IMMERSE 2010 ALGEBRA SESSION COURSE NOTES

The following are notes taken during the 2010 IMMERSE Algebra Session. Each chapter represents oneday and is written by a different participant.

1. Groups and Basic Properties (by Derrek Yager)

• An operation is a function from a set A to a set B and is a rule that assigns each element of A toexactly one element of B.

For example, in the operation ∗ : G×G 7−→ G, if we let a, b ∈ G, then (a, b) 7−→ a ∗ b.• A group G is simply a nonempty set G under a closed operation ∗ where the following properties

are satisfied:(1) Associativity. For a, b, c ∈ G, a ∗ (b ∗ c) = (a ∗ b) ∗ c.(2) Identity. There exists e ∈ G such that a ∗ e = e ∗ a = a.(3) Inverses. For all a ∈ G, ∃ a−1 ∈ G such that a ∗ a−1 = a−1 ∗ a = e.

Examples of Groups

Group Form of Element Identity Inverse

〈Z,+〉 a 0 −a

〈Q,+〉 a/b 0 −a/b

〈Q− {0}, ∗〉 a 1 1/a

〈Zn,+〉 0 ≤ a ≤ n− 1 0 n− a

• For an example of the 〈Zn,+〉 group, consider the following Cayley table representing 〈Z5,+〉 .

0 1 2 3 40 0 1 2 3 41 1 2 3 4 02 2 3 4 0 13 3 4 0 1 24 4 0 1 2 3

• A group G is considered abelian if a ∗ b = b ∗ a ∀ a, b ∈ G. So far, all of the group examples havebeen abelian.

One non-abelian group is GL(2,R), which is all 2 × 2 matrices with real entries and nonzerodeterminants.

Another non-abelian group consists of the Quaternions {1,−1, i,−i, j,−j, k,−k} where i2 =j2 = k2 = −1. Also, they must satisfy

i · j = k j · i = −kj · k = i k · j = −ik · i = j i · k = −j

Yet another non-abelian group consists ofDn.When n = 4, this group represents the symmetrieson a square.

1

• There are also four key properties of groups.(1) In every group there exists one unique identity element.

Proof. By contradiction, let e and e′ be two distinct identity elements in a group G. By defi-nition, e′e = e′ since e is an identity element, but we also have e′e = e since e′ is an identityelement. Thus, we have e′ = e′e = e which contradicts our statement that e and e′ are distinct.Hence, the identity element is unique. (Fun Fact! The tendency to use e to represent theidentity element roots from the German word ”Einheit”) �

(2) In every group, right and left cancellation laws hold.Proof. Let a, b, c ∈ G such that a−1, b−1 ∈ G by definition of groups. If ab = ac, then

a−1ab = a−1aceb = ecb = c

By symmetry, the right cancellation law holds such that ab = cb implies a = c. �(3) If G is a group, with a, b ∈ G and ab = e, then a = b−1 and b = a−1.

Proof. For showing a = b−1, we see that

ab = ea−1ab = a−1e

eb = a−1

b = a−1.

For showing b = a−1, we also see that

ab = eabb−1 = eb−1

ae = b−1

a = b−1.

�(4) If G is a group with a, b ∈ G, then (ab)−1 = b−1a−1.

Proof. We see this to be obviously true with

(ab)(ab)−1 = (ab)(b−1a−1)= a(bb−1)a−1

= aea−1

= aa−1

= e.

Thus, the given inverse achieves its purpose. �

Stay tuned for the next exciting part of the series... Permutations!

∼⊙⊙^./

2. Permutations (by Anne Ho)

Definition. A function f : X → Y is injective if for every x1, x2 ∈ X, when f(x1) = f(x2), x1 = x2. Afunction f : X → Y is surjective if for every y ∈ Y , there exists an x ∈ X such that f(x) = y. A functionf :→ Y is bijective if it is both injective and surjective.

Definition. A function is a permutation if it is a bijection onto itself f : X → X. Notation: X = {1, 2, ...n}.

α =(

1 2 ... nα(1) α(2) ... α(n)

).

Definition. Sn is the group of permutations called the symmetric group.

Example. Suppose X = {1, 2, 3, 4} and α =(

1 2 3 43 1 4 2

).

α ∈ S4, the set of permutations on X, and α(1) = 3, α(2) = 1, α(3) = 4, α(4) = 2.

Suppose X = {1, 2, 3, 4} and β =(

1 2 3 42 3 1 4

).

β ∈ S4 as well.Composing α ◦ β(1), we get α(2) = 1.

Example. Let X = {1, 2, 3}. We list the elements of S3:

I =(

1 2 31 2 3

)α =

(1 2 33 1 2

)β =

(1 2 33 2 1

)γ =

(1 2 31 3 2

)δ =

(1 2 32 1 3

)µ =

(1 2 32 3 1

)Example. Consider the Dihedral Group, D4.

The identity is I =(

1 2 3 41 2 3 4

)

A rotation of 90-degrees would be: R90 =(

1 2 3 42 4 1 3

)We can do this for the other rotations and flips about various axes too: R180, R270, H, V,D1, D2.

Definition. If α ∈ Sn and i ∈ {1, 2, ..., n}, then α fixes i if α(i) = i. α moves i if α(i) 6= i.

Definition. Let i1, i2, ..., ir be elements in X = {1, 2, ..., n}. If α ∈ Sn fixes the other integers, and ifα(i1) = i2, α(i2) = i3, ..., α(ir−1) = in and α(ir) = i1, then α is called an r-cycle. A 2-cycle is called atransposition. A 1-cycle is just the identity.

Example. α =(

1 2 3 4 54 3 1 5 2

)=(1 4 5 2 3

)Example. α =

(1 2 3 4 52 3 1 4 5

)=(1 2 3

) (4) (

5)

=(1 2 3

)Example. Write as a product of disjoint cycles. α =

(1 2 3 4 5 6 7 8 96 4 7 2 5 1 8 9 3

)=(1 6

) (2 4

) (3 7 8 9

) (5)

Example. Product of Permutations. α =(1 2

) (1 3 4 2 5

) (2 5 1 3

)We see that:1→ 3→ 42→ 5→ 1→ 23→ 2→ 54→ 2→ 15→ 1→ 3So we have: =

(1 4

) (2) (

3 5)

=(1 4

) (3 5

)Definition. Two cycles α = (a1a2...am) and β = (b1b2...bm) are disjoint if ai 6= bj for all i, j.1

Lemma 2.1. Disjoint cycles α, β ∈ Sn commute.

