Algebra II

Algebra II: Rings and Modules Kevin McGerty Typed up by Alex McKenzie April 28, 2014

Contents 1 Rings 1.1 Definitions and Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1.2 Basic Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1.3 The Field of Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2 2 4 5

2 Homomorphisms and Ideals 2.1 Basic Properties of Ideals . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2.2 Quotients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2.3 The Isomorphism Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . .

6 7 8 9

3 Prime Ideals, Maximal Ideals and Construction of Rings

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4 An Introduction to Fields

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5 Unique Factorisation 5.1 Irreducible Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

15 20

6 Modules 6.1 Definition and Examples . . . . . . . . . . . . . . . . . .

6.2 Submodules, Generation and Linear Independence . . . .

6.3 Quotient Modules and Isomorphism Theorems . . . . . .

6.4 Free, Torsion and Torsion-Free Modules . . . . . . . . . .

6.5 Presentations of Finitely Generated Modules over a PID

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7 Matrices over R and Normal Forms

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8 The Canonical Form for Finitely Generated Modules

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9 Rational and Jordan Canonical Forms

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1

1

Rings

1.1

Definitions and Examples

Definition 1.1. A ring is a datum (R, +, ×, 0, 1) where R is a set, 1, 0 ∈ R and +, × are binary operations on R such that 1. R is an abelian group under + with identity element 0.

2. The binary operation × is associative and 1 × x = x × 1 = x for all x ∈ R. 3. Multiplication distributes over addition: x × (y + z) = (x × y) + (x × z), ∀x, y, z ∈ R. We say R is a commutative ring if × is commutative.

Remark 1.1. We will suppress the symbol for the operation × and write · or nothing instead.

Example 1.2.

• The integers Z form the fundamental example of a ring. Similarly if n ∈ Z then Z/n, the integers modulo n, form a ring.

• The subset Z[i] = {a + ib ∈ C : a, b ∈ Z}, known as the Gaussian Integers, is a ring under the normal operations on complex numbers.

• Any field is a ring.

• If k is a field and n ∈ N then the set Mn (K) of n × n matrices with entries in k is a ring, with the usual operations of addition and multiplication.

• If V is a vector space over a field k then End(V ), the space of linear maps from V to itself, is a ring, where multiplication is given by composition. This ring is not commutative.

• Given any ring R and n ∈ N, Mn (R) is a ring.

• Polynomials in any number of indeterminates with coefficients in a field form a ring.

Given the field k and indeterminates t1 , . . . , tn we write k[t1 , . . . , tn ] for said ring.

• Given a ring R the set of polynomials R[t] is a ring.

• If R and S are rings, then R ⊕ S = {(r, s) : r ∈ R, s ∈ S} is a ring, with operations defined component-wise.

• For any set X and ring R, the set R× = {f : X → R} is a ring, with multiplication and addition as usual for functions.

• In particular, the set of all functions f : R → R is a ring. Moreover, the set of all continuous (or differentiable, or infinitely differentiable) functions also forms a ring.

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Definition 1.3. If R is a ring, a subset S ⊆ R is said to be a subring if 0, 1 ∈ S and S is closed under addition and multiplication.

Lemma 1.1. (Subring Criterion). Let R be a ring and S a subset of R. Then S is a subring if and only if 1 ∈ S and for all s1 , s2 ∈ S we have s1 s2 , s1 − s2 ∈ S. Definition 1.4. A map f : R → S between rings R and S is said to be a ring homomorphism if 1. f (1R ) = 1S , 2. f (r1 + r2 ) = f (r1 ) + f (r2 ), 3. f (r1 ¡ r2 ) = f (r1 ) ¡ f (r2 ) Remark 1.2. Strictly speaking we might have written +R and +S for the addition operations in the two different rings R and S, and similarly for the multiplication operation. Note that from 2 it follows that f (0) = 0.

Example 1.5.

1. For each positive integer n, there is a natural ring homomorphism from Z to Z/nZ which an integer to its equivalence class modulo n.

2. LetPV be a k-vector and let Îą ∈ Endk (V ). Then φ : k[t] → Endk (V ) given by Pn space n i i φ( i=0 ai t ) = i=0 ai Îą is a ring homomorphism.

3. The inclusion map i : S → R of a subring S into a ring R is a ring homomorphism.

a −b 4. Let A = : a, b ∈ R . It is easy to check this A is a subring of M2 (R). b a

a −b The map φ : C → A given by a + ib 7→ is a ring isomorphism. (This b a homomorphism arises by sending a complex number z to the map of a plane to itself given by multiplication by z.)

Definition 1.6. For a ring R, the characteristic of R, denoted Char(R), is the smallest integer d such that 1R added to itself d times equals 0R . If this repeated sum never reaches 0R , the ring is said to have characteristic 0.

Definition 1.7. If R is a ring, we define R[t], the ring of polynomials in R, to be the set {(an ) : n ∈ N} where for each (an ) there is some N ∈ N such that an = 0 for all n ≼ N . The operations are defined thusly: • (an ) + (bn ) = (an + bn ); P • (an ) ¡ (bn ) = ( nk=0 ak bn−k ).

We define R[[t]], the ring of formal power series, as above but without the restriction that only finitely many an s are nonzero.

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Proposition 1.2. The above operations are well-defined, such that R[t] and R[[t]] are indeed rings. Also, given (an ) ∈ R[t] and N as above and setting t = (0, 1, 0, . . .) gives: (an ) =

N X

an tn

n=0

Lemma 1.3. (Evaluation homomorphisms) Let R, S be rings and φ : R → S a ring homomorphism. If s ∈ S then there is a unique ring homomorphism Φ : R[t] → S such that Φ ◦ i = φ (where i : R → R[t] is the inclusion of R into R[t]) and Φ(t) = s.

P Proof. Any element of R[t] has the form ni=0 ai ti , hence if Θ is any homomorphism satisfying Θ ◦ i = φ and Θ(t) = s we see that ! n n n n X X X X i i i Θ ai t = Θ(ai t ) = Θ(ai )Θ(t ) = φ(ai )si . i=0

i=0

i=0

i=0

Hence Θ is uniquely determined. P To check there P is indeed such a homomorphism we just have to check that the function Φ ( ni=0 ai ti ) = ni=0 φ(ai )si is indeed a homomorphism, but this is straight-forward from the definitions.

1.2

Basic Properties

Remark 1.3. From now on, unless stated otherwise, all rings are commutative.

Definition 1.8. If R is a ring, then an element a ∈ R \ {0} is said to be a zero-divisor if there is some b ∈ R \ {0} such that a · b = 0. A ring which is not the zero ring and has no zero-divisors is called an integral domain.

Example 1.9. If R is a ring, then R2 is a again a ring, and (a, 0) · (0, b) = (0, 0) so that (a, 0) and (0, b) are zero-divisors. The ring Mn (k) of n × n matrices over a field k also has lots of zero divisors. The integers modulo n have zero-divisors whenever n is not prime.

On the other hand, all fields are integral domains. Z is an integral domain. If R is an integral domain, then so are R[t] and R[[t]].

Lemma 1.4. Suppose that R is an integral domain. Then any subring S of R is also an integral domain. Moreover, char(R), the characteristic of R, is either zero or a prime p ∈ Z. Proof. It is clear from the definition that a subring of an integral domain must again be an integral domain. Now from the definition of the characteristic of a ring, if char(R) = n > 0 then Z/nZ is a subring of R. Clearly if n = a · b where a, b ∈ Z are both greater than 1, then aR · bR = 0 in R with neither aR nor bR zero, thus both are zero divisors. It follows that if R is an integral domain then char(R) is zero or prime.

Definition 1.10. Let R be a ring. The subset R× = {r ∈ R : for some s ∈ R, r · s = 1} is called the group of units in R - it’s a group under the multiplication operation with identity element 1.

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Example 1.11.

1. The units in Z form the group {±1}.

2. If k is a field, then k × = k \ {0}.

3. If R = Mn (k) then the group of units is GLn (k).

Lemma 1.5. Let R be an integral domain which has finitely many elements. Then R is a field.

Proof. We need to show that if a ∈ R \ {0} then a has a multiplicative inverse - that is, we need to show there are a, b ∈ R with a · b = 1.

Consider the map ma : R → R given by left multiplication by a, so that ma (x) = a · x.

We claim that ma is injective: if ma (x) = ma (y) then we have a · x = a · y ⇒ a · (x − y) = 0 and since R is an integral domain and a 6= 0 it follows that x − y = 0, that is, x = y. But now since R is finite, an injective map must be surjective, and hence there is some b ∈ R with ma (b) = 1, that is, a · b = 1 as required.

1.3

The Field of Fractions

Lemma 1.6. For a field R, The relation ∼ on R × R \ {0}, defined by (a, b) ∼ (c, d) iff a · d = b · c, is an equivalence relation.

Proof. Symmetry and reflexivity are immediate. For transitivity, suppose that (a, b) ∼ (c, d) and (c, d) ∼ (e, f ). Then we have ad = bc and cf = de and need to check that af = be. af − be = 0 ⇐⇒ d · (af − be) = 0 ⇐⇒ (ad) · f − b · (de) = 0 ⇐⇒ (bc) · f − b · (cf ) = 0 ⇐⇒ bcf = bcf As required.

Lemma 1.7. the binary operations (R × R \ {0}) × (R × R \ {0}) → R × R \ {0} given by: ((a, b), (c, d)) 7→ (ad + bc, bd) ((a, b), (c, d)) 7→ (ac, bd) induce binary operations on F (R), the set of equivalence classes as above.

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Proof. Note first that since R is an integral domain and b, d are nonzero, hence the above formulas do indeed define binary operations R × R \ {0}. To check that they induce binary operations on F (R) we need to check that the equivalence classes of the pairs on the righthand side depends only on the equivalence classes of the two pairs on the left-hand side. We check this for the first operation (the second one being similar but easier).

Suppose that (a1 , b1 ) ∼ (a2 , b2 ) and (c1 , d1 ) ∼ (c2 , d2 ) so that a1 b2 = a2 b1 and c1 d2 = c2 d1 .

Then we need to show that (a1 d1 + b1 c1 , b1 d1 ) ∼ (a2 d2 + b2 c2 , b2 d2 ) which holds if and only if (a1 d1 + b1 c1 )(b2 d2 ) = (a2 d2 + b2 c2 )(b1 d1 ) which holds if and only if (a1 b2 d1 d2 ) + (b1 b2 c1 d2 ) = (a2 b1 d1 d2 ) + (b1 b2 c2 d1 ).

But a1 b2 = a2 b1 so a1 b2 d1 d2 = a2 b1 d1 d2 and b2 c1 = b1 c2 so b1 b2 c1 d2 = b1 b2 c2 d1 and we are done.

Remark 1.4. We let + and × denote the binary operations the first and second operations above induce on F (R), and we write ab for the equivalence class of a pair (a, b). Thus we have ad + bc a c ac a c + = ; × = b d bd b d bd Theorem 1.8. The above formulas give well-defined addition and multiplication operations on F = F (R) the set of equivalence classes { ab : a, b ∈ R, b 6= 0}, and F is a field with respect to these operatios with additive identity 01 and multiplicative identity 11 . Moreover there is an injective homomorphism ι : R → F (R) sending a 7→ a1 .

Remark 1.5. Proof of Theorem 2.5 is not examinable.

Definition 1.12. The field F (R) is known as the field of fractions of R. Proposition 1.9. Let k be a field and let θ : R → k be an injective homomorphism. Then there is a unique injective homomorphism θ˜ : F (R) → k extending θ (in the sense that ˜ k = θ). θ| Proof of this proposition is not examinable.

2

Homomorphisms and Ideals

Definition 2.1. Let f : R → S be a ring homomorphism. The kernel of f is ker(f ) = {r ∈ R : f (r) = 0} and the image of f is im(f ) = {s ∈ S : ∃ r ∈ R, f (r) = s} Definition 2.2. Let R be a ring. A subset I ⊆ R is called an ideal if it is a subgroup of (R, +) and moreover for any a ∈ I and r ∈ R we have a · r ∈ I. Lemma 2.1. If f : R → S is a homomorphism, then ker(f ) is an ideal.

Proof. Immediate from definitions.

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2.1

Basic Properties of Ideals

Lemma 2.2. Let R be a ring, and I, J ideals in R. Then I + J, I ∩ J and IJ are ideals, where I + J = {i + j : i ∈ I, j ∈ J}; ( n ) X IJ = ik jk : ik ∈ I, jk ∈ I, n ∈ N k=1

Moreover we have IJ ⊆ I ∩ J and I, J ⊆ I + J. Proof. For I + J it is clear that this is an abelian subgroup of R, while if i ∈ I, j ∈ J and r ∈ R, then r(i + j) = r · i + r · j ∈ I + J as both I and J are ideals, hence I + J is an idea.

It is easy to see that an arbitrary intersection of ideals is an ideal.

To see that IJ is an ideal, note P that it is clear that the sum of two elements of IJ is clearly of the same form, and if nk=1 xk yk ∈ IJ then −

n X

xk yk =

k=1

n X (−xk )yk ∈ IJ k=1

since if xk ∈ I then −xk ∈ I. Thus IJ is an abelian subgroup. It is also straightforward to check the multiplicative condition. The containments are all clear once you note that if i ∈ I and j ∈ J then ij is in I ∩ J because both I and J are ideals.

Definition 2.3. Given any subset T of R, we define \ hT i = I T ⊆I; I ideal

to be the ideal generated by T .

Lemma 2.3. Let T ⊆ R. Then we have ( n ) X hT i = rk tk : rk ∈ R, tk ∈ T, n ∈ N i=1

Proof. Let I denote the right hand side.

I is an ideal. P The multiplicative Pn We first check thatP n property is immediate: if r ∈ R and k=1 rk xk ∈ I then r ( k=1 rk xk ) = nk=1 (r · rk )xk ∈ I. Moreover the sum of twoP such elements Pis certainly of the same form, and I is closed under additive inverses since − nk=1 rk xk = nk=1 (−rk )xk , so that it is an additive subgroup of R. It remains to show that if J is an ideal containing X then J contains I. If {xP 1 , . . . , xk } ⊆ T ⊆ J and r1 , . . . , rk ∈ R, then since J is an ideal certainly rk xk ∈ J and hence nk=1 rk xk ∈ J. Since the xk , rk and n ∈ N were arbitrary it follows that I ⊆ J as required.

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Remark 2.1. If T ⊂ R is a subset of a ring R we can also consider the subring which it generates: the intersection of subrings is again a subring, so we may set \ hT iS = S T ⊆S

Where the subscript ’s’ denotes ’subring’.

Definition 2.4. If an ideal is generated by a single element we say it is principal. Two elements a, b ∈ R are said to be associates if there is a unit u ∈ R× such that a = u · b. (This is an equivalence relation on elements of R.) Lemma 2.4. Let R be a domain. Then if I is a principal ideal, the generators of I are associates, and any associate of a generator is again a generator. Thus the generators of a principal ideal form a single equivalence class of associate elements of R. Proof. If I = {0} = h0i the claim is immediate, so assume I 6= {0} and hence any generator is nonzero. Let a, b ∈ R be generators of I, so I = hai = hbi. Since a ∈ hbi, there is some r ∈ R with a = r · b, and similarly as b ∈ hai there is some s ∈ R with b = s · a. It follows that a = r · b = r · (s · a) = (r · s) · a, hence a(1 − r · s) = 0, and so since a 6= 0 and R is an integral domain, r · s = 1, that is r and s are units.

Finally, if I = hai and b = u · a where u ∈ R× then certainly b ∈ hai = I so that hbi ⊆ I, but also if x ∈ I then x = r · a for some r ∈ R and hence x = r · (u−1 · b) = (r · u−1 ) · b so that x ∈ hbi, and hence I ⊆ hbi. It follows that I = hbi.

2.2

Quotients

Theorem 2.5. The datum (R/I, +, ×, 0 + I, 1 + I) defines a ring structure on R/I and moreover the map q : R → R/I given by q(r) = r + I is a surjective ring homomorphism.

Proof. Checking each axiom is a consequence of the fact that the binary operations +, × on R/I are defined by picking arbitrary representatives of the cosets, computing up in the ring R and then taking the coset of the answer (the important part of the definitions being that this last step is well-defined).

Finally, the map q : R → R/I is clearly surjective, and that it is a homomorphism is also immediate from the definitions.

Remark 2.2. The map q : R → R/I is called the quotient homomorphism.

Corollary 2.6. Any ideal is the kernel of a ring homomorphism.

Proof. If I is an ideal and q : R → R/I is the quotient map then clearly ker(q) = I. Lemma 2.7. Let R be a ring, I an ideal in R and q : R → R/I the quotient map. If J is an ideal then q(J) is an ideal in R/I, and if K is an ideal in R/I then q −1 (K) = {r ∈ R : q(r) ∈ K} is an ideal in R which contains I. Moreover, these correspondences give a bijection between the ideals in R/I and the ideals in R which contain I. 8

Proof. There are a number of things to check but all follow from the definitions.

2.3

The Isomorphism Theorems

Theorem 2.8. (First Isomorphism Theorem) Let f : R → S be a homomorphism of rings, and let I = ker(f ). Then f induces an isomorphism f : R/I → im(f ) given by: f (r + I) = f (r) Proof. First note that if r − s ∈ I, then f (r − s) = 0 and hence f (r) = f (s), so that f takes a single value on each coset r + I, and hence f is well-defined.

Clearly from the definition of the ring structure on R/I it is a ring homomorphism, so it only remains to check that it is an isomorphism. It is clearly surjective, as if s ∈ im(f ) then s = f (r) for some r ∈ R and hence s = f (r + I) = 0. To see that f is injective it is enough that ker(f ) = 0, but this is also obvious, as if f (r + I) = 0 then f (r) = 0 hence r ∈ I so that r + I = I. Theorem 2.9. (Second Isomorphism Theorem) Let R be a ring and A a subring of R and B an ideal of R. Then A + B is a subring of R and the natural map R/I ∩ J → R/I induces an isomorphism A/(A ∩ B) ∼ = (A + B)/B Proof. We checked in Problem Set 1 that A + B is a subring of R, and it is easy to see that A ∩ B is an ideal in A + B, so the two quotients A/(A ∩ B) and (A + B)/B certainly exist.

Let q : R → R/B be the quotient map. It restricts to a homomorphism p from A to R/B, whose image is clearly (A + B)/B so by the first isomorphism theorem it is enough to check that the kernel of p is A ∩ B. This is clear: let a ∈ A. Then p(a) = 0 ⇔ a + B = 0 ⇔a∈B ⇔a∈A∩B completes the proof.

Theorem 2.10. (Universal property of quotients) If f : R → S is a homomorphism and I ⊆ ker(f ) is an ideal, then there is a unique homomorphism f : R/I → S such that f = f ◦q where q : R → R/I is the quotient map.

Proof. Note that if r1 − r2 ∈ I then since I ⊆ ker(f ), we have 0 = f (r1 − r2 ) = f (r1 ) − f (r2 ) and hence f (r1 ) = f (r2 ). It follows f is constant on the I-cosets, and so induces a map f : R/I → S. it is then immediate from the definitions of the ring structure on R/I that f is a homomorphism, and by construction we have f ◦ q = f as required.

To see that f is unique, suppose g : R/I → S is a homomorphism such that f = g ◦ q.

Then for all r + I ∈ R/I we have g(r + I) = g(q(r)) = f (r), hence g(r + I) = f (r + I) and g = f as required.

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Theorem 2.11. (Third Isomorphism Theorem) If I ⊆ J are ideals, then J/I = {j + I : j ∈ J} is an ideal in R/I and (R/I)/(J/I) ∼ = R/J Proof. Let qI : R → R/I and qJ : R → R/J be the two quotient maps. By the universal property for qJ , we see that there is a homomorphism qI : R/J → R/I induced by the map qI : R → R/I and qI ◦ qJ = qI . Clearly qI is surjective (since qI is) and if qi (r + J) = 0 then since r + J = qJ (r) so that qI (r + J) = qI (r) we have r ∈ I, so that ker(qI ) = I/J and the result follows by the first isomorphism theorem.

Example 2.5. Suppose that V is a k-vector space and α ∈ End(V ). Then we saw before φ : k[t] → End(V ) given by φ(f ) = f (α). It is easy to see that this map is a homomorphism, and hence we see that im(φ) is isomorphic to k[t]/I where I = ker(f ) is a principal ideal.

The monic polynomial generating I is the minimal polynomial of α as was studied in Algebra I. Theorem 2.12. (General Chinese Remainder Theorem) Let R be a ring, and I, J ideals of R such that I + J = R. Then R/(I ∩ J) ∼ = R/I ⊕ R/J Proof. We have quotient maps q1 : R → R/I and q2 : R → R/J. Define q : R → R/I ⊕ R/J by q(r) = (q1 (r), q2 (r)). By the first isomorphism theorem, it is enough to show that q is surjective and that ker(q) = I ∩ J. The latter is immediate: if q(r) = 0 then q1 (r) = 0 and q2 (r) = 0, whence r ∈ I and r ∈ J, that is r ∈ I ∩ J. To see that q is surjective, suppose (r + I, s + J) ∈ R/I ⊕ R/J. Then since R = I + J we may write r = i1 + j1 and s = i2 + j2 , where i1 , i2 ∈ I and j1 , j2 ∈ J. But then r + I = j1 + I and s + J = i2 + J, so that q(j + 1 + i2 ) = (r + I, s + J).

Remark 2.3. We proved in a problem sheet that this reduces to the classical statement of the Chinese Remainder Theorem.

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Prime Ideals, Maximal Ideals and Construction of Rings

Definition 3.1. Let R be a ring, and I an ideal of R. We say that I is a maximal ideal if it is not strictly contained in any proper ideal of R. We say I is a prime ideal if I 6= R and for all a, b ∈ R, whenever a · b ∈ I then either a ∈ I or b ∈ I. Lemma 3.1. An Ideal I in a ring R is prime if and only if R/I is an integral domain. It is maximal if and only if R/I is a field. In particular, a maximal ideal is prime.

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Proof. Suppose that a · b ∈ R. Note that (a + I)(b + I) = 0 + I if and only if a · b ∈ I. Thus if R/I is an integral domain, (a + I)(b + I) = 0 forces either a + I = 0 or b + I = 0, that is, a or b lies in I, which shows I is prime. The converse is similar.

For the second part, note that a field is a ring which has no nontrivial ideals. The claim then follows immediately from the correspondence between ideals in the quotient ring and the original ring given in Lemma 2.7. Since fields are obviously integral domains, the in “particular” claim follows immediately.

Example 3.2. Let R = Z. It is easy to see that every ideal is of the form nZ. The ideal nZ is prime if and only if n is 0 or prime, and since a finite integral domain is a field, the nonzero prime ideals are also maximal.

P Lemma 3.2. (Division Algorithm) Let R be a ring and f = ni=0 ai ti ∈ R[t], where an ∈ R× .

Then if g ∈ R[t] is any polynomial, there are unique polynomials q, r ∈ R[t] such that either r = 0 or deg(r) < deg(f ) and g = q · f + r.

Proof. By induction on deg(g). Since an ∈ R× , if h ∈ R[t] \ {0} it is easy to see that deg(f · h) = deg(f ) + deg(h). It follows that if deg(g) ) we must take q = 0 and thus Pm < deg(f j r = g. If m = deg(g) ≥ n = deg(f ), then if g = j=0 bj t where bm 6= 0 the polynomial m−n h = g − a−1 f n bm t

has deg(h) < deg(g), and so there are unique q 0 , r0 with h = q 0 · f + r0 . Setting q = m−n a−1 + q 0 and r = r0 it follows that g = q · f + r. Since q and r are clearly uniquely n bn t determined by q 0 and r0 they are also unique as required.

Lemma 3.3. Let I be a nonzero ideal in k[t]. Then there is a unique monic polynomial f such that I = hf i. Thus all ideals in k[t] are principal.

Proof. Since I is nonzero we may pick an f ∈ I of minimal degree, and rescale it if necessary to make it monic. We claim I = hf i. Indeed if g ∈ I, then using the division algorithm, we may write g = q · f + r where either r = 0 or deg(r) < deg(f ). But then r = g − q · f ∈ I, and thus by minimality of degree of f ∈ I we must have r = 0 and so g = q · f as required.

The uniqueness follows from the fact that if I = hf i = hf 0 i then we would have f = a · f 0 and f 0 = b · f , for some polynomials a, b ∈ k[t]. But then f = a · f 0 = (ab) · f so that a and b must have degree zero, that is, a · b ∈ k. Since we required f and f 0 to be monic, it follows that a = b = 1 and so f = f 0 as required.

Lemma 3.4. Let f, g ∈ k[t] be nonzero polynomials. Then there is a unique monic polynomial d ∈ k[t] such that d = a · f + b · g. We write d = hcf(f, g), the highest common factor of f, g as if c divides f and g then c divides a · f + b · g.

Proof. Let I = hf, gi. By Lemma 3.3 it follows that I = hdi for a unique monic polynomial d. Then certainly f ∈ hf i ⊆ I = hdi so that d divides f , and similarly d divides g. Since I = hf, gi = {r · f + s · g : r, s ∈ k[t]} and d ∈ I it is also clear that we may find polynomials a, b with d = a · f + b · g as required.

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Lemma 3.5. Let k be a field and I = hf i a non-zero ideal in k[t]. Then I is prime if and only if f ∈ k[t] is an irreducible polynomial. Moreover such an ideal is in fact maximal, and all maximal ideals are of this form.

Proof. If hf i is a nonzero prime ideal (so that f 6= 0) and f = g · h then f | g or f | h; WLOG let f | g. But then g = f · k and so f = (f · k) · h, and f (1 − k · h) = 0. But k[t] is a domain, so it follows that k · h = 1 and h is a unit. Thus f is irreducible as claimed.

Conversely, suppose that f is irreducible and suppose that f divides a product g · h. We must show that it divides one of g or h. But if f does not divide g, then the highest common factor of f and g must be 1. But then by Bezout’s Lemma we have 1 = af + bg for some a, b ∈ k[t] and so h = h · 1 = h(af + bg) = f · (ah) + b(gh) so that f clearly divides h as required.

To see the moreover part, suppose that M is a maximal ideal. Then it is certainly prime, and so by Lemma 3.3 and the above, M = hf i for some irreducible f . On the other hand, if I = hf i is a prime ideal, then suppose that I ⊂ J for some proper ideal J. Then the ideal J must be principal by Lemma 3.3 again, so that J = hgi. But then f = g · h for some h ∈ k[t], where since J is proper, deg(g) > 0. But then since f is irreducible, we must have h ∈ k, and so h is a unit, and I = J as required.

Remark 3.1. As noted before, if R is an integral domain then R[t] is also (this is easy to see by considering the degree function). It follows that {0} is a prime ideal in k[t]. The above Lemma (Lemma 3.5) then shows that this is the only prime ideal which is not maximal.

4

An Introduction to Fields

Definition 4.1. Let E and F be fields with F ⊂ E, such that E is finite-dimensional as an F -vector space. Then • We say this vector space is a field extension, and write denote it E/F .

• the degree of the field extension E/F is defined as dimF (E) and written [E : F ].

Lemma 4.1. Let E/F be a field extension and let d = [E : F ] < ∞. Then if V is an E-vector space, we may view V as an F -vector space, and V is finite dimensional as an F -vector space if and only if it is finite dimensional as an E-vector space, and moreover dimF (V ) = [E : F ] dimE (V ).

Proof. Certainly if V is an E-vector space then by restricting the scalar multiplication map to the subfield F it follows that V is an F -vector space. Moreover, if V is finite dimensional as an F -vector space it is so as an E-vector space (a finite F -spanning set will certainly be a finite E-spanning set).

Conversely, suppose that V is a finite dimensional E-vector space. Let {x1 , x2 , . . . , xd } be an F -basis of E, and let {e1 , e2 , . . . , en } be an E-basis of V . To finish the proof it is enough to check that B := {xi ej : 1 ≤ i ≤ d, 1 ≤ j ≤ n} is an F -basis of V .

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First, we check that B spans V . If vP ∈ V , then since {e1 , . . . , en } is an E-basis of V there are λi ∈ E for 1 ≤ i ≤ n such that v = ni=1 λi ei . Moreover, since {x1 , . . . , xd } is an F -basis P of E then for each λi there are elements µij for 1 ≤ j ≤ d such that λi = dj=1 µij xj . Thus we have ! n n d X X X X v= λi ei = µij xj ei = µij (xj ei ). i=1

i=1

j=1

1≤i≤n 1≤j≤d

Hence the set {xj ei : 1 ≤ i ≤ n, 1 ≤ j ≤ d} spans V as an F -vector space (and in particular we have already established that V is finite dimensional as an F -vector space).

To see that B is linearly independent, and hence establish the dimension formula, just notice that in the above equation, v = 0 if and only if each λi = 0 by linear independence of the vectors {e1 , . . . , en }, and λi = 0 if and only if each µji = 0 for 1 ≤ j ≤ d by the linear independence of {x1 , . . . , xd }.

Example 4.2. Let V be a C-vector space with basis {e1 , . . . , en }. Then {e1 , . . . , en , ie1 , . . . , ien } is an R-basis of V .

Corollary 4.2. (Tower Law) Let F ⊂ E ⊂ K be fields. Then [K : F ] is finite if and only if both degrees [E : F ], [K : E] are, and when they are finite we have [K : F ] = [E : F ][K : E].

Proof. Apply Lemma 4.1 to the E-vector space K. Example 4.3. Let E be a finite field. Then E has characteristic p for some prime p ∈ N (since otherwise E contains a copy of Z and is hence infinite). Thus E contains the subfield Fp ∼ = Z/pZ. In particular we can view it as an Fp -vector spce, and since it is finite, it must certainly be finite-dimensional. But then if d = dimFp (E), clearly there are pd elements in E. Thus we see that a finite field must have prime-power order. In fact there is exactly one finite field (up to isomorphism) of order pd for every prime p and positive integer d.

Definition 4.4. Let α ∈ C. • We say that α is algebraic over Q if there is a field E which is a finite extension of Q containing α.

• We say that α is transcendental otherwise.

• Given any set T ⊆ C, the smallest subfield containing T is called the field generated

by T , and is denoted Q(T ) (recall that any subfield of C contains Q, since it contains Z and hence Q because it is the field of fractions of Z).

• In the case where T has just a single element α we write Q(α) and we say the field extension is simple.

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• Slightly more generally, if F is any subfield of C and α ∈ C we let F (α) = Q(F ∪ {α}) be the smallest subfield of C containing both F and α, and one says α is algebraic over F if F (α)/F is a finite extension.

Lemma 4.3. Suppose that E/F is a finite extension of fields (both subfields of C) and let α ∈ E/F . Then there is a unique monic irreducible polynomial f ∈ F [t] such that F (α) ∼ = F [t]/hf i.

Proof. The field K = F (α) is a finite extension of F since it is a subfield of E (and hence a subspace of the finite dimensional F -vector space E). Let d = [K : F ] = dimF (K). Since the set {1, α, α2 , . . . , αd } has d + 1 elements, it must therefore be linearly dependent, and so P there exist λi ∈ F for 0 ≤ i ≤ d, not all zero, such that di=0 λi αi = 0.

P Suppose that g = di=0 λi ti ∈ F [t] \ {0} so that g(α) = 0. It follows that the kernel I of Pm Pm i j the homomorphism φ : F [t] → E given by φ c t = j j=0 j=0 cj α is nonzero. Now any nonzero ideal in F [t] is generated by a unique monic polynomial, thus I = hf i for some such polynomial, say f . By the first isomorphism theorem, we have im(S) ∼ = F [t]/I = F [t]/hf i.

Now im(S) is a subring of a field, so certainly an integral domain, hence hf i must be a prime ideal, and by our description of prime ideals in F [t] thus in fact maximal, so that im(S) is therefore a field. Finally, any subfield of C containing F and α must clearly contain im(S) (as the elements of im(S) are F -linear combinations of powers of α) so it follows im(S) ⊆ F (α).

But F (α) is the smallest field containing F and α, so F (α) ⊆ im(S) and hence we have the required isomorphism.

Definition 4.5. Given α ∈ C, the irreducible polynomial f for which Q(α) ∼ = Q[t]/hf i is called the minimal polynomial of α over Q. Remark 4.1. Note that our description of the quotient Q[t]/hf i shows that [Q(α) : Q] = deg(f ), hence the degree of the simple field extension Q(α) is just the degree of the minimal polynomial of α. √ Example 4.6.

1. Consider 3 ∈ C. There is a unique ring homomorphism φ : Q[t] → C √ such that φ(t) = 3. Clearly the ideal ht2 −3i lies in ker φ, and since t2 −3 is irreducible in Q[t] so that ht2 − 3i is a maximal ideal, we see that ker φ = ht2 − 3i, and hence im φ ∼ = Q[t]/ht2 − 3i.

Now the quotient √ Q[t]/ht2 − 3i is a field, hence im φ is also. Moreover, √any subfield of C which contains 3 clearly contains im φ, so we see that im φ = Q( 3). In particular, the images of {1, t} form a basis of the quotient Q[t]/ht2 − 3i by our description of quotients of polynomial rings in√the previous section, and under the isomorphism induced by φ, these map to 1 and 3 respectively, so we see that √ √ Q( 3) = {a + b 3 : a, b ∈ Q} a degree two extension of Q. 14

2. Exactly the same strategy shows that Q(21/3 ) is isomorphic to Q[t]/ht3 − 2i, and hence Q(21/3 ) is a 3-dimensional Q-vector space with basis {1, 21/3 , 22/3 }, again given by the image of the standard basis we defined in the quotient Q[t]/ht3 − 2i. √ us figure out what E = Q(T ) looks like. Certainly it 3. Now let T = { 3, 21/3 }. Let √ contains the subfields E1 = Q( 3) and E2 = Q(21/3 ). Now √ [E : E2 ] = [E2 ( 3) : E2 ] and hence by √ Lemma 4.3 we see that this √ degree is just the degree of the minimal polynomial of 3 over Q(21/3 ). We know 3 is a root of t2 − 3 ∈ Q[t] ⊂ Q(21/3 )[t], so the degree of this minimal polynomial is at most 2, and hence [E : E2 ] ≤ 2. It follows from the tower law that [E : Q] = [E : E2 ][E2 : Q] = [E : E2 ] · 3 ≤ 6.

On the other hand, we then also have [E : Q] = [E : E1 ][E1 : Q] = 2[E : E1 ].

It follows that [E : Q] is divisible by 2 and 3 and hence by 6, and moreover [E : Q] ≤ 2 · 3 = 6, so that [E : Q] = 6. With a little more work you can then check that √ √ √ {1, 21/3 , 3, 22/3 3, 22/3 3} is actually a Q-basis of E.

5

Unique Factorisation

Remark 5.1. Throughout this section, unless otherwise explicitly stated all rings are integral domains.

Definition 5.1. Let R be an integral domain. If a, b ∈ R we say that a divides b if there is some c ∈ R such that b = a · c, and we denote this a | b.

Definition 5.2. Let R be an integral domain.

1. We say c ∈ R is a common factor of a, b ∈ R if c | a and c | b.

2. We say that c is the highest common factor, and write c = hcf(a, b) if, whenever d is a common factor of a and b we have d | c.

3. We say k ∈ R is a common multiple of a, b ∈ R if a | k and b | k.

4. We say k ∈ R is a least common multiple if, whenever d is a common multiple of a and b, we have that k | d.

Remark 5.2. These definitions can be rephrased in terms of principal ideals: c is a common factor of a, b if and only if a, b ∈ cR. An element g is the highest common factor of a and b if and only if a, b ∈ cR implies gR ⊆ cR. 15

Lemma 5.1. Let a, b ∈ R, where R is an integral domain. If a highest common factor hcf(a, b) exists, it is unique up to units.

Proof. This is immediate from our description of the highest common factor in terms of ideals. Indeed if g1 , g2 are two highest common factors, then we must have g1 R ⊆ g2 R ⊆ g1 R so g1 R = g2 R, and since R is an integral domain this implies they are associates, i.e. they differ by a unit.

Remark 5.3. Note that although the definition makes sense in any integral domain, it does not follow that the highest common factor necessarily exists. We will see shortly that it does always exist for a number of interesting classes of integral domain.

Definition 5.3. Let R be an integral domain. We say that R is a principal ideal domain (or P.I.D.) if every ideal in R is principal.

We say R is a Euclidean domain (or E.D.) if there is a function N : R \ {0} → N (the norm) such that given any a ∈ R, b ∈ R \ {0} there exist q, r ∈ R such that a = bq + r and either r = 0 or N (r) < N (b).

Remark 5.4. Some texts require that the norm N satisfies additional properties.

Example 5.4.

1. The integers are an example of a Euclidean domain where N (a) = |a|.

2. If k is a field, then k[t] is a Euclidean domain where N (p) = deg(p). (This is a case where it’s easier not to have to define N (0)).

3. Let R = {a + ib ∈ C : a, b ∈ Z}. Then N (a + ib) = a2 + b2 can be shown to be a norm on R. This shows that this ring is a principal ideal domain.

Lemma 5.2. If R is a Euclidean domain, then R is a principal ideal domain.

Proof. Let I be an ideal. If I = {0} there is nothing to show. Otherwise, pick a ∈ I with N (a) minimal. We claim I = hai. To see this suppose that s ∈ I. Using the property of the norm, we may write s = a · q + r where r = 0 or N (r) < N (a). Since r = s − aq ∈ I it follows from the minimality of N (a) that we must have r = 0 and so s ∈ hai and thus I ⊆ hai. Since clearly hai ⊆ I it follows I is principal as required.

Lemma 5.3. Let R be a PID. Then if a, b ∈ R their highest common factor hcf(a, b) exists.

Proof. Consider I = ha, bi. Since R is a PID, this ideal is principal, say I = hdi. Then clearly d divides a and b and moreover there are elements r, s ∈ R such that d = ra + sb, hence any element which divides a and b also divides d.

Remark 5.5. This proof can be easily modified for n elements of R rather than just 2.

Definition 5.5. Let R be an integral domain.

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• A nonzero element a ∈ R \ {0} is said to be prime if aR is a prime ideal, that is, aR 6= R and whenever a divides r · s it divides at least one of r and s.

• A nonzero element a ∈ R \ {0} is said to be irreducible if it is not a unit, and whenever a = b · c either b or c is a unit.

Lemma 5.4. If R is an integral domain then any prime element is irreducible.

Proof. Suppose that p ∈ R \ {0} is prime and p = a · b. Then p divides a · b so that p divides one of a or b, say a. But then we can write a = r · p and hence p = (r · b)p, so that cancelling we see 1 = r · b, i.e. b is a unit.

Lemma 5.5. Let R be an integral domain and let a ∈ R. Then a is irreducible if and only if aR ⊆ bR implies either b is a unit or a = u · b for some unit u ∈ R. In particular, if R is a PID any irreducible element is prime and all nonzero prime ideals are maximal.

Proof. If a, b ∈ R and aR ⊆ bR then we have a = b · c for some c ∈ R. If c is a unit then aR = bR since a and b are associates.

Now if a ∈ R is irreducible it follows that any principal ideal containing aR is either aR or all of R. Conversely, if aR is maximal among proper principal ideals and a = b · c then aR is contained in bR and cR. Since a is not a unit, one of b or c must not be a unit, say b. But then aR ⊆ bR 6= R and so by maximality aR = bR. Hence a = b · u for some unit u ∈ R, and so b · c = b · u and hence c = u is a unit as required.

If R is a PID then every ideal is principal, so that if a ∈ R is irreducible hai is (by above) a maximal ideal, and hence prime. Moreover, any nonzero prime ideal in R is of the form hpi for some prime element p ∈ R \ {0}. But since R is a domain, prime elements are irreducible and hence by the above hpi is in fact maximal. It follows all nonzero prime ideals are maximal as required.

Definition 5.6. An integral domain R is said to be a unique factorisation domain (written UFD) if every element of R \ {0} is either a unit or can be written as the product of prime elements, and the factorisation is unique up to reordering and units.

Lemma 5.6. Suppose R is an integral domain and that every element a ∈ R can be written as a product of prime factors, and suppose that a = p1 p2 . . . pk = q1 q2 . . . ql where p1 , . . . , pk , q1 , . . . , ql are prime. Then k = l and there are units ui ∈ R× such that, after reordering the qi s, pi = ui qi (1 ≤ i ≤ k). (i.e. a prime factorisation is unique up to reordering and units) Proof. We use induction on the minimal number M (a) of primes in an expression for a ∈ R as a product of primes.

Base case: M (a) = 1. Then a is prime and uniqueness follows from the fact that primes are irreducible in an integral domain.

Inductive case: Let M = M (a) > 1 and a = p1 p2 . . . pM = q1 q2 . . . qk for primes pi , qj and k ≥ M . Now it follows that p1 | q1 . . . qk , and so since p1 is prime there is some qj with p1 | qj .

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Since qj is prime and hence irreducible, this implies that qj = u1 p1 for some unit u1 ∈ R. Thus we see that (u−1 1 p2 ) . . . pM = q2 q2 . . . qk , and by induction it follows that k − 1 = M − 1, i.e. k = M , and moreover the irreducibles occurring are equal up to reordering and units as required.

Proposition 5.7. Let R be a PID and S suppose that {In : n ∈ N} is a sequence of ideals such that In ⊆ In+1 . Then the union I = n≥0 In is an ideal and there exists an N ∈ N such that In = IN = I for all n ≥ N .

S Proof. Let I = n≥0 In . Given any two elements p, q ∈ I, we may find k, l ∈ N such that p ∈ Ik and q ∈ Il . It follows that for any r ∈ R we have r · p ∈ Ik ⊂ I, and taking n = max{k, l} we see that r, s ∈ In so that r + s ∈ In ⊂ I. It follows that I is an ideal.

Since R is a PID, we have I = hci for some c ∈ R. But then there must be some N such that c ∈ IN , and hence I = hci ⊆ IN ⊆ I so that I = IN = In for all n ≥ N as required.

Theorem 5.8. Every PID is a UFD. Proof. It follows from the fact that irreducibles are prime in a PID and Lemma 5.6 that we need only show that any element has some factorisation into prime elements, or equivalently for PIDs, some factorisation into irreducibles.

Assume, for a contradiction, that there is some a = a1 ∈ R which is not the product of irreducible elements. Clearly a cannot be irreducible, so we may write it as a = b · c where neither b nor c is a unit. If both b and c can be written as a product of prime elements, then multiplying these expressions together we see that a is also, hence at least one of b or c cannot be written as a product of prime elements. Pick one, and denote it a2 .

Note that if we set Ik = hak i (for k = 1, 2) then I1 ⊂ I2 . As before a2 cannot be irreducible, so we may find an a3 such that I2 = ha2 i ⊂ ha3 i = I3 . Continuing in this fashion we get a nested sequence of ideals Ik each strictly bigger than the previous one. But by above proposition this cannot happen if R is a PID, thus no such a exists.

Definition 5.7. If f ∈ Z[t] then we denote the content c(f ) of f to be the highest common factor of the coefficients of f .

Lemma 5.9. (Gauss) Let f, g ∈ Z[t]. Then c(f · g) = c(f ) · c(g).

Proof. Suppose first f, g ∈ Z[t] have c(f ) = c(g) = 1. Then letP p ∈ N be prime.

Pn We ihave for n i each such prime a homomorphism Z[t] → Fp [t] given by φp ( i=0 ai t ) = i=0 ai t , where ai denotes ai + pZ ∈ Fp . It is immediate that ker(φp ) = pZ[t], so that we see p | c(f ) if and only if φp (f ) = 0. But since Fp is a field, Fp [t] is an integral domain, and so as φp is a homomorphism we see that p | (f · g) ⇔ φp (f · g) = 0 ⇔ φp (f ) = 0 or φp (g) = 0 ⇔ p | c(f ) or p | c(g)

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Whence it is clear that c(f · g) = 1 if c(f ) = c(g) = 1.

Now let f, g ∈ Z[t] and write f = a · f 0 and g = b · g 0 where f 0 , g 0 ∈ Z[t] have c(f 0 ) = c(g 0 ) = 1 (so that c(f ) = a and c(g) = b). Then clearly f · g = (a · b) · (f 0 · g 0 ) and since c(f 0 · g 0 ) = 1 it follows that c(f · g) = c(f ) · c(g) as required.

Lemma 5.10. Suppose f ∈ Q[t] is nonzero. Then there is a unique α ∈ Q>0 such that f = αf 0 where f 0 ∈ Z[t] and c(f 0 ) = 1. We write c(f ) = α. Moreover, if f, g ∈ Q[t] then c(f · g) = c(f ) · c(g).

Pn i Proof. Let f = i=0 ai t where ai = bi /ci for bi , ci ∈ Z with hcf(bi , ci ) = 1 for all i = 0, 1, . . . , n. Then pick d ∈ Z>0 such that dai ∈ Z for Qnall i = 1, . . . , n so that df ∈ Z[t] (for example you can take d = lcm{ci : 0 ≤ i ≤ n} or i=0 ci ). Let c = c(d · f ) ∈ Z so that if α = d/c we have f = αf 0 where f 0 ∈ Z[t] and c(f 0 ) = 1 as required.

To show uniqueness, suppose that f = α1 f1 = α2 f2 , where α1 , α2 ∈ Q≥0 and f1 , f2 ∈ Z[t] such that c(f1 ) = c(f2 ) = 1. Then α1 α2−1 f1 = f2 ∈ Z[t]. But since f1 has content 1, we must −1 then have α1 alpha−1 2 ∈ Zα2 Z (since an irreducible p occuring in α1 /α2 to a negative power cannot divide all the coefficients of f1 because c(f1 ) = 1). Similarly writing α2 α1−1 f2 = f1 we see that α2 α1−1 ∈ Z so that α1−1 α2 ∈ Q>0 ∩ ZX = {1} and hence α1 = α2 as required.

The moreover part follows immediately from Gauss’ Lemma: if f, g ∈ Q[t] writing f = αf 0 and g = βg 0 we see that f · g = (α · β)(f 0 · g 0 ) and by Gauss’ Lemma c(f 0 g 0 ) = [1], hence c(f · g) = αβ = c(f ) · c(g).

Lemma 5.11.

1. Suppose that f ∈ Z[t] ⊂ Q[t] is nonzero, and that f = g · h where g, h ∈ Q[t]. Then there exists α ∈ Q such that (α · g), (α−1 · h) ∈ Z[t]. Thus f = (α · g)(α−1 h) is a factorisation of f in Z[t].

2. Suppose that f ∈ Q[t] is irreducible and c(f ) = 1. Then f is a prime element of Z[t].

3. Let p ∈ Z be a prime number. Then p is a prime element in Z[t].

Proof.

1. For the first part, by Lemma 5.10 we may write g = c · g 0 and h = d · h0 where 0 0 g , h ∈ Z[t] have content [1]. Then c(f ) = c · d so that as f ∈ Z[t] we have c · d ∈ Z. Setting α = d we see that f = (α · g)(α−1 h) where α · g = (c · d) · g 0 and α−1 h = h0 both lie in Z[t] as required.

2. For the second part, first note that if f ∈ Q[t] has c(f ) = 1 then by definition f must lie in Z[t] (and has content 1). To see that such an f is prime, we need to show that if g, h ∈ Z[t] and f | g ·h in R[t] then f | g or f | h in Z[t]. Now if f | g ·h in Z[t], certainly it does so in Q[t]. Since Q[t] is a PID, irreducibles are prime and some k ∈ Q[t]. Now

by Lemma 5.10 we may write k = c(k) · k 0 where k 0 ∈ Z[t]. Moreover by the same Lemma, c(h) = c(f ) · c(k) since c(f ) = 1. But h ∈ Z[t], hence c(h) = c(k) ∈ Z and in fact k ∈ Z[t] so that f divides g in Z[t] as required.

3. For the final part, we have already seen that the homomorphism φp : Z[t] → Fp [t] has kernel pZ[t], and so since Fp [t] is an integral domain, the ideal pZ[t] is prime, that is, p is a prime element of Z[t].

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Theorem 5.12. The ring Z[t] is a UFD. Proof. Since Z[t] is an integral domain (as Z is), by Lemma 5.6 it is enough to show that any element of Z[t] is a product of primes. Let f ∈ Z[t]. We may write f = a · f 0 where c(f 0 ) = 1, and since Z is a UFD we may factorise a into a product of prime elements of Z which we have just seen are prime in Z[t]. Thus we may assume c(f ) = 1. But then viewing f as an element of Q[t] we can write it as a product of prime elements in Q[t], say f = p1 p2 · · · pk . Now using Lemma 5.10, each pi can be written as ai qi where ai ∈ Q and qi ∈ Z[t] and c(qi ) = 1. But then by Lemma 5.11, qi is prime in Z[t], and f = (a1 · · · ak )q1 · · · qk . Comparing contents we see that (a1 · · · ak ) must be a unit in Z, and so we are done.

Remark 5.6. It can be deduced from this that al primes in Z[t] are either primes in Z or primes in Q[t] which have content 1.

Remark 5.7. One can show directly that if R is a UFD then the highest common factors exist. This and the fact that R, because it is a domain, has a field of fractions F and that F [t] is a PID for any field F , is all we need to run the above proof that Z[t] is a UFD. It follows that if R is any UFD, then so is R[t].

Hence, for example, Q[t] is a UFD (and in fact by induction, so is the polynomial ring k[t1 , . . . , tn ] for any n ∈ N and any field k). It is not hard to see that these rings are not PIDs.

5.1

Irreducible Polynomials

Remark 5.8. Recall that for any prime p ∈ Z we have the homomorphism φp : Z[t] → Fp [t]. This allows us to transport questions about factorisation in Z[t] to questions about factorisation in Fp [t]: clearly if f ∈ Z[t] is reducible in Z[t] its image in Fp [t] is nonzero (which will always be the case if c(f ) = 1 say) it will be reducible in Fp [t]. Since the rings Fp [t] are “smaller” than either Z[t] or Q[t] this can give us ways of testing irreducibility.

Example 5.8. Suppose that f = t3 −349t+19 ∈ Z[t]. If f is reducible in Q[t], it is reducible in Z[t] and hence its image under φp in Fp [t] will be reducible. But since f has degree 3 it follows it is reducible if and only if it has a degree 1 factor, and similarly for its image in Fp [t], which would therefore mean it has a root in Fp . But taking p = 2 we see that φ2 (f ) = f = t3 + t + 1 ∈ F2 [t] and so it is easy to check that f (0) = f (1) = 1 ∈ F2 , so f does not have a root, and hence f must be irreducible. Note on the other hand t2 + 1 is irreducible in Z[t] but in F2 [t] we have t2 + 1 = (t + 1)2 , so φp (f ) can be reducible even when f is irreducible.

Lemma 5.13. (Eisenstein’s criterion) Suppose that f ∈ Z[t] and f = tn + an−1 tn−1 + · · · + a1 t + a0 . Then if there is a prime p ∈ Z such that p | ai for all 0 ≤ i ≤ n − 1, but p2 does not divide a0 , then f is irreducible in Z[t] and Q[t].

Proof. Clearly c(f ) = 1, so irreducibility in Z[t] and Q[t] are equivalent. Let φp : Z[t] → Fp [t] be the quotient map. Suppose that f = gh was a factorisation of f in Z[t] where say deg(g) = k > 0. Then we have φp (f ) = φp (g) · φp (h). By assumption φp (f ) = tn , hence 20

since Fp [t] is a UFD and t is irreducible, we must have φp (g) = tk , φp (h) = tn−k . But then it follows that the constant terms of both g and h must be divisible by p, and hence a0 must be divisible y p2 , contradicting our assumption.

Example 5.9. This gives an easy way to see that 21/3 ∈ / Q: if it was t3 − 2 would be reducible, but we see this is not the case by applying Eisenstein’s criterion with p = 2.

Example 5.10. Suppose that p ∈ N is prime, and f = 1 + t + · · · + tp−1 ∈ Z[t]. Then we claim f is irreducible. We cannot apply Eisenstein’s criterion to f (t), but can to f (t + 1), as we will see.

Let g(t) = f (t + 1). Then if g was reducible, say g = h1 · h2 it would follow that f (t) = g(t − 1) = h1 (t − 1)h2 (t − 1) is reducible, and similarly if g is irreducible so is f . Thus f is irreducible if and only if g is. But as f = (tp−1 )/(t − 1) we see that

p−1 X p g = t ((t + 1) − 1) = ti , i+1 −1

p

i=0

p i+1

But it is well known that p divides for 0 ≤ i ≤ p − 2, while the constant term

p = p is not divisible by p2 , so Eisenstein’s Criterion shows g (and hence f ) is irreducible.

1

6

Modules

6.1

Definition and Examples

Definition 6.1. Let R be a ring with identity 1R . A module over R is an abelian group (M, +) together with a multiplication action of R on M satisfying, for all r1 , r2 ∈ R and m1 , m2 ∈ M : 1. 1R · m1 = m1 2. (r1 · r2 ) · m = r1 · (r2 · m) 3. (r1 + r2 ) · m = r1 · m + r2 · m 4. r1 · (m1 + m2 ) = r1 · m1 + r1 · m2 Remark 6.1. Just as with vector spaces, we write the addition in M and in R as the same symbol ’+’, and similarly the multiplication action of R on M is written in the same way as the multiplication in the ring R, since the axioms ensure that there is no ambiguity in doing so.

Example 6.2.

1. If R is a field, the definition is exactly that of a vector space over R, so modules over a field are just vector spaces over that field.

21

2. if A is an abelian group, then it has a natural structure of a Z-module: if n is a positive integer, then set   a + ... + a when n ≥ 0 and  | {z } n times n·a=  −(a + . . . + a) when n < 0  | {z } -n times

It’s easy to check this makes A a Z-module, and moreover, the conditions (1), (2) and (3) in fact force this definition on us, so that this Z-module is unique. Thus we see that Z-modules are just abelian groups.

3. Suppose that R is a ring. Then R is a module over itself.

4. If R is a ring and I is an ideal in R, then it follows directly from the definitions that I is an R-module.

5. If I is an ideal in R then R/I is an R-module where the multiplication action is given via the quotient homomorphism q : R → R/I. 6. If φ : R → S is a homomorphism of rings, and M is an S-module, then we can give M the structure of an R-module by setting r · m = φ(r) · m.

7. If R is a ring and n a positive integer, we may consider the module Rn = {(r1 , . . . , rn ) : ri ∈ R} of n-tuples of elements of R (written as row or column vectors), where the addition and multiplication by scalars is done componentwise.

8. Suppose that V is a vector space and φ : V → V is a linear map. Then we can make V into a C[t]-module by setting p(t) · v = p(φ)(v) for any v ∈ V and p(t) ∈ C[t].

Conversely, if we are given a C[t]-module M , we can view it as a complex vector space where the multiplication by scalars is given to us by viewing complex numbers as degree zero polynomials. The action of multiplication by t is then a C-linear map from M to itself. Thus C[t]-modules are just C-vector spaces equipped with an endomorphism.

6.2

Submodules, Generation and Linear Independence

Definition 6.3. If M is an R-module, a subset N ⊆ M is called a submodule if it is an abelian subrgoup of M and whenever r ∈ R and n ∈ N then r · n ∈ N .

T Remark 6.2. If {Ni : i ∈ I} is a collection of submodules then their intersection i∈I Ni is also a submodule. This allows us to make the following definition.

Definition 6.4.

• Given a subset X of a module M , the submodule generated by X is given as follows: \ hXi = N N ⊇X

Explicitly, hXi =

nP k

o r x : r ∈ R, x ∈ X . i i i=1 i i 22

• We say that X generates or spans the submodule hXi. • We say a module is finitely generated if there is a finite set which generates it.

Definition 6.5.

• If M is a module over R, we say a set S ⊆ M is linearly independent if whenever we have an equation r1 s1 + r2 s2 + · · · + rk sk = 0 for ri ∈ R and si ∈ S we have r1 = . . . = rk = 0.

• A set S is a basis for M if it is linearly independent and it spans M . Any module which has a basis is called a free module.

6.3

Quotient Modules and Isomorphism Theorems

Remark 6.3. All rings R are integral domains from now on.

Definition 6.6.

• If N is a submodule of M , then we can define the quotient module M/N and the associated operations exactly as quotient spaces were defined for vector spaces.

• If M1 , M2 are modules, we define a module homomorphism exactly as linear transformations were defined for vector spaces.

• We define isomorphisms and the kernel and image of a module homomorphism exactly as linear isomorphisms, kernels and images were defined for vector spaces.

It is easy to see that, for a module homomorphism φ, ker(φ) and im(φ) are both submodules.

Example 6.7. When R is a field, module homomorphisms are exactly linear maps. When R is Z a Z-module homomorphism is just a homomorphism of the abelian groups. As another important example, it is easy to see that if M is an R-module and N is a submodule of M then the definition of the module structure on M/N ensures precisely that the map q : M → M/N given by q(m) = m + N is a surjective module homomorphism.

Lemma 6.1. (Submodule Correspondence) Let M be an R-module and N a submodule. Let q : M → M/N be the quotient map. If S is a submodule of M then q(S) is a submodule of M/N , while if T is a submodule of M/N then q −1 (T ) is a submodule of M . Moreover the map T 7→ q −1 (T ) gives an injective map from submodules of M/N to the submodules of M which contain N , thus submodules of M/N correspond bijectively to submodules of M which contain N .

Proof. To check that q(S) and q −1 (T ) are submodules of N and M respectively follows directly from the definitions, we give the argument for q −1 (T ), since the argument for q(S) follows exactly the same pattern. If m1 , m2 ∈ q −1 (T ) then q(m1 ), q(m2 ) ∈ T and it follows since T is a submodule that q(m1 ) + q(m2 ) = q(m1 + m2 ) ∈ T which says precisely that m1 + m2 ∈ q −1 (T ). Similarly if r ∈ R then q(r · m1 ) = r · q(m1 ) ∈ T since q(m1 ) ∈ T and T is a submodule, so that r · m1 ∈ q −1 (T ). Thus q −1 (T ) is a submodule of M as required.

Now if T is any subset of M/N we have q(q −1 (T )) = T simply because q is surjective.

Since we have just checked q −1 (T ) is always a submodule in M , this immediately implies that 23

the map S 7→ q(S) is a surjective map from submodules in M to submodules in M/N and that T 7→ q −1 (T ) is an injective map, and moreover since q(N ) = 0 ∈ T for any submodule T of M/N we have N ⊆ q −1 (T ) so that the image of the map T 7→ q −1 (T ) consists of submodules of M which contain N . Hence it only remains to check that the submodules of M of the form q −1 (T ) are precisely these submodules. To see this suppose that S is an arbitrary submodule of M , and consider q −1 (q(S)). By definition this is q −1 (q(S)) = {m ∈ M : q(m) ∈ q(S)} = {m ∈ M : ∃s ∈ S such that m + N = s + N } = {m ∈ M : m ∈ s + N } =S+N But if S contains N then we have S + N = S and hence q −1 (q(S)) = S and any submodule S which contains N is indeed the preimage of a submodule of M/N as required.

Theorem 6.2. (Universal Property of Quotients) Suppose that φ : M → N is a homomorphism of R-modules, and S is a submodule of M with S ⊂ ker(φ). Then there is a unique homomorphism φ : M/S → N such that φ = φ ◦ q where q : M → M/S is the quotient homomorphism. Moreover ker(φ) is the submodule ker(φ)/S. Proof. Since q is surjective, the formula φ(q(m)) = φ(m) uniquely determines the values of φ, so that φ is unique if it exists. But if m − m0 ∈ S, then since S ⊆ ker(φ) it follows that 0 = φ(m − m0 ) = φ(m) − φ(m0 ) and hence φ is constant on the S-cosets, and therefore induces a map φ(m + S) = φ(m). The fact that φ is a homomorphism then follows directly from the definition of the module structure on the quotient M/S, and clearly φ = φ ◦ q by definition. To see what the kernel of φ is, note that φ(m + S) = φ(m) = 0 if and only if m ∈ ker(φ), and hence m + S ∈ ker(φ)/S as required.

Corollary 6.3. Let M be an R-module. We have the following isomorphisms: 1. (First Isomorphism Theorem) If φ : M → N is a homomorphism then φ induces an isomorphism φ : M/ ker(φ) → im(φ).

2. (Second Isomorphism Theorem) If M is an R-module and N1 , N2 are submodules of M then (N1 + N2 )/N2 ∼ = N1 /(N1 ∩ N2 ) 3. (Third Isomorphism Theorem) Suppose that N1 ⊆ N2 are submodules of M . Then we have (M/N1 )/(N2 /N1 ) ∼ = M/N2 Proof. The proofs are exactly the same as for rings.

1. For the first isomorphism theorem, apply the universal property to K = ker(φ). Since in this case ker(φ) = ker(φ)/ ker(φ) = 0 it follows φ is injective and hence induces an isomorphism onto its image which from the equation φ ◦ q = φ must be exactly im(φ).

24

2. For the second isomorphism theorem, let q : M → M/N2 be the quotient map. It restricts to a homomorphism p from N1 to M/N2 , whose image is clearly (N1 +N2 )/N2 , so by the first isomorphism theorem it is enough to check that the kernel of p is N1 ∩N2 .

To do this note that if n ∈ N1 is such that p(n) = 0 then n + N2 = 0 + N2 so that m ∈ N2 , and so n ∈ N1 ∩ N2 .

3. For the third isomorphism theorem, let qi : M → M/Ni , for i = 1, 2. By the universal property for q1 we see that there is a homomorphism q2 : M/N1 → M/N2 induced by the map q2 : M → M/N2 , with kernel ker(q2 )/N1 = N2 /N1 and q2 ◦ q1 = q2 . Thus q2 is surjective (since q2 is) and hence the result follows by the first isomorphism theorem.

6.4

Free, Torsion and Torsion-Free Modules

Definition 6.8. Let M be an R-module and suppose that m ∈ M .

• The annihilator of m, denoted AnnR (m) is {r ∈ R : r · m = 0}. A direct check shows that AnnR (m) is an ideal in R. • m ∈ M is a torsion element if AnnR (m) 6= {0}.

• M is torsion if every m ∈ M is a torsion element.

• M is torsion-free if M has no nonzero torsion elements.

Lemma 6.4. Let M be an R-module and m ∈ M , and let R · m be the submodule of M generated by m. Then R · m is isomorphic to AnnR (m).

Proof. The map R → M given by r 7→ r · m defines an R-module homomorphism whose image is exactly R · m. Since the kernel of the map is evidently AnnR (m) the isomorphism follows from the first isomorphism theorem.

Definition 6.9. A cyclic module is a module generated by a single element.

Remark 6.4. It follows from what we have just said that any cyclic module is isomorphic to a module of the form R/I where I is an ideal of R (corresponding to the annihilator of a generator of the cyclic module).

Example 6.10. Let R = C[x, y]. Then the ideal I = hx, yi is a module for R. It is torsionfree because R is an integral domain (and I is a submodule of R) but it is a good exercise to see that it is not free.

Lemma 6.5. Let M be a finitely generated free R-module. Then the size of a basis for M is uniquely determined.

25

Proof. Let X =P{x1 , . . . , xn } be a basis of M . Pick a maximal ideal I ⊆ R. Let NI = { ni=1 ri xi : ri ∈ I}. Since I is an ideal it follows that NI is a submodule of M . In fact, since X generates M , any element of the form r · m where r ∈ I and m ∈ M lies in NI , and so the submodule IM generated by the set of all such elements must lie in NI .

On the other hand, certainly IM contains ri xi for any ri ∈ I, and so all sums of the form P n i=1 ri xi , so that NI ⊆ IM . Hence we see NI = IM . Notice that in particular this means the submodule NI = IM does not depend on the choice of a basis of X. Let q : M → M/(IM ) be the quotient map. The quotient module M/(IM ) is a module for not just R but in fact for the quotient field k = R/I, via the action (r+I)·q(m) = r·q(m).

To see this we just need to check this definition does not depend on the choice of r +I ∈ R/I: if r − r0 ∈ I then r · m − r0 · m = (r − r0 ) · m ∈ IM and so r0 · q(m) = q(r0 · m) = r · q(m) as claimed.

We now claim that if X is a basis for M then q(X) is a basis for the k-vector space M/(IM ). Note that if we assume the claim then |X| = dimk (M/(IM )) and the righthand side is clearly independent of X (since we have checked that the submodule IM is) so this will finish the proof of the Lemma. To prove the claim first note that since X generates Pn M and q is surjective it follows that q(X) generates M/(IM ). Now suppose we have i=1 ci q(xi ) = 0 ∈ M/(IM ), where ci ∈ k. Picking any representatives ri ∈ R for the ci ∈ R/I we see that ! n n n n X X X X 0= ci q(xi ) = ri q(xi ) = q(ri xi ) = q ri xi i=1

i=1

i=1

i=1

P It follows that y = ki=1 ri xi ∈ ker(q) = IM . But since IM = NI , this means that ri ∈ I for each i, so ci = 0 for each i. It follows that q(X) is linearly independent and hence a k-basis of M/IM as required.

Definition 6.11. The size of a finitely generated free R-module M is called the rank rk(M ) of M .

Proposition 6.6. Let R be a PID, and let F be a free module with finite basis X. Then any nonzero submodule N of F is free on a basis with at most |X| elements.

Proof. We prove this by induction on k = |X|.

Base case: k = 1, so the free module F is isomorphic to R itself, and a submodule of R is just an ideal. Since R is a PID, any ideal (submodule) I of R is principal, i.e. is generated by a single element, d say, and since R is a domain so that d is not a zero divisor, it follows R∼ = Rd via the map r 7→ r · d so that I = Rd is a free module.

Inductive case: k > 1. Using the coordinates given by the basis we may assume that F = Rk and we have N ⊂ Rk . Consider the projection π : Rk → Rk−1 (where we view Rk−1 as a submodule of Rk consisting of the elements whose final entry is zero). Then clearly π is a surjective homomorphism, and the image π(N ) is a submodule of Rk−1 . By induction, we know that π(N ) ⊆ Rk−1 is a free module whose bases all have the same size which is at most k − 1.

26

Now if L = ker(π) then the kernel of π restricted to N is just N ∩ L, so that by the first isomorphism theorem, N/(N ∩ L) ∼ = π(N ). Now if N ∩ L = {0} we see that N ∼ = π(N ) and so we are done by induction. Otherwise note that L is just the submodule of elements of Rk which are zero except for their last entry which is clearly a copy of the free module R itself, and so N ∩ L, by the k = 1 case, is free on one generator, m say. Since π(N ) is free, we may pick a basis S = {π(n1 ), π(n2 ), . . . , π(ns )} where again by induction 0 ≤ s ≤ k − 1 and n1 , n2 , . . . , ns ∈ N (if π(N ) = 0 then we just take S = ∅, so that s = 0).

We claim that {n1 , . . . , ns , m} is a basis for N . Let n ∈ N . Since S is a basis for π(N ) we may write s X π(n) = ci π(ni ) i=1

uniquely forP some c1 , . . . , cs ∈ R. But then Ps using the fact that π is a homomorphism Ps we see s that π (n − i=1 ci nP i ) = 0, that is n − i=1 ci ni ∈ ker(π). Thus we see that n − i=1 ci ni ∈ N ∩ L, and so n − si=1 ci ni = d · m for some d ∈ R. Thus we see that N is generated by {n1 , . . . , ns , m}.

Ps+1 For convenience let n = m.

Then if we have s+1 i=1 ai ni = 0, applying π we see that Ps a π(n ) = 0, so by linear independence of {n , . . . , ns } we have that a1 = · · · = as = 0. i 1 i=1 i Then we have as+1 ns+1 = 0 and since F is torsion-free it follows ns+1 = 0. Thus the set {n1 , . . . , ns+1 } is linearly independent as required.

Example 6.12. Suppose that M ⊂ Z3 is the submodule M = {(a, b, c) ∈ Z3 : a + b + c ∈ 2Z} The above proposition tells us that M must be finitely generated, but let’s use the strategy of proof to actually find a basis. Let π : Z3 → Z2 be the projection (a, b, c) 7→ (a, b). Now in our case it is clear that π(M ) is all of Z2 . Thus if we pick say n1 = (1, 0, 1) and n2 = (0, 1, 99) then n1 , n2 ∈ M and π(n1 ) = (1, 0) and π(n2 ) = (0, 1) certainly form a basis for π(M ).

Thus given any m ∈ M , we can find a, b ∈ Z so that m − an1 − bn2 ∈ ker(π): explicitly if m = (a, b, c) then clearly we have m−an1 −bn2 ∈ ker(π). Now if L = ker(π) then L is the copy {(0, 0, z) : z ∈ Z} of Z inside Z3 , and clearly M ∩ L = {(0, 0, 2n) : n ∈ Z}, so that (0, 0, 2) is a generator of L ∩ M , and so putting everything together {(1, 0, 1), (0, 1, 99), (0, 0, 2)} must be a basis of M as required.

Lemma 6.7. Let M be an R-module, and let M tor = {m ∈ M : AnnR (M ) 6= {0} be a submodule of M . Moreover, the quotient module M/M tor is a torsion-free module.

Proof. Let x, y ∈ M tor . Then there are nonzero r, s ∈ R such that s · x = t · y = 0. But then s · t ∈ R \ {0}, since R is an integral domain, and (s · y)(x + y) = t(s · x) + s(t · y) = 0, and clearly if r ∈ R then s(r · x) = r(s · x) = 0, so that it follows M tor is a submodule of M as required.

To see the moreover part, suppose that x + M tor is a torsion element in M/M tor . Then there is a nonzero r ∈ R such that r(x + M tor ) = 0 + M tor , that is, r · x ∈ M tor . But then by definition there is an s ∈ R such that s(r · x) = 0. But then s · r ∈ R is nonzero (since R is an integral domain) and (s · r)x = 0 so that x ∈ M tor and hence x + M tor = 0 + M tor so that M/M tor is torsion free as required.

27

6.5

Presentations of Finitely Generated Modules over a PID

Lemma 6.8.

1. Let F be a finitely-generated free R-module with basis X = {x1 , . . . , xn } and M any R-module. If f : X → M is any function, there is a unique homomorphism φ : F → M such that φ(x) = f (x) for all x ∈ X. 2. Let M be a finitely generated R-module. Then there is an n ∈ N and a surjective homomorphism φ : Rn → M .

3. Let M be a finitely generated R-module. Then there is an n ∈ N such that M is isomorphic to an Rn /N for some submodule N of Rn .

Proof.

1. Given fPwe define φ as follows:for every m ∈ F we P may write m uniquely in the form m = ni=1 ri xi where ri ∈ R. Then define φ(m) = ni=1 ri f (xi ).

2. Given any finite subset {m1 , m2 , . . . , mn } of M , by the first part if {e1 , . . . , en } is a basis for Rn the map f (ei ) = mi extends to a homomorphism φ : Rn → M . Clearly the condition that {m1 , . . . , mk } is a generating set is equivalent to the map φ being Pn both assert that any element of M can be written in the form Pn surjective, since i=1 ri ei ) i=1 ri mi = φ ( 3. Note that if φ : Rn → M is a surjective homomorphism as in part 2, then by the first isomorphism theorem we have Rn / ker(φ) ∼ = M. Definition 6.13. Let M be a finitely generated R-module. A presentation of M is an injective homomorphism ψ : Rm → Rk such that Rk / im(φ) ∼ = M . We often specify the k isomorphism φ : R / im(φ) → M , and then view the presentation as a chain of maps ψ

φ

Rm − → Rk → − M where ψ is injective and φ is surjective, and the image of φ is exactly the kernel of φ = φ ◦ q with q the quotient map Rk → Rk / im(ψ). Remembering the whole chain remembers the module M on the nose, whereas remembering just the part ψ : Rm → Rk remembers M up to isomorphism.

7

Matrices over R and Normal Forms

Remark 7.1. All rings R in this section are Euclidean domains.

Definition 7.1. Let A ∈ Mm×k (R) be a matrix, and let r1 , . . . , rm be the rows of A, which are row vectors in Rk . An elementary row operation on a matrix A ∈ Mm×k (R) is an operation of the form: 1. Swap two rows ri and rj .

2. Replace one row i with a new row ri0 = ri + crj for some c ∈ R and j 6= i.

28

In the same way, viewing A as a list of column vectors, we define elementary column operations.

Remark 7.2. Note that the row operations correspond to multiplying A by elementary matrices ont he left and column operations corresponding to multiplying A by elementary matrices on the right. Indeed if we let Eij denote the matrix with (i, j)-th entry equal to 1 and all other entries zero, then the matrix corresponding to the first row operation is Sij = Ik − Eii − Ejj + Eij + Eji while the second elementary row operation is given by Xij (c) = Ik + cEij .

The column operations are given by multiplying  1  0   1  Sij =     1      Xij (c) =    

on the right by these matrices:  1 ..

        1

.

0



1 ...

1

c ..

.

1

        1

Definition 7.2. If A, B ∈ Mk×m (R) we say that A and B are equivalent if B = P AQ where P ∈ Mk×k (R) and Q ∈ Mm×m (R) are invertible matrices. We will say that A and B are ERC equivalent if one can be obtained from the other by a sequence of row and column operations.

Remark 7.3. Since row and column operations correspond to pre- and post-multiplying a matrix by elementary matrices, it is clear that two ERC equivalent matrices are equivalent.

Remark 7.4. Recall that we write N : R \ {0} → N for the norm function of our Euclidean domain R. Theorem 7.1. Suppose that A ∈ Mk×m (R). Then A is ERC equivalent to a diagonal matrix D where   d1   d2   .   ..     dm  D=   0 ··· ··· 0  . ..   .. .  0 ··· ··· 0 29

and each successive di divides the next.

Proof. We claim that by using row and column to A which is of the form  b11 0  0 b22  B =  .. ..  . .

0 bk2

operations we can find a matrix equivalent  ··· 0 · · · b2m   (1) ..  ... .  · · · bkm

where b11 divides all the entries bij in the matrix. Factoring out b11 from each entry, we may then apply induction to the submatrix B 0 = (bij /b11 )i,j≥2 to obtain the proposition.

(Note that row and column operations on B 0 correspond to row and column operations on B because b11 is the only nonzero entry in the first row and column of B.) To prove the claim, we use a sequence of steps which either put A into the required form or reduce the degree of the (1, 1)-th entry, hence iterating them we must stop with a matrix of the required form.

Step 1: by using row and column operations, ensure the entry a11 has N (a11 ) ≤ N (aij ) for all i, j. By the division algorithm, we can write aij = qij a11 + rij where rij = 0 or N (rij ) < N (a11 ). Subtract qqj × column 1 from column i, and qi1 × row 1 from row i for each i and j (in any order, we only care at the moment about what hapens to the first row and column).

Step 2: the resulting matrix now has all nonzero entries in the first row and column of strictly smaller norm than then (1, 1)-th entry. Thus either we have a matrix of the form (1), or we can repeat step 1.

Now since at each iteration of steps 1 and 2 the minimum norm of the nonzero entries of our matrix strictly decreases, the process must terminate with a matrix of the required form, at least in the sense that all the entries in the first row and column are zero except for the (1, 1)-th entry. Let C denote this matrix. If c11 does divide all the entries of C then we are done. Otherwise: Step 3: Pick an entry cij not divisible by c11 so that cij = qij c11 + rij where rij 6= 0 and N (rij ) < N (c11 ). but adding qij × column 1 to column j and then subtracting row 1 from row i, we get a new matrix C 0 with (i, j)-th entry equal to rij , and hence the minimum norm of the nonzero entries of C 0 has again strictly decreased.

Repeat steps 1 and 2 now on it until we obtain a matrix again of the form (1). If its (1, 1)-th entry still does not divide all the entries of the matrix we may repeat step 3.

Ultimately since the minimum norm of the nonzero entries of the matrix keeps strictly decreasing we must terminate at a matrix of the required form.

Example 7.3. The above theorem is really taking R = Z. Let  2 A = 8 3

an algorithm, so let’s use it in an example,

30

 5 3 6 4 1 0

The entry of smallest norm is the (3, 2)-th entry, so we swap it to the (1, 1)-th entry (by swapping rows 1 and 3 and then columns 1 and 2 say) to get   1 3 0 A1 = 6 8 4 5 2 3 Now since the (1, 1)-th entry is a unit, there  1  A2 = 0 0

will be no remainders when dividing so we get  0 0 −10 4  4 −10

Next we must swap the (3, 3)-th entry to the  1  A3 = 0 0

(2, 2)-th entry to get  0 0 3 −13 4 −10

Dividing and repeating our row and column operations now on the second row and column (this time we do get remainders) gives:     1 0 0 1 0 0 A4 = 0 3 −13 ∼ A5 = 0 3 2 0 1 3 0 1 8 (where ∼ is to denote ERC equivalence). Now moving the (3, 2)-th entry to the (2, 2)-th entry and dividing again gives       1 0 0 1 0 0 1 0 0 A6 = 0 1 8 ∼ A7 = 0 1 0  ∼ A8 = 0 1 0  0 3 2 0 3 −22 0 0 −22 which is in the required normal form.

8

The Canonical Form for Finitely Generated Modules

Remark 8.1. In this section, all rings are Euclidean Domains, although all results stated here actually also hold more generally for PIDs. Theorem 8.1. Suppose that M is a finitely generated module over a Eucliean domain R. Then there is an integer s and nonzero nonunits c1 , . . . , cr ∈ R such that c1 | c2 | · · · | cr and such that: ! r M R/(ci R) M∼ = Rs ⊕ i=1

31

Proof. Since R is a PID we may find a presentation for M , that is an injection ψ : Rm → Rk (so that m ≤ k) and a surjection φ : Rk → M , so that M ∼ = Rk / im(ψ). Now if A is the matrix of ψ with respect to the standard bases of Rk and Rm , by Theorem 7.1 which gives a normal form for matrices over a Euclidean domain, we know we can transform A into a diagonal matrix D with diagonal entries d1 | d2 | · · · | dm . But since row and column operations correspond to pre- and post-multiplying A by invertible matrices, and these correspond to changing bases in Rk and Rm respectively, it follows that there are bases of Rk and Rm respectively with respect to which ψ has matrix D. But then if {f1 , . . . , fk } denotes the basis of Rk , we see that the image of ψ has basis {d1 f1 , . . . , dm fm }. But now define a map ! m M k θ:R → R/di R ⊕ Rk−m i=1

θ

k X

! ai f i

= (a + 1 + d1 R, . . . , am + dm R, am+1 , . . . , ak )

i=1

It is then clear that θ is surjective and ker(θ) is exactly the submodule generated by {di fi : 1L≤ i ≤ m}, that is, im(θ), and so by the first isomorphism theorem M ∼ = Rk / im(θ) ∼ = m k−m . Now since ψ is injective it follows that each of the di are nonzero. i=1 (R/di R) ⊕ R On the other hand if di is a unit (and so all dj for j ≤ i are also) then R/di R = 0, so this summand can be omitted from the direct sum. The result now follows.

Remark 8.2. The elements {c1 , c2 , . . . , cm } are in fact unique up to units. We won’t show this here.

Corollary 8.2. Let M be a finitely generated torsion-free module over R. Then M is free.

In general if M is a finitely generated R-module, the rank s of the free part of M given in the structure theorem is M/M tor and hence it is unique. s Proof.

Lr by the above structure theorem, M is isomorphic to the module of the form R ⊕s ( i=1 R/cL i R), thus we can assume M is actually equal to a module of this form. Let F = R and N = ri=1 R/ci R, so that M = F ⊕N . We claim that N = M tor . Certainly if a ∈ R/ci R then since ci | ck we see that ck (a) = 0. But then if m ∈ N , say m = (a1 , . . . , ak ) where ai ∈ R/ci R it follows ck (a1 , . . . , am ) = (ck a1 , . . . , ck ak ) = (0, . . . , 0) so N is torsion. On the other hand if m = (f, n) where f ∈ F and n ∈ N then r(f, n) = (r · f, r · n) = (0, 0) we must have f = 0 since a free module is torsion-free. Thus M tor = N as claimed. It follows that M is torsion free if and only if M = F is free. Moreover, by the second isomorphism theorem F ∼ = M/M tor so that s = rk(F ) = rk(M/M tor ).

Corollary 8.3. (Structure theorem for finitely generated abelian groups) Let A be a finitely generated abelian group. Then there exist uniquely determined integers r, c1 , . . . , ck ∈ Z≥0 , ci 6= 0 such that c1 | c2 | · · · | ck and A∼ = Zr ⊕ (Z/c1 Z) ⊕ · · · ⊕ (Z/ck Z) 32

Proof. This is simply a restatement of the previous theorem, except that once we insist the ci are positive the ambiguity caused by the unit group Z× = {±1} is removed.

Theorem 8.4. (Structure theorem in primary decomposition form) Let R e a Euclidean domain and suppose that M is a finitely generated R-module. Then there are irreducibles p1 , . . . , pk ∈ R, unique up to units, and integers ri , 1 ≤ i ≤ k, such that M∼ =

k M (R/pri i R) i=1

Moreover, the integers ri are uniquely determined. (Note that the pi s are not necessarily distinct.)

Proof. Suppose that a, b ∈ R are coprime, that is, hcf(a, b) = 1 and hence R = Ra + Rb and aR ∩ bR = (a · b)R. Then the Chinese Remainder Theorem shows that R/(a · b)R ∼ = /aR ⊕ R/bR. If a1 , . . . , am are pairwise coprime so that hcf(ai , aj ) = 1 for all i 6= j, then hcf(ai , ai+1 , . . . , am } = 1, since if p is a prime dividing ai and ai+1 , . . . , am then p divides some aj for j > i and hence hcf(ai aj ). Thus if c = a1 a2 . . . am , iteratively we see that R/cR ∼ = R/a1 R ⊕ R/(a2 · · · am )R ∼ = R/a1 R ⊕ R/a2 R ⊕ R/(a3 · · · am )R ... ∼ = R/a1 R ⊕ R/a2 R ⊕ · · · ⊕ R/am R In particular if c = pn1 1 · · · pnr r is the prime factorisation of c where the pi are distinct primes we see that r M R/cR ∼ R/pri R. = i

i=1

The result follows.

Example 8.1. Suppose that A ∼ = Z/44Z ⊕ Z/66Z. Then the first structure theorem would write A as A∼ = Z/22Z ⊕ Z/132Z Indeed the generators corresponding to the direct sum decomposition give a presentation of A as Z2 → Z2 → A where the first map is given by the matrix

44 0 0 66 and as 66 = 1 · 44 + 22 we see that row and column operations allow us to show this matrix is equivalent to:

44 0 44 44 44 44 22 −44 22 0 ∼ ∼ ∼ ∼ .

0 66 0 66 −44 22 44 44 0 132 33

On the other hand, for the primary decomposition we write A as A∼ = (Z/2Z) ⊕ (Z/22 Z) ⊕ (Z/3Z) ⊕ (Z/11Z)2

Notice that the prime 2 appears twice to two different powers. Intuitively you should think of the primary decomposition as decomposing a module into as many direct summands as possible, while the canonical form decomposes the module into as few direct summands as possible.

Remark 8.3. Note that the first structure theorem gives a canonical form which can be obtained algorithmically, while the second one requires one to be able to factorise elements of the Euclidean domain, which for example in C[t] is not automatically computable.

9

Rational and Jordan Canonical Forms

Remark 9.1. The structure theorem allows us to recover structure theorems for linear maps: if V is a k-vector space and T : V → V is a linear map, then we view V as a k[t]-module by setting t · v = T (v) for all v ∈ V .

Lemma 9.1. Let M be a finitely generated k[t]-module. Then M is finite dimensional as a k-vector space if and only if M is a torsion k[t]-module. Moreover, a subspace U of V is a k[t]-submodule if and only if U is T -invariant, i.e. T (U ) ⊆ U .

Proof. Given M we can apply the structure theorem to see that: V ∼ = k[t]2 ⊕ k[t]/hc1 i ⊕ · · · ⊕ k[t]/hck i.

Now k[t] is infinite dimensional as a k-vector space while k[t]/hf i is deg(f )-dimensional as a k-vector space, so it follows that M is torsion if and only if s = 0 which holds if and only if M is finite dimensional as a k-vector space. For the final statement, notice that a subspace U of V is T -invariant if and only if it is p(T )-invariant for every p ∈ k[t].

Definition 9.1. For a monic polynomial n

f ∈ k[t], f = t +

n−1 X

,

i=0

the matrix 

0 1   0 .  ..

0

0 ··· 0 ···

0 0

−a0 −a1 .. .

     −ak−2 

1 ··· 0 .. . . .. . . .

0 · · · 1 −ak−1

is called the companion matrix of f .

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

Theorem 9.2. Suppose V is a finite dimensional k-vector space and T : V → V is a linear map. Then there are nonconstant polynomial sf1 , . . . , fk ∈ k[t] such that f1 | f2 | · · · fk and a basis B of V with respect to which T has matrix which is block diagonal with blocks C(fi ):   C(f1 ) 0 ··· 0 ..   C(f2 ) 0 .   0 B [T ]B =  . ..  . ..  ..

0 .  0 ··· 0 C(fk ) Proof. by the canonical form theorem and Lemma 9.1, there is an isomorphism θ : V → L k are nonunits (hence i=1 k[t]/hfi i of k[t]-modules, where f1 | f2 | · · · | fk and the fiP i nonconstant polynomials). But now for any monic polynomial g = tn + n−1 i=0 ai t , the division algorithm shows that the k[t]-module k[t]/hgi has k-basis {1 + hgi, t + hgi, . . . , tn−1 + hgi}, and matrix for multiplication by t with respect to this basis is just C(g), since t · tn−1 = Pthe i − n−1 i=0 ai t mod hgi. The union of these bases for each submodule k[t]/hfi i gives a basis for the direct sum, and the matrix of T with respect to the basis of V corresponding to the basis via the isomorphism θ−1 has the asserted form.

Remark 9.2. This matrix form for a linear map given by the previous theorem is known as the Rational Canonical Form of T . Notice that this form, unlike the Jordan canonical form, makes sense for a linear map on a vector space over any field, not just an algebraically closed field like C. Remark 9.3. We can also recover the Jordan canonical form for linear maps of C-vector spaces from the second, primary decomposition, version of our structure theorem, which expresses each module in terms of cyclic modules k[t]/hf k i where f is irreducible. The monic irreducibles over C are exactly the polynomials t − λ for λ ∈ C. Thus the second structure theorem tells us that, for V a finite dimensional complex vector space and T : V → V , we may write V = V1 ⊕ V2 ⊕ · · · ⊕ Vk , where each Vi is isomorphic to C[t]/h(t − λ)r i for some λ ∈ C, and r ∈ N. It is thus enough to pick a standard basis for a vector space V with T : V → V where as C[t]-modules we have an isomorphism φ : V → C[t]/h(t − λ)r i. Here we pick the basis {1 + h(t − λ)r i, (t − λ) + h(t − λ)r i, . . . , (t − λ)r−1 + h(t − λ)r i} of C[t]/h(t − λ)r i and take its preimage {v1 , . . . , vk } numbered so that φ(vi ) = (t − λ)k−i + h(t − λ)r i.

Clearly (T − λ)vi = vi−1 or T (vi ) = λvi + vi−1 if i > 1 and T (v1 ) = λv1 . Thus the matrix of T with respect to this basis is just the Jordan block of size k, and we recover the Jordan normal form.

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