = vpi

dnj=1 x

djj . Therefore

xy = uvpie+dni=1

xdi+eii ,

hence (i) is true.

(ii) Let x = rs , pi | r, pi | s and y = pietw , pi | t, pi | w, e 1. Then

x+ y =r

s+piet

w=rw + piets

ws.

Note that pi - s, pi - w pi - sw (as irreducibles are prime in a UFD).Furthermore, if pi | (rw+piets) then pi | rw pi | r or pi | w, contradiction.Hence pi - (rw + piets) = valpi(x+ y) = 0, so (ii) is also true.

This extends to polynomials. Say 0 6= f F [x], f = di=0 aixi.Define valpi(f) =mini valpi(ai) (or valpi(f) = mini:ai 6=0 valpi(ai)). Note that factorsation of eachof the ai only involves finitely many pis and f has only finitely many ais.Therefore valpi(f) = 0 for all but finitely many pi.

Definition 5.8. The content of 0 6= f F [x] is

cont(f) =piX

pivalpi(f) = a finite product.

We say 0 6= f F [x] is primitive if cont(f) = 1.

41

Example. Let R = Z, X = {2, 3, 5 . . . } and f(x) = 94 + 45x+ 21x2. Note that94 =

32

22 , 45 = 32 5, 21 = 3 7. Therefore the only irreducibles that show up

are 2,3,5,7 valp(f) = 0 p > 7. Now

p valp(94 ) valp(45) valp(21) valp(f)

2 -2 0 0 -23 2 2 1 15 0 1 0 07 0 0 1 0

Hence cont(f) = 22 3 = 34 .Note.

f(x)/cont(f) =

(9

4+ 45x+ 21x2

)/(

3

4) = 3 + 60x+ 28x2

which is a polynomial with all coefficients in Z and gcd = 1.Remark. The content of f(x) is only defined up to a unit - for example if wehad chosen X = {2,3,5, . . . }, then cont(f) = (2)2 (3)1 = 34 . Sothink of cont(f) as only defined up to a unit.Note. In general, cont(f) F.When is the content in R? Well,

cont(f) R xX xvalx(f) R valx(f) 0 x X mini valx(ai) 0 x, i valx(ai) 0 x, i ai R i f R[x].

Furthermore,

f primitive cont(f) = 1 minivalx(ai) = 0 x x, i s.t. x - ai

f R[x] and gcdof all coefficients is 1.Lemma 5.9. If 0 6= f F [x] and F, then

cont(f) = ucont(f) u a unit.

Proof. Let f =aix

i, then

valx(f) = minivalx(ai) = min

i(valx() + valx(ai))

by 5.6(i). So

valx(f) = valx() + mini

(valx(ai)) = valx() + valx(f)

and the lemma follows asx x

valx() = u, where u is a unit..

Lemma 5.10. If R is a UFD and F = Frac(R) and if f, g F [x] are bothnon-zero, then

cont(fg) = cont(f)cont(g).

42

Proof. Let c = cont(f), d = cont(g). Then f = cf1, g = dg1 and Lemma5.9 implies that cont(f1) = cont(g1) = 1. Again by Lemma 5.9, STP thatcont(f1g1) = 1. STP cont(f) = cont(g) = 1 = cont(fg) = 1, and nowcont(f) = 1 = f = aixi where ai R andfor pi irreducible i such thatpi - ai. Similarly, g =

bjx

j , bj R and no pi divides all of them. Certainly,then, fg R[x] cont(fg) R (as opposed to F ), i.e.cont(f) = cont(g) = 1 = f, g R[x] = fg R[x] = cont(fg) R

(i.e. cont(h) R h R[x]). Now, we want to prove that cont(fg) = 1.We will prove this by contradiction. Say cont(fg) 6= 1. Choose an irreduciblepi R such that pi | cont(fg). Now pi - cont(f), and so if f = i0 aixi, thenwe know that i such that pi | ai. Choose the biggest such i, so pi - ai butpi | ai+1, pi | ai+2, . . . . Similarly g =

j0 bjx

j , and choose the biggest j suchthat pi - bj and pi | bj+1pi | bj+2, . . . . Now say fg =

k0 ckx

k. As pi | cont(fg),we must have pi | ck k. But I claim that pi - ci+j . Forci+j = a0bi+j + a1bi+j1 + + ai1bj+1 + aibj + ai+1bj1 + + ai+jb0.

Notice that by the choice of i and j, pi | bn if n > j and pi | an if n > i. Henceall terms in ci+j are multiples of pi except aibj . So pi - ci+j (by Lemma 5.7 ifyou like: vpi(ai) = vpi(bj) = 0 = vpi(aibj) = 0 and vpi 1 for all other termsin ci+j). This proves the lemma.

Corollary 5.11. (Gauss Lemma) Let R be a UFD and 0 6= f R[x] withdeg f 1. If f is irreducible in R[x], then f is irreducible in F [x].

Proof. Say f R[x],deg f 1 and say f is irreducible in F [x] and in R[x]. Letus arrive at a contradiction. First note that f must be primitive (i.e. cont(f) =1), because if f =

aix

i and pi is irreducible and pi | ai i, then f = pig, wherepi, g are not units and so f is factored. Say f = gh in F [x],deg(g),deg(h) > 0.By 5.10, if a = cont(g), b = cont(h), then ab = 1. Now set g = ga , h

= hb . Thenf = gh and cont(g) = cont(h) = 1. Therefore g, h R[x]. Hence f is notirreducible in R[x]. Contradiction. This proves the corollary.

in fact, one can greatly generalise 5.11 to really, fully understand factorisationin R[x].

Proposition 5.12. (Factorisation in R[x]) Let R be a UFD and f R[x], f 6=0. Then f can be written as f = cg1g2 . . . gn with c R, gi R[x] where gi isprimitive and gi is irreducible in F [x], where F = Frac(R). Furthermore, thisfactorisation is unique up to

(1) changing factors by units

(2) reordering the gi.

Proof. Let F be a field. Then F [x] is a UFD (by 3.8 and 3.11), hence wecan factor f = u

i gi, where gi F [x] is irreducible and u F =units in

F [x]. Set gi = gi/cont(gi). Then gi is primitive by 5.8 and f = ci gi(x) with

c F. Taking contents, we see that cont(f) = cont(c) = c, a unit in R.This proves existence. Now, for uniqueness, say f = cg1g2 . . . gm = cg1g2 . . . gn.

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We know that F [x] is a UFD, hence n = m and after reordering, we havegi = igi, F. Hence

cont(gi)=1

= unit in R i cont(gi)=1

= i R.

Finally, taking contents of

cg1 . . . gm = cg1 . . . g

m = c = uxc, u R.

Theorem 5.13. If R is a UFD, then so is R[x]. Furthermore, the irreducibleelements of R[x] are of two kinds:

(1) The irreducible elements of R.

(2) Primitive polynomials in R[x] which are irreducible in F [x].

Moreover, the units in R[x] are the units in R (easy to show).

Proof. Let R be a UFD. Then R is an ID = R[x] is an ID (by 1.13). Byconsidering degrees

= (deg f > 0 = f not a unit) = R[x] = R.

The axioms UF1 and UF2 of a UFD are precisely 5.12.

Corollary 5.14. R a UFD = R[x1, x2, . . . , xn] is a UFD.

Proof. Trivial.

Example. We know that Z[x] is a UFD, hence Z[x1, . . . , xn] is a UFD. Also, ifk is a field, then k[x1, . . . , xn] is a UFD.

Remark. Z[x] is not a PID (the ideal (2, x) is not principal) and k[x, y] is alsonot a PID (the ideal (x, y) is not principal).

44

Chapter 6

Modules

A module can be viewed as a vector space over a ring. In this chapter, R willdenote a ring (not necessarily commutative).

Definition 6.1. An Rmodule (or more precisely a left-module), is an abeliangroup M equipped with a map RM M, traditionally denoted (r,m) 7 rmsuch that

1) r(m+ n) = rm+ rn r R,m, n M2) (r + s)m = rm+ sm r, s R,m M3) (rs)m = r(sm) r, s R,m M4) 1m = m m M .Remark. If R is a field, then these are precisely the axioms for a vector spaceover a field.

Remark. It is called a left Rmodule, because the action of R on M is on theleft: rm, not mr.

Remark. If R is non-commutative, the notions of a left Rmodule and a rightRmodule do not concide. If R is commutative andM is a left Rmodule, thenwe can makeM into a right Rmodule by defining mr := rm and checking thatthe axioms are still met.

Exercise. Let R be a non-commutative ring and M be a left Rmodule. Whydoes the trick mentioned above not make M into a right Rmodule?Basic consequences of the axioms: (rs)m = r(sm). Set r = 1 and obtain(s)m = (1)(sm) and (1)x = x because

(1)x+ x = (1)x+ (1)x = (1 + 1)x = 0x = 0because

(1 + 0)x = x = 1x+ 0x = x+ 0x.

In other words, 0x = 0 for all x M and (1)x = x. Hence 0 is the identityfor M.

45

Example.

1. If R is a field, an Rmodule is a vector space over R.2. R is naturally an Rmodule: R is an abelian group, and define a mapR R R by multiplication in R. Then axioms 1. and 2. follow fromdistributivity in R and 3. and 4. are the axioms for multiplication. Moregenerally, set M = Rn = column vectors in R. Then

r

12...n

=

r1r2...

rn

makes Rn into an Rmodule.

3. Let R be a ring and I be a left ideal. By definition, a left ideal is anabelian group and if r R and i I, then ri I. Define a map RI Iby (r, i) 7 ri. This makes I into an Rmodule (all four axioms follow asabove, in the case where I = R).

4. R = Z. Say M is an abelian group. We will turn M into a Zmodule.For m M , define

- 0m = 0- 1m = m- 2R

mM

= (1 + 1)m = m+m

- if t Z0, tm = m+m+ +mt times

and (t)m = (tm).

These definitions were forced upon us by the axioms and it is a relativelyeasy check to show that the axioms are satisfied. Conclusion: If M isan abelian group, then M can be given the structure of a Zmodulein a unique way. More vaguely: abelian groups = Zmodules. Nowwe clearly see that it is not the case that a general (finitely-generated)Rmodule is isomorphic to Rn for some n. For example, Z/nZ is aZmodule. Clearly not Zm for any m.

5. Let F be a field, R = Mn(F ) = set of n n matrices over F . LetM = Fn = column vectors of length n. Then R acts on M in an obviousway. We check the axioms and conclude that M is an Rmodule.

6. Let F be a field and let R = F [x]. Let M be any vector space over F . Mis an F -module, but it does not have an action yet. Let : M M beany F linear map. Define (m) = xm. More generally, define

xnm = ((. . . (m) . . . ))n times

= nm

and if f =i aix

i F [x], define

fm =i

aiFimM

.

46

This makes M into an F [x]module. F [x]modules = F - vectorspaces equipped with a linear map .

7. Recall that a representation of a group G is simply a group homomorphism : G GL(V ) where V is a finite dimensional vector space over, say, C.If R is the group ring of G, i.e. R = Cg1 Cg2 Cgn is a vectorspace with basis G, where G = {g1, . . . , gn} and g h = gh and extendslinearly. Then R becomes a ring and V becomes an Rmodule.

Definition 6.2. Let R be a ring andM1 andM2 be Rmodules. An Rmodulehomomorphism f : M1 M2 is a homomorphism of abelian groups f : M1 M2 such that f(rm1) = rf(m1). An endomorphism of an Rmodule M is anRmodule homomorphism f : M M . An isomorphism of Rmodules isan Rmodule homomorphism f : M1 M2 such that there exists g : M2 M1 also an Rmodule homomorphism such that f g = g f = identity.Equivalently, an Rmodule homomorphism is an isomorphism if it is bijective.An automorphism is a bijective endomorphism.

Example.

1. Let R be a field and V,W be Rmodules (i.e. vector spaces) then anRmodule homomorphism V W is just a linear map V W.

2. R = Z. An Rmodule is an abelian group and an Rmodule homo-morphism f : A B is an abelian group homomorphism (because, e.g.f(3a) = 3f(a) automatically as f(3a) = f(a+a+a) = f(a)+f(a)+f(a) =3f(a)).

3. (obscure) If f : A B is a ring homomorphism, then B becomes anAmodule. Thus:

a b = f(a)b B(easy exercise: show that the axioms are met and f is an Amodulehomomorphism).

4. Let R be a commutative ring and M1 = Rn and M2 = Rm. Let X bethe m n matrix with entries in R. Then X induces an Rmodulehomomorphism M1 M2 in an obvious way: XV is the result of matrixmultiplication.

Definition 6.3. LetM be anRmodule. A subgroup P ofM is anRsubmoduleor just a submodule, if r R,p P, rp P.Example.

1. Let R be a field = M a vector space and P is just a subspace.2. Let R be a ring and M = R. A submodule of R is a subgroup I which

is closed under left multiplication by R, i.e. r R, i I = ri I.Therefore, a submodule of R is a left ideal of R.

3. Say that R is a ring and M1,M2 are Rmodules and f : M1 M2is an Rmodule homomorphism. The kernel of f (as an abelian group

47

homomorphism) is a submodule of M1 if the image of f is a submodule ofM2.Proof. Set K = ker f = {m1 M1 : f(m1) = 0}. By group theory, weknow that K is an abelian subgroup. Say k K and r R, then

f(rk) = rf(k) = r0 = 0 = rk K

so K is a submodule. Now let I = Imf = {f(m1) : m1 M1} M2. Bygroup theory, I is a subgroup. Now say i I and r R. Then by definitioni = f(m1) for some m1 M1 and note that f(rm1) = rf(m1) = ri whichis in I. Hence ri Iand I is indeed a submodule.

Quotient Modules

Let R be a ring, M an Rmodule and P M be a Rsubmodule. By grouptheory, we know that the quotient M/P is an abelian group. Now, we want togive M/P the structure of an Rmodule. A general element of M/P is of theform m+ P. Define

r(m+ P ) = rm+ P.

Is this well-defined? Well, say m+P = m+P, i.e. m = m+ p for some p P .To be well-defined, we need rm+ P = rm + P. But this is true, as

rm rm = rm+ rp rm = rpand P is a submodule: rp P = rm + P = rm + P. So M/P has anatural Raction. From here, it is easy to check the axioms.Definition 6.4. The quotient module M/P is an abelian group with the aboveaction (1st isomorphism for vector spaces (?) : V W, V/ ker = Im.Therefore, dimV dim(ker) = dim(Im)).Proposition 6.5. Let R be a ring (not necessary, actually) and f : M Nan Rmodule homomorphism. Then the natural map from M/ ker f to Imf isan isomorphism of Rmodules.

Proof. f is an abelian group homomorphism. By the first isomorphism theoremfor groups, there exists a natural map M/ ker f Imf given by m + ker f 7f(m). This map is an isomorphism of abelian groups. It is only left to checkthat this is an Rmodule homomorphism, i.e. STP that r R,m M,r(m+ ker f) rf(m), i.e. STP f(rm) = rf(m). But this is true by definitionof an Rmodule.Example. The abelian group {0} is an Rmodule. r0 = 0 for all r. It isR/I if I = R. Further, N, ! an Rmodule homomorphism 0 N given by0 7 0 and M, ! Rmodule homomorphism M 0 given by m 7 0 m. Thecomposition of these maps is the zero map 0 : M N such that m 7 0 for allm.

Definition 6.6. IfM,N are Rmodules, their direct sumMN is the abeliangroup M N with R acting diagonally, i.e. r(m,n) = (rm, rn). More generally,if Mi are modules for 1 i n, then

ni=1Mi is too: r(m1,m2, . . . ,mn) =

(rm1, rm2, . . . , rmn).

48

We have already seen an example for this: Rn = R R.Definition 6.7. A moduleM for a ring R is finitely generated if m1, . . . ,mn M such that any m M can be written as

m = r1m1 + + rnmn.Remark. {r1m1+ +rnmn : ri R} is checked to be the smallestRsubmoduleof M containing {m1, . . . ,mn}. Hence m1, . . . ,mn generate M the small-est submodule of M containing m1, . . . ,mn is M .

Remark. As opposed to the study of vector spaces, bases are much rarer in thetheory of modules - most modules are not free, i.e. have no basis.

Remark. If R is a commutative ring and I is an ideal, then I is f.g. as an ideal I is f.g. as Rmodule.Lemma 6.8. A module M is finitely-generated n 0 and a surjectionpi : Rn M of Rmodules (i.e. pi is an Rmodule homomorphism).Remark. If R is a field, an Rmodule V is f g dimV < , and thenV = Rn for some n. Lemma 6.8. is some weak analogy of this for general R.

Proof.( = ) Say M is generated by m1, . . . ,mn. Define pi : Rn M by

pi(r1, . . . , rn) =rimi. It is easy to check that pi is an Rmodule homo-

morphism. By the definition of finitely generated, pi is a surjection.

(=) Say pi : Rn M is a surjection of Rmodules. Set

mi = pi(0, 0, . . . 1ith place

, 0, . . . 0).

I claim that the mi for 1 i n generate M. For if m M, pi isa surjection. Therefore there exist r1, . . . , rn such that pi(r1, . . . , rn) =m. But if ei = (0, 0, . . . 1

ith place, 0, . . . 0), then (r1, . . . , rn) =

riei and

applying pi, we getm =

ripi(ei) =

rimi

where we used the fact that pi is a Rmodule homomorphism.

Remark. Q is not a f.g. Zmodule. For if m1, . . . ,mn Q,mi = cidi withci, di Z, di 6= 0, and if D = gcd(di), then each mi is of the form aiD , ai Z andhence

rimi is also of this form if ri R = Z. Therefore

1

2D6=

rimi, ri Z.

Therefore Q is not finitely generated as Zmodule.Definition 6.9. Say M1, . . . ,Mn are Rmodules and if fi : Mi Mi+1 areRmodule homomorphisms, 1 i n 1. We say that this collection ofmodules and maps is a complex if fi fi1 = 0 for 2 i n 1. Equivalently,it is a complex iff

49

Imfi1 ker fi 2 i n 1.The image of fi1 and the kernel of fi are both submodules of M.

M1f1 // M2

f2 // M3f3 // . . . fn2 // Mn1

f1 // Mn

need Imf1 ker f2

OO

Imf2 ker f2

OO

Imfn2 ker fn1

OO

The definition needs to be checked at Mi iff whenever Mi has an arrow in andan arrow out (definition continued). We say that this collection of modules andmaps is an exact sequence if Im(fi1) = ker(fi) for 2 i n 1. A morphismof exact sequences (i.e. a homomorphism of exact sequences) from

M1 // M2 // . . . // Mn

toN1 // N2 // . . . // Nn

is a collection of maps fi : Mi Ni such that the resulting diagram

M1d1 //

f1

M2d2 //

f2

. . .dn1 //

fn1

Mn

fn

N1

e1 // N2e2 // . . . e3 // Nn

commutes.

To check that the diagram commutes it is necessary and sufficient to check thateach square

Midi //

fi

Mi+1

fi+1

Ni

ei // Ni+1

commutes for 1 i n 1. The same definition (for commutative) holds forcomplexes in general.Remark. If

0 // A // B // C // 0

and0 // A // B // C // 0

are two short exact sequences, then a map between them is a collection of threemaps: A A, B B, C C as there is a unique map 0 0 and similarly,we only need to check that the two middle squares commute. Again, somethingthat has to be checked at Mi Mi has an arrow in and an arrow out.

50

Note. One also sees examples where the sequence goes off to infinity in either,or both, directions. We can make an (obvious) modification to the definition:

M1 // M2 // . . . // Mn // . . .

// M2 // M1 // M0 // . . .

Example.

1. Af // B // 0 with B 0 the zero map. This sequence is exact iff

Imf = ker(zero map) Imf = B f is injective.

2. 0 // Af // B is exact iff

ker f = Im(zero map) ker f = 0 f is injective.

3. 0 // Af // B // 0 is exact iff it is exact at A and B f is

injective and surjective f is an isomorphism.

The next example is so ubiquitous, it has its own special name.

Definition 6.10. An exact sequence of the following form

0 // A // B // C // 0

is called a short exact sequence. Alternatively, it is called an extension of C byA.

Definition 6.11. A commutative diagram of R-modules and R-module mapsis a collection of R-modules and maps between them such that any two inducedmaps between any two R-modules in the diagram coincide.

What does it actually mean for a complex

0 // Af // B

g // C // 0

to be a short exact sequence?Well, exactness at A implies that Im 0 = ker f , i.e. f is injective. Exactness atC implies that g is surjective. Exactness at B says that Imf = ker g.Say we only know g : B C. Can we reconstruct f and A?The answer is yes. f is injective and hence identifies A with ker g = Imf . Hencegiven g, the only way (up to isomorphism) of building a short exact sequencewith g : B C is

0 // ker gInclusion// B

g // C // 0

51