Note. Every permutation is a product of disjoint cycles.

Theorem 2.2. (1) The inverse of (i1, i2, ..., ir) is the r-cycle (ir, ir−1, ..., 1).(2) If α = β1β2...βt is a product of disjoint cycles, then α−1 = β−1

1 β−12 ...β−1

t .

Proof. (1) Consider

(i1, i2, ..., ir)(ir, ir−1, ..., 1).

We see that:i1 → ir → i1i2 → i1 → i2...ir−1 → ir−2 → ir−1

ir → ir−1 → irSo (i1, i2, ..., ir)(ir, ir−1, ..., 1) = (1)Similarly, (ir, ir−1, ..., 1)(i1, i2, ..., ir) = (1).

(2) Consider (β1β2...βt)(β−1t β−1

t−1...β−11 ).

We see that βtβ−1t = 1, βt−1β

−1t−1 = 1 and so on. Thus we have 1.

Similarly, (β−1t β−1

t−1...β−11 )(β1β2...βt) = 1.

Since α = β1β2...βt, then α−1 = β−1t β−1

t−1...β−11 . From the Lemma, we know that α−1 =

β−11 β−1

2 ...β−1t .

�

Note. If n ≥ 2, then every α ∈ Sn is a product of transpositions. To check this, consider:(1 2 ... r

)=(1 r

) (1 r − 1

)...(1 3

) (1 2

)We see that:1→ 22→ 1→ 33→ 1→ 4...r − 1→ 1→ r

Definition. A permutation is even if it can be factored into a product of an even number of transpositions;otherwise, it is odd. The parity of a permutation is whether it is even or odd.

Example. α =(1 2 3

)=(1 3

) (1 2

)So it is even.

1Due to some confusion during the lecture, this definition was later added to these notes and is from Durbin’s Modern

Algebra, Fifth Edition.

3. Subgroups, Order and Cyclic Groups (by Kirsten Hogenson)

Definition. A subset H of a group G is a subgroup of G if the following axioms hold:(i) 1 ∈ H(ii) If x,y ∈ H, then xy ∈ H (closure)(iii) If x ∈ H, then x−1 ∈ H (inverses).

Example. Let G = R∗ = R− {0} and H = {2n : n ∈ Z}. Show that < H, • > is a subgroup of < G, • >.(i) 20 = 1, so 1 ∈ H

(ii) Let x, y ∈ H such that x = 2n and y = 2m. Then xy = 2n2m = 2n+m. Since n+m ∈ Z, xy ∈ H.(iii) Let x ∈ H such that x = 2n. Then x−1 = 2−n because 2n2−n = 1.

Theorem 3.1. Let G be a group and H a nonempty subset of G. If a, b ∈ H implies ab−1 ∈ H, then H isa subgroup of G.

Proof. (i) Since H is nonempty, choose x ∈ H. Let a = b = x. Then ab−1 = xx−1 = 1. So 1 ∈ H.(ii) Let a = e ∈ H, b = x. Then ab−1 = ex−1 = x−1. So x−1 ∈ H.(iii) Let a = x, b = y−1 for x, y ∈ H. Then xy = x(y−1)−1 = ab−1 ∈ H.

�

Example. Let < G, • > be an abelian group. Show H = {x2|x ∈ G} is a subgroup of G.Let a2, b2 ∈ H. Then a2(b2)−1 = a2b−2 = ab−1ab−1 = (ab−1)2 ∈ H.

Definition. The number of elements in a group is called its order.

Definition. The order of an element g ∈ G, denoted |g|, is the smallest positive integer n such that gn = 1.If no such integer exists, we say that g has infinite order.

Example. Z6 = {0, 1, 2, 3, 4, 5}|0| = 1 |1| = 6 |2| = 3 |3| = 2 |4| = 3 |5| = 6

Example. U(10) = {k| gcd(k, 10) = 1} = {1, 3, 7, 9}. With ∗ = • mod 10,

|1| = 1 |3| = 4 |7| = 4 |9| = 2

Definition. Let < a > denote the set {an : n ∈ Z}.

Theorem 3.2. Let G be a group, and let a ∈ G. Then < a > is a subgroup of G.

Proof. Prove in homework. �

Example. Z6 = {0, 1, 2, 3, 4, 5}< 0 >= {0} < 1 >= Z6 < 2 >= {0, 2, 4} < 3 >= {0, 3}

The order of the element is equal to the order of the subgroup generated by the element. |a| = | < a > |.

Definition. The center Z(G) of a group G is the subset of elements in G that commute with every otherelement in G. Symbolically, Z(G) = {a ∈ G|ag = ga ∀g ∈ G}.

Theorem 3.3. The center of a group G is a subgroup of G.

Proof. Prove in homework. �

Definition. A group G is called cyclic if there is an element a such that

G = {an|n ∈ Z}.The element a is called the generator.

Example. Zn = {0, 1, · · · , n− 1} is always generated by 1 or n− 1, and may be generated by other numbersalso. For instance,

Z8 =< 1 >=< 7 >=< 3 >=< 5 > .

Example. U(8) = {1, 3, 5, 7}|1| = 1 |3| = 2 |5| = 2 |7| = 2

Theorem 3.4. Let G be a group and a ∈ G.(i) If a has infinite order, then ai = aj iff i = j.

(ii) If a has finite order n, then < a >= {e, a, · · · , an−1} and ai = aj iff n divides i− j.

Proof. (i) Suppose a has infinite order. Then there is no n such that an = e.(⇒) Assume ai = aj . Then aia−j = e. So ai−j = a0. Thus i− j = 0 and i = j.(⇐) If i = j, then ai = aj follows directly.

(ii) Suppose a has finite order n. Then |a| = n. Choose x ∈ {e, a, a2, · · · , an−1}. Then x = ak for somek ∈ Z so by definition, x ∈< a >. Next choose x ∈< a >. Then x = ak for some k ∈ Z. By thedivision algorithm, k = nq + r with 0 ≤ r < n. So

ak = anq+r = anqar = ear = ar.

Since 0 ≤ r < n and r ∈ Z, x ∈ {e, a, · · · , an−1}.So, < a >= {e, a, · · · , an−1}.Now, we’ll show that ai = aj iff n divides i− j.

(⇒) Suppose ai = aj . Then ai−j = e = a0. By the division algorithm, i− j = nq + r for 0 ≤ r < n.So

e = ai−j = anq+r = anqar = ear = ar.

This implies that r = 0, so i− j = nq, which is divisible by n.(⇐) Suppose n divides i − j. Then i − j = nk for some k ∈ Z. So ai−j = ank = ek = e. Thus,

ai−j = e so ai = aj .�

Corollary 3.5. For any group element a, |a| = | < a > |.

Corollary 3.6. Let G be a group and let a have order n. If ak = e, then n divides k.

Definition. If H is a subgroup of G and a ∈ G, then the left coset aH is the subset

aH = {ah : h ∈ H}.

Example. G = S3 = {(1), (1, 2), (1, 3), (1, 2, 3), (1, 3, 2), (2, 3)} and H = {(1), (1, 2)}.

(1, 3)H = {(1, 3), (1, 2, 3)} (2, 3)H = {(2, 3), (1, 3, 2)}.

Example. G = Z12 = {0, 1, · · · , 11} and H = {0, 3, 6, 9}.

1 +H = {1, 4, 7, 10} 2 +H = {2, 5, 8, 11}.

4. Cosets, Lagrange’s Theorem, and Normal Subgroups (by Lilith Ciccarelli)

Lemma 4.1. Let H be a subgroup of G with a, b ∈ G.(1) aH = bH if and only if b−1a ∈ H. In particular, aH = H if and only if a ∈ H.(2) If aH ∩ bH 6= ∅, then aH = bH.(3) |aH| = |H| for all a ∈ G

Proof. (1) (=⇒) Assume aH = bH. Then a ∈ bH, which implies that a = bh0 for some H.⇒ b−1 = h0 ⇒ b−1a ∈ H(⇐=) Assume b−1 ∈ H, which implies b−1a = h0 ∈ H which then implies that a = bh0 and b = ah−1

0 ,since H is a subgroup.We want to show aH = bH. Begin by choosing x ∈ aH. Then x = ah′ = bh0h

′. Since h0h′ ∈ H

(through closure), x ∈ bH ≤ aH.Now choose y ∈ bH. This implies that y = bh′ = ah−1

0 ∈ aH, and thus bH ≤ aH. Throughcontainment, then aH = bH.Now we prove that aH = H if and only if a ∈ H.(=⇒) Let aH = H = 1 ∗H. Then a−1 ∈ H and a ∈ H.(⇐=) Let a ∈ H. Then 1 ∗ a ∈ H, which implies that aH = H.

(2) Let aH ∩ bH 6= ∅ Then there exists an x such that x ∈ aH ∩ bH. x = ah for some h ∈ H = bh′

for some h′ ∈ H. Then b−1ah = h′, so b−1a = h′h−1 ∈ H. Now we know from part 1 that, sinceb−1a ∈ H, aH = bH.

(3) To show that |aH| = |H|, define f : H → aH such that f(h) = ah for each h ∈ H.�

Theorem 4.2 (Lagrange’s Theorem). If H is a finite subgroup of G, then |H| is a divisor of |G|.

Proof. Let {a1H, a2H, ..., atH} be the family of distinct cosets of H. Then G = a1H ∪ a2H ∪ ... ∪ atH. Forevery g ∈ G, gH = aiH for some i. By part 2 of the previous theorem, for some i 6= j, aiH and ajH aredisjoint.

|G| =t∑i=1

|aiH|

Since |aiH| = |H|, |G| = t|H|.t = [G : H] ([G : H] = |G|

|H| is called the index of H in G.) �

Corollary 4.3. If G is a finite group, and H is a subgroup of G, then [G : H] = |G||H| .

Corollary 4.4. In a finite group G, the order of each element divides the order of the group. I.e. |a| = | <a > | = |G|

[G:H] .

Corollary 4.5. A group of prime order is cyclic.

Proof. Suppose G has prime order p. Then let a ∈ G, with a 6= e. Then |a| divides |G|. Since a 6= e,| < a > | 6= 1. Then | < a > | = |G| = P , since that’s the only other possibility. Thus, G is generated bya. �

Side note: aH = Ha does not imply ah = ha. Instead, it means ah = h′a, where h may or may not equalh′.

Definition. A subgroup H of a group G is called a normal subgroup of G if aH = Ha for all a in G.

Theorem 4.6 (Normal Subgroup Test). A subgroup H of G is a normal subgroup of G if and only ifxHx−1 ⊆ H for all x ∈ G.

Proof. (=⇒)Let H be a normal subgroup of G. Then aH = Ha. For x ∈ G, and h ∈ H, there is anotherelement h′ ∈ H such that xh = h′x. This means xhx−1 = h′ ∈ H and thus, xHx−1 ⊆ H.(⇐=) Let xHx−1 ⊆ H for all x. Then for x = a, aHa−1 ⊆ H, aH ⊆ Ha and xH ⊆ Hx.For x = a−1, a−1H(a−1)−1 ⊆ H,Ha ⊆ aH and Hx ⊆ xH. This implies that xH = Hx and therefore, H isa normal subgroup. �

Example. Every subgroup of an abelian group is normal, since ah = ha implies aH = Ha.

Example. The center Z(G) is a normal subgroup, since for h ∈ Z(G) and a ∈ G, ah = ha implies aH = Ha.

Example. Let An be the group of even permutations of n. S3 = {(1), (1, 2), (1, 3), (2, 3), (1, 2, 3), (1, 3, 2)}A3 = {(1, 2, 3), (1, 3, 2), (1)} (This is the alternating group of 3 elements) (1, 2)(1, 2, 3) = (1, 3, 2)(1, 2) = (2, 3)H = A3 is a normal subgroup, since ah = h′a for some a ∈ G, h, h′ ∈ H.

Definition. A Hamiltonian group is a group in which all its subgroups are normal.

5. Quotient Groups and Homomorphisms (by Linda Mummy)

Example. Dihedral GroupG = D4 = {r0, r90, r180, r270, h, v, d, d

′}H = {r0, r180}

The cosets areH = {r0, r180}

r90H = {r90, r270}

hH = {h, v}

dH = {d, d′}

There are no more cosets (recall that 2 cosets are either equal or their intersection is ∅.)

Is H normal, i.e., H / G? We need to check to see if gH = Hg ∀ g ∈ G. (Equivalently, xHx−1 ⊆ ∀x ∈ G.)We know

xr180x−1 = xx−1 = r0 ∈ H ∀ x.

Now we only need to check that xr180x−1 ∈ H ∀x. A quick check shows that it is a normal subgroup since

xr180x−1 ⊆ H ∀x ∈ D4.

Theorem 5.1. Let [G, ?] be a group and H / G. Then the set of left cosets G/H = {gH|g ∈ G} forms agroup under the operation (aH)(bH) = (a ? b)H.

Proof. Prove that the operation is well defined and has closure. Suppose aH = a′H and bH = b′H wherea, a′, b, b′ ∈ G. Then there exists h1, h2 ∈ H such that a′ = ah1 and b′ = bh2

(a′H)(b′H) = (ah1H)(bh2H) = (aH)(bH)

There is closure. The identity is eH = H. Inverses exist: (aH)−1. The elements are also associative:(aH)(bH)cH = (abH)cH = abcH = aH(bcH) = aH(bH)(cH). This is called the quotient group of G by H(aka the factor group of G by H). �

Example. G = D4, H = {r0, r180} = {H, r90H, vH, dH}

MULTIPLICATION TABLE for D4

◦ H r90H vH dHH H r90H vH dHr90H r90H H dH vHvH vH dH H r90HdH dH vH r90H H

Example. G = ZH = nZ = 〈n〉 = {0,±n,±2n, . . .}.

H is normal because G is abelian. By the theorem, G/H exists. The elements of G/H are:

0 + nZ = {0, 0± n, . . .}

1 + nZ = {1, 1± n, . . .}

2 + nZ = {2, 2± n, . . .}

...

n− 1 + nZ = {n− 1, n− 1± n, . . .}

(a+ nZ) + (b+ nZ) = (a+ b) + nZ

Say n = 5. So G/H = Z/(5Z) and thus

(1 + 5Z) + (2 + 5Z) = 3 + 5Z

(2 + 5Z) + (4 + 5Z) = 1 + 5Z

This is just Z5 (and in general Zn).

QUESTION: What is the order of G/H? If |G| <∞ then Lagrange’s Theorem implies

|G/H| = [G : H] =|G||H|

.

If |G| =∞, anything is possible. For example,

|Z/(nZ) = n| for n > 0

|Z/0| =∞

Example. The set of even permutations A4 has no subgroup of order 6. |A4| = 12 a subgroup of order 6 hasindex 2. If there exists a subgroup H of order G, then [G : H] = 12

6 = 2 and so H /G. Use this fact to checkthat A4 does not in fact have such a subgroup.

Definition. Let G, G be groups. We say a mapping φ : G→ G is a homomorphism if

φ(gh) = φ(g)φ(h) for all g, h ∈ G.We define the kernel of φ as

kerφ : {g ∈ G|φ(g)} = 1Gand the image of φ a

imφ = {φ(g)|g ∈ G}.We say φ is an isomorphism if φ is 1-1 and onto (bijective).

Note: φ is 1-1 ⇐⇒ kerφ is trivial.

Proof. (⇒) If φ is 1-1 then φ(a) = φ(b) ⇐⇒ a = b. Notice φ(1) = 1 since

1G · φ(1) = φ(1 · 1) = φ(1)φ(1)

1G = φ(1G)Thus 1G is the only thing that maps to 1G which implies that kerφ = 1G.(⇐) Suppose φ(a) = φ(b) for a, b ∈ G. Notice that φ(b−1) = φ(b)−1 as

φ(b)φ(b−1) = φ(b · b−1) = φ(1) = 1.

φ(b)φ(b−1) = φ(b · b−1) = φ(1) = 1.Multiply both sides of φ(a) = φ(b) by φ(b−1) to get

φ(ab−1) = φ(a)φ(b−1) = φ(b)φ(b−1) = 1

However,kerφ = {1} ⇒ ab−1 = 1⇒ a = b.

�

Note: φ is onto ⇐⇒ im φ = G. (This follows directly from the definition of image)

Examples. (1) φ : Z→ Znm 7→ m mod n (φ(m) = m mod n)

φ is a homomorphism as it is well defined and φ(ab) = φ(a)φ(b) for all a, b ∈ Z(2) φ : G→ G/H, H / G

g → gH

Is φ(ab) = φ(a)φ(b)? Yes, by how we defined quotient groups.(3) φ : Z→ Z

x 7→ x2

φ(a+ b) = (a+ b)2

φ(a)φ(b) = a2 + b2

Since φ(a+ b) 6= φ(a) + φ(b), φ is not a homomorphism.

(4) φ : Zn → Z/(nZ)m 7→ m+ nZ

This is a homomorphism (and an isomorphism!)

PROPERTIES OF HOMOMORPHISMS(1) φ(1G) = 1G(2) φ(gn) = (φ(g))n since φ(gg . . . g︸ ︷︷ ︸

n

) = φ(g)φ(g) . . . φ(g)︸ ︷︷ ︸n

(3) If |g| <∞, |φ| | |g|(4) ker φ ≤ G(5) If H ≤ G, then φ(G) ≤ G.

φ(a)φ(b−1) = φ(ab−1) ⊆ φ(H) for all a, b ∈ H

(6) H cyclic ⇒ φ(H) is cyclic(7) H abelian ⇒ φ(H) is abelian(8) H / G, then φ(H) / φ(G)(9) If |H| = n, then |φ(H)| | n

Theorem 5.2 (First Isomorphism Theorem). Let G, G be groups and φ : G → G be an isomorphism.Then

τ : G/ kerφ→ imφ

g kerφ 7→ φ(g)

is an isomorphism, writtenG/ kerφ ∼= imφ

Proof. τ is well defined. Let H = kerφ. Suppose xH = yH for x, y ∈ G. We want to show that φ(x) = φ(y),because we want τ(xH) = τ(yH). We know x−1yH = H Therefore

x−1y ∈ H = kerφ

φ(x−1y) = 1

φ(x−1)φ(y) = 1

φ(y) = φ(x)

τ is a homomorphism.τ(xy,H) = φ(xy) = φ(x)φ(y) = τ(xH)τ(yH)

Is this map onto? Let g′ ∈ imφ. Then g′ = φ(g) for some g ∈ G. Then τ(gH) = φ(g) = g′. Is τ 1-1? Wewant to show that τ(xH) = τ(yH) thenxH = yH. Suppose

τ(xH) = τ(yH)⇒ φ(x) = φ(y)

⇒ 1 = φ(y)φ(x−1) = φ(yx−1).

⇒ yx−1 ∈ kerφ = H

yx−1H = H ⇒ yH = xH

�

Definition. For two groups G,H define the external direct product of G and H as

G⊕H = {(g, h)|g ∈ G, h ∈ H}

where the operations are componentwise.

6. Definition of a Ring (by Rob Davis)

Definition. A ring R is a set with two binary operations, addition (a+ b) and multiplication (ab) such thatfor all a, b ∈ R, the following hold:

(1) a+ b = b+ a

(2) a+ (b+ c) = (a+ b) + c

(3) There exists 0 ∈ R such that a+ 0 = a

(4) For every a ∈ R, there exists −a such that a+ (−a) = −a+ a = 0These four properties tell us that R is an abelian group over addition. A ring further requires that

(5) a(bc) = (ab)c(6) a(b+ c) = ab+ ac and (b+ c)a = ba+ ca

If ab = ba, then R is a commutative ring. If R has an identity for multiplication, then it is a ring withunity. Further, a ring with unity does not necessarily have a multiplicative inverse for each element. Thoseelements that do have an inverse are called units.

Examples. • Z is a commutative ring with a unity element 1 and units {1,−1}

• R,C,Q are commutative rings with unity and, and units are all nonzero elements of their respectivesets

• Zn is a commutative ring with unity and units U(n)

• nZ is a commutative ring and without unity for n > 1

• GL(2,Z) is a noncommutative ring with unity element[1 00 1

]• GL(2, nZ) is a noncommutative ring, and without unity for n > 1

• GL(2, Zn) is a finite, noncommutative ring with unity

• R = the set of real-valued functions is a ring with unity f(x) = 1

We again emphasize that rings do not necessarily have a multiplicative inverse for each element, thereforecancellation is not implied. That is,

ab = ac⇒ b = c

anda2 = a⇒ a = 0 or a = 1.

If we do have the cancellation property, then R is an integral domain.

Rules of Multiplication:(1) a · 0 = 0

(2) a(−b) = (−a)b = −ab

(3) (-a)(-b) = ab

(4) a(b− c) = ab− ac and (b− c)a = ba− ca

If R has unity 1,(5) (−1)a = −a

(6) (-1)(-1) = 1Partial Proof :

(1) We see that 0 + a0 = a0 = a(0 + 0) = a0 + a0, hence 0 + a0 = a0 + a0. Subtracting a0 from bothsides leaves us with 0 = a0.

(2) Notice that a(−b) + ab = a(−b + b) = a0 = 0, so a(−b) + ab = 0. Subtracting ab from both sidesyields a(−b) = ab. A similar argument works to show (−a)b = −ab.

(3) The equations 0 = 0(−b) = (−a + a)(−b) = (−a)(−b) + a(−b) imply 0 = (−a)(−b) + a(−b). If weadd ab to each side, we see ab = (−a)(−b) + a(−b) + ab = (−a)(−b). Therefore, ab = (−a)(−b).

Definition. A subset S of a ring R is a subring of R if S is itself a ring with the operations on R.

Theorem 6.1. A nonempty subset S of a ring R is a subring if it is closed under subtraction and multipli-cation (Check to see if a− b and ab are in S when a, b ∈ S).

Examples. • {0}, R

• R = Z6 has unity 1, S = {0, 2, 4} has unity 4

• R = Z10, S = {0, 2, 4, 6, 8} has unity 6

• R = Z, S = nZ

• Each of the following is a subring of the other: 6Z ⊂ 3Z ⊂ Z ⊂ Q ⊂ RC. We can also writeZ[i] = {a+ bi|a, b ∈ Z} ⊂ C.

7. Integral Domains and Fields (by Sarah Behrens)

Definition. A zero-divisor is a nonzero element a in a commutative ring R such that there is a nonzeroelement b with ab = 0.

Definition. An integral domain is a commutative ring with unity that has no zero-divisors. (A productequals zero only when one of the factors is zero.)

Example. Z,Z[i],Zp(p is prime),C,R,QNote. The existence of zero-divisors in a ring makes it hard to find zeros of polynomials whose coefficientsare ring elements.

Example. x2 − 4x+ 3 = 0Solve in Z and Z12.Z: 0 = x2 − 4x+ 3 = (x− 3)(x− 1). Thus x = {1, 3}.Z12: (Method: check each element) x = {1, 3, 7, 9}.

Theorem 7.1. Let a, b, c belong to an integral domain. If a 6= 0 and ab = ac, then b = c. (CancellationProperty)

Proof. ab = acab− ac = 0a(b− c) = 0Since a is not a zero-divisor:b− c = 0b = c

�

Definition. A field is a commutative ring with unity in which every nonzero element is a unit (has aninverse).

Note. (1) A field is a type of integral domain.(Let a, b belong to a field F . Choose a 6= 0 withab = 0a−1ab = a−10b = 0⇒ no zero divisors.)

(2) A field is an algebraic system that is closed under +,−, ·,÷ (except 0).(3) Example. C,R,Q,Zp (p is prime).

Theorem 7.2. A finite integral domain is a field.

Proof. Let D be a finite integral domain with unity 1. Let a be any nonzero element of D. (Goal: show ais a unit.)If a = 1, then a is its own inverse, so we can assume that a 6= 1. Consider the sequence of elementsa, a2, a3, · · · . Since D is finite, there must be ai = aj for some i > j.

ai = aj

aj+k − aj = 0aj(ak − 1) = 0⇒ ak − 1 = 0

ak = 1But since a 6= 1

⇒ k > 1aak−1 = 1⇒ a−1 = ak−1

�

Corollary 7.3. Zp is a field.

Proof. Suppose a, b ∈ Zp and ab = 0. Then ab = pk for some k. By Euclid’s Lemma (below) p divides a orp divides b so either a = pm or b = pn in Zp. Hence a = 0 or b = 0 and a, b are not zero-divsors. Thus Zp isan integral domain. By the previous theorem, Zp is a field. �

Lemma 7.4 (Euclid’s Lemma). If p is a prime and p divides ab, then p divides a or p divides b.

Example. (Another finite field example)Z3[i] = {a+ bi : a, b ∈ Z3} = {0, 1, 2, i, 2i, 1 + i, 1 + 2i, 2 + i, 2 + 2i}Q[√

2] = {a+ b√

2 : ab ∈ Q}

Definition. The characteristic of a ring R is the least positive integer n such that nx = 0 ∀x ∈ R

• If no such integer exists we say R has characteristic zero.• We denote the characteristic by charR.

Example. Z3[i], char Z3[i] = 3 x+ x+ x = 0Zn has characteristic n{0, 3, 6, 9}12 has characteristic 4

Theorem 7.5. Let R be a ring with unity 1.

• If 1 has infinite order under addition, then charR = 0.• If 1 has order n under addition, then charR = n.

Proof. • If 1 has infinite order, then there is no n such that n · 1 = 0, so R has characteristic 0.

• If 1 has order n, then n · 1 = 0 and n is the least such n with this property. But for x,n · x = x+ x+ · · ·+ x

= 1x+ 1x+ · · ·+ 1x= (1 + 1 + · · ·+ 1) · x= (n · 1) · x= 0 · x= 0

Thus R has characteristic n.�

8. Ideals and Factor Rings (by Annette Honken)

Definition. A subring A of a ring R is called an ideal of R if for every r ∈ R and a ∈ A, ar ∈ A and ra ∈ A.rA = {ra|a ∈ A} ⊂ A (left ideal)

Ar = {ar|a ∈ A} ⊂ A (right ideal)

Theorem 8.1. A nonempty subset A of a ring R is an ideal of R if(1) a− b ∈ A whenever a, b ∈ A(2) ra and ar are in A whenever a ∈ A and r ∈ R

Examples of Ideals:(1) R and {0} are ideals of R(2) R = Z and nZ = {0,±n,±2n, ...} is an ideal of Z

Definition. Let R be a commutative ring with unity and let a ∈ R. The set < a >= {ra|r ∈ R} is an idealof R called the principal ideal generated by a.

Example. (1) R[x] = ring of polynomials with real coefficients. A =< x >= { set of polynomials witha0 = 0} (A is an ideal of R[x]).

(2) I =< a1, a2, ..., an >= {r1a1 + r2a2 + ...+ rnan}

Definition. Since R is a group under addition and A is a normal subgroup of R, we can form the factorring R/A = {r +A|r ∈ R}.

Theorem 8.2. Existence of Factor Rings - Let R be a ring and let A be a subring of R. The set of cosets{r+A|r ∈ R} is a factor ring under the operations (s+A)+(t+A) = (s+ t)+A and (s+A)(t+A) = st+Aif and only if A is an ideal of R.

Proof. Know {r+A} forms a group under addition. (Assoc. and dist. are trivial). Check: Multiplication isa binary operation. Show A is an ideal if and only if multiplication is well-defined.

(⇒) A is an ideal. Let s + A = s′ + A and t + A = t′ + A. Since A is an ideal, s = s′ + a and t = t′ + bfor a, b ∈ A.

st = (s′ + a)(t′ + b) = s′t′ + at′ + s′b+ ab

st+A = s′t′ + at′ + s′b+ ab+A = s′t′ +A

Coset multiplication is well-defined.(⇐) Now, show well-defined implies that A is an ideal. We will show the contrapositive. Thus, we will

show that A not an ideal implies that it is not well-defined. Suppose A is a subring that is not an ideal.Then, there exists a ∈ A and r ∈ R such that ar /∈ A (or ra /∈ A). Assume it is ar that is not in A. Considera + A = 0 + A and r + A. We see that (a + A)(r + A) = ar + A, but (0 + A)(r + A) = 0 + A = A. Wenote that ar +A 6= A. Then, multiplication is not well-defined, which implies that the set of cosets is not aring. �

Examples. • Z/4Z = {0 + 4Z, 1 + 4Z, 2 + 4Z, 3 + 4Z}• 2Z/6Z = {0 + 6Z, 2 + 6Z, 4 + 6Z}

• R ={[a1 a2

a3 a4

]|ai ∈ Z

}, I =

{[b1 b2b3 b4

]|bi ∈ 2Z

}, R/I =

{[c1 c2c3 c4

]+ I|ci ∈ {0, 1}

}

• R[x] = ring of polynomials with real coefficients. < x2 + 1 >= principal ideal generated by x2 + 1 ={f(x)(x2 +1)|f(x) ∈ R[x]}. Then, R[x]/ < x2 +1 >= {g(x)+ < x2 +1 > |g(x) ∈ R[x]} = {ax+b+ <x2 + 1 > |a, b ∈ R}. g(x) = q(x)(x2 + 1) + r(x) by division algorithm - degree of r(x) < 2⇒ r(x) =ax+b. g(x)+ < x2 +1 >= q(x)(x2 +1)+r(x)+ < x2 +1 >= r(x)+ < x2 +1 >= ax+b+ < x2 +1 > .

• R = Z[i]/ < 2− i > Elements have the form a+bi+ < 2− i > . Since 2− i+ < 2− i >= 0+ < 2− i >,2− i = 0⇒ i = 2⇒ i2 = −1 = 4⇒ −1 = 4⇒ 0 = 5.

Definition. A prime ideal of a commutative ring R is a proper ideal of R such that a, b ∈ R and ab ∈ Aimply a ∈ A or b ∈ A.

Example. R = Z, I = 5Z, I = 6Z

Definition. A maximal ideal A of a commutative ring R is a proper ideal of R such that whenever B is anideal of R containing A, A ⊆ B ⊆ R, then either A = B or B = R. (Can’t have A ⊂ B ⊂ R).

Theorem 8.3. Let R be a commutative ring with unity with A an ideal of R. R/A is an integral domain ifand only if A is prime.

Proof. (⇒) Suppose R/A is an integral domain. Let ab ∈ A. Then, (a+A)(b+A) = ab+A = A. Since R/Ais an integral domain, either a+A = A or b+A = A, so either a ∈ A or b ∈ A. Therefore, A is prime.

(⇐) Know R/A is a commutative ring with unity. Our goal is to show that when A is prime, R/A has nozero divisors. Let A be prime. (ab ∈ A⇒ a ∈ A or b ∈ A). Let (a+A)(b+A) = A (want either a+A = Aor b + A = A). ⇒ ab + A = A ⇒ ab ∈ A Since A is prime, a ∈ A or b ∈ A. (a + A = A or b + A = A). So,R/A is an integral domain. �

Theorem 8.4. Let R be a commutative ring with unity with A an ideal of R. R/A is a field if and only ifA is maximal.

Proof. (⇒) Suppose R/A is a field. let B be an ideal that properly contains A. A ⊂ B ⊆ R Choose b ∈ Bsuch that b /∈ A. Then, b + A 6= 0 + A. Since R/A is field, there exists c + A such that (b + A)(c + A) =1 +A⇒ bc+A = 1 +A⇒ 1− bc ∈ A. 1 = (1− bc) + bc ∈ B so B = R and A is maximal.

(⇐) Suppose A is maximal. Let b ∈ R, but b /∈ A. Show b+A has an inverse. B = {br+A|r ∈ R, a ∈ A}.This B is an ideal that properly contains A. A ⊂ B ⊆ R Since A is maximal, B = R. Then, 1 ∈ B and1 = bc+ a′. 1 +A = bc+ a′ +A = bc+A = (b+A)(c+A), so (b+A) has an inverse and R/A is a field. �

9. Ring Homomorphisms (by Nathan McNew)

Definition. A ring homomorphism φ from a ring R to a ring S is a mapping that preserves ring operations.Specifically, for all a, b ∈ R

φ(a+ b) = φ(a) + φ(b)

φ(ab) = φ(a)φ(b)A ring homomorphism that is both 1-1 and onto is a ring isomorphism.

Examples. • φ : Z→ Zn, φ(k) = k mod n• φ : C→ C, φ(a+ bi) = a− bi• φ : R[x]→ R, φ(f(x)) = f(1)• φ : Z4 → Z10, φ(x) = 5x

Proof. Let x, y in Z. By the division algorithm let x+ y = 4q1 + r1 and xy = 4q2 + r2. Then we canwrite r1 = x+ y − 4q1 and r2 = xy − 4q2. Then working first within Z4 we have:

φ(x+ y) = φ(r1) = 5(x+ y − 4q1)= 5x+ 5y − 20q1

= 5x+ 5y (in Z10)

= φ(x) + φ(y)

and

φ(xy) = φ(r2) = 5(xy − 4q2)= 5xy − 20q1

= 25xy (in Z10)

= (5x)(5y)

= φ(x)φ(y)

�

Note. Group isomorphic sets are not necessarily ring isomorphic. Take for example Z and 2Z.

PROPERTIES OF RINGS

Theorem 9.1. Let φ be a ring homomorphism from a ring R into a ring S. Let A be a subring of R andlet B be an ideal of S.

(1) For any r ∈ R and positive integer n, φ(nr) = nφ(r) and φ(rn) = φ(r)n).(2) φ(A) = {φ(a)|a ∈ A} is a subring of S.(3) If A is an ideal and φ is onto then φ(A) is an ideal.(4) φ−1(B) is an ideal in R.(5) If R is commutative then φ(R) is commutative.(6) if R has unity 1, S 6= {0} and φ is onto then φ(1) is the unity.(7) φ is an isomorphism if and only if φ is onto and kerφ = 0.(8) If φ is an isomorphism then φ−1 is an isomorphism.

Theorem 9.2. Let φ be a ring homomorphism from a ring R to a ring S. Then kerφ is an ideal of R.

Proof. Let a, b ∈ kerφ, r ∈ R. then consider φ(a − b) = φ(a) + φ(−b) = φ(a) − φ(b) = 0 − 0 = 0. Thusa− b ∈ kerφ. Also, φ(ra) = φ(r)φ(a) = φ(r)0 = 0, so ra ∈ kerφ. Thus kerφ is an ideal. �

Theorem 9.3 (First Isomorphism Theorem for Rings). Let φ be a ring homomorphism from R to S. Thenthe mapping from R/ kerφ to φ(R) given b r + kerφ→ φ(r) is an isomorphism. (R/ kerφ ∼= φ(R))

THE FIELD OF QUOTIENTSRecall Z is not a field, but Q is.

Theorem 9.4. Let D be an integral domain. Then there exists a field F called the field of quotients of Dthat contains a subring isomorphic to D.

Proof. Let S = {(a, b)|a, b ∈ D, b 6= 0} Define an equivalence relation on S by (a, b) ≡ (c, d) if ad = bc. LetF be the set of equivalence classes of S under the relation ≡ and denote the equivalence class containing(x, y) by x

y . Define addition and multiplication by

a

b+c

d=ad+ bc

ad(ab

)( cd

)=ac

bd

We now show that these relations are well-defined on F . Let ab = a′

b′ and cd = c′

d′ .

Addition:(

Want to show ad+bcad = a′d′+b′c′

a′d′

)(ad+ bc)b′d′ = adb′d′ + bcb′d′

= ab′dd′ + cd′bb′

= a′bdd′ + c′dbb′

= a′d′bd+ b′c′bd

= (a′d′ + b′c′)bd

Multiplication:(

Want to show acbd = a′c′

b′d′

)acb′d′ = ab′cd′

= ba′dc′

= bd(a′c′)

For F to be a field, let 1 be the unity element for D. Then 01 is the additive identity for F , is the multiplicative

identity, −ab is the additive inverse and ba is the multiplicative inverse. �

Example. D = Z[x], F = { f(x)g(x) |f(x), g(x) ∈ D, g(x) 6= 0}

10. Polynomial Rings (by Nathan Salazar)

Definition. LetR be a commutative ring. R[x] = {anxn+an−1xn−1+. . . a1x+a0|ai ∈ R, n is a non-negative integer}

is called the ring of polynomials over R in the indeterminate x. Two elements anxn+an−1xn−1 + . . . a1x+a0

and bmxm + bm−1x

m−1 + . . . b1x+ b0 are equal if ai = bi ∀i.

• In Z3[x], if f(x) = x3 and g(x) = x5, then f(x) 6= g(x), even though, in Z3, f(0) = 0 = g(0),f(1) = 1 = g(1), and f(2) = 2 = g(2).

• In Z3[x], let f(x) = 2x3 + x2 + 2x + 2 and g(x) = 2x2 + 2x + 1. Then f(x) + g(x) = 2x3 + x, andf(x)g(x) = x5 + 2x3 + 2.

Division Algorithm Let F be a field and let f(x), g(x) ∈ F [x], with g(x) 6= 0. Then there exist uniquepolynomials q(x) and r(x) in F [x] such that f(x) = g(x)q(x) + r(x), with either r(x) = 0 or deg(r(x)) <deg(g(x)).

Example. Let f(x) = 3x4 + x3 + 2x2 + 1 and let g(x) = x2 + 4x+ 2 in Z5[x]. Then f(x)÷ g(x) =

3x2 + 4x

x2 + 4x+ 2)

3x4 + x3 + 2x2 + 0x+ 1−(3x4 + 2x3 + x2)

4x3 + x2 + 0x+ 1− (4x3 + x2 + 3x)

0x2 + 2x+ 1

Thus, we can write f(x) = (x2 + 4x+ 2)(3x2 + 4x) + (2x+ 1).

Theorem 10.1. If D is an integral domain, then D[x] is an integral domain.

Proof. We know D[x] is a ring, because D is a ring. Also, if D is commutative, then D[x] is commutative,and f(x) = 1 is the multiplicative identity. Now, suppose f(x) = anx

n + an−1xn−1 + . . . a1x + a0 and

g(x) = bmxm + bm−1x

m−1 + . . . b1x + b0, where an 6= 0 and bm 6= 0. Then f(x)g(x) has leading coefficientanbm 6= 0, since D is an integral domain. Therefore, the product of two non-zero polynomials cannot give 0,and so D[x] is an integral domain. �

Theorem 10.2 (The Remainder Theorem). Let F be a field with a ∈ F and f(x) ∈ F [x]. Then f(a) isthe remainder when f(x) is divided by x− a.

Theorem 10.3 (The Factor Theorem). Let F be a field, a ∈ F , and f(x) ∈ F [x]. Then a is a zero off(x) if and only if x− a is a factor of f(x).

Definition. A principle ideal domain is an integral domain R in which every ideal has the form 〈a〉 ={ra|r ∈ R} for some a ∈ R.

Theorem 10.4. Let F be a field. Then F [x] is a principal ideal domain.

Proof. We already know that F [x] is an integral domain. Let I be an ideal in F [x]. If I = {0}, thenI = 〈0〉. If I 6= {0}, then choose an element g(x) ∈ I, where g(x) has minimum degree in I, and g(x) 6= 0.We want to show that I = 〈g(x)〉. We know 〈g(x)〉 ⊆ I. We show that I ⊆ 〈g(x)〉. Choose f(x) ∈ I.Using the division algorithm, f(x) = g(x)q(x) + r(x), with either r(x) = 0 or deg(r(x)) < deg(g(x)). Now,r(x) = f(x)−g(x)q(x) ∈ I. Since g(x) is an element of minimal degree, and deg(r(x)) < deg(g(x)), we musthave r(x) = 0. Therefore f(x) ∈ 〈g(x)〉, and so I ⊆ 〈g(x)〉. �

11. Polynomial Factorization (by Lisa Moats)

Definition. Let D be an integral domain. A polynomial f(x) ∈ D[x] that is neither the zero polynomial nora unit in D[x] is said to be irreducible over D, if, whenever f(x) is expressed as the product f(x) = g(x)h(x),with g(x), h(x) ∈ D[x], then g(x) or h(x) is a unit in D[x].A nonzero, nonunit element of D[x] that is not irreducible over D is called reducible.

Example. f(x) = 2x2 + 4 = 2(x2 + 2) is irreducible over Q and R. However, it is reducible over Z since 2and x2 + 2 are not units in Z.In C, f(x) = 2x2 + 4 = 2(x+ i

√2)(x− i

√2) is reducible over C.

Definition. When the integral domain is a field, f(x) ∈ F [x] is irreducible if f(x) cannot be written as aproduct of two polynomials of lower degree.

Example. f(x) = x2 − 2 = (x−√

2)(x+√

2) is reducible over R but not over Q.

Theorem 11.1. Let F be a field. If f(x) ∈ F [x] and deg f(x) is 2 or 3, then f(x) is reducible over F ifand only if f(x) has a zero in F .

Proof. (⇒) Let f(x) = g(x)h(x) where both g(x) and h(x) are in F [x], and have degree less than that off(x). Then, since deg f(x) = deg g(x) + deg h(x) and deg f(x) = 2 or 3, then at least one of g(x) or h(x)has degree 1. Without loss of generality, assume deg g(x) = 1. Then, g(x) = ax+ b for some a, b ∈ F . SinceF is a field, a−1 ∈ F , so −a−1b ∈ F . Then, g(−a−1b) = a(−a−1b) + b = −b + b = 0 so −a−1b is a zero ofg(x). Thus, −a−1b is a zero of f(x).(⇐) Suppose f(a) = 0 where a ∈ F . Then x − a is a factor of f(x). Therefore, f(x) is reducible, sincef(x) = (x− a)h(x) where h(x) ∈ F [x]. �

Example. In Z5, f(x) = x2 + 1 is reducible since f(2) = 0.In Z3, f(x) = x2 + 1 is irreducible since f(0) = 1, f(1) = 2, and f(2) = 2.

Note. Sometimes a polynomial with degree greater than 3 may be reducible over a field, but have no zerosin the field. For example, f(x) = x4 + 2x2 + 1 = (x2 + 1)2 is reducible over R but has no zeros in R.

Definition. The content of a nonzero polynomial anxn + an−1xn−1 + ...+ a1x+ a0, where ai ∈ Z for each

i with 1 ≤ i ≤ n, is the greatest common divisor of the integers an, an−1, ..., a1, a0.

Definition. A primitive polynomial is an element of Z[x] with content 1.

Theorem 11.2 (Gauss’ Lemma). The product of two primitive polynomials is primitive.

Proof. Let f(x) and g(x) be primitive. Suppose f(x)g(x) is not primitive. Let p be a prime divisor of thecontent of f(x)g(x). Let f(x), g(x) and f(x)g(x) be the polynomials obtained from reducing the coefficientsof f(x), g(x) and f(x)g(x) mod p, respectively. Then, f(x), g(x) ∈ Zp[x], and f(x)g(x) = 0 by how p wasdefined. Recall Zp[x] is an integral domain. So f(x)g(x) = f(x)g(x) = 0. Then either f(x) = 0 or g(x) = 0.This means that p either divides every coefficient of f(x) or of g(x). So, either f(x) is not primitive or g(x)is not primitive. This is a contradiction. Therefore, f(x)g(x) must be primitive. �

Theorem 11.3. Let f(x) ∈ Z[x]. If f(x) is reducible over Q, then f(x) is reducible over Z.

Irreducibility Tests.

Theorem 11.4. Let p be a prime and suppose f(x) ∈ Z[x] with deg f(x) ≥ 1. Let f(x) be the polynomial inZp[x] obtained from f(x) by reducing all the coefficients of f(x) modulo p. If f(x) is irreducible over Zp[x]and deg f(x) = deg f(x), then f(x) is irreducible over Q.

Example. Let f(x) = 21x3 − 3x2 + 2x + 9. Over Z2, f(x) = x3 − x2 + 1. Notice f(0) = 1, f(1) = 1, anddeg f(x) = deg f(x) = 3. Thus, by reducibility in degrees 2 and 3, f(x) is irreducible over Z2. So by themod p theorem, f(x) is irreducible over Q.

Theorem 11.5 (Eisenstein’s Criterion). Let f(x) = anxn + an−1x

n−1 + ... + a1x + a0 ∈ Z[x]. If thereis a prime p such that p 6 |an, p|an−1, ..., p|a0 and p2 6 |a0, then f(x) is irreducible over Q.

12. Extension Fields (by Robert Arn)

Show R[x]/〈x2 + 1〉 ∼= C φ : R[x]→ C by φ(f(x)) = f(c). Then x2 + 1 ∈ kerφ. x2 + 1 is a polynomial ofminimum degree in kerφ. Then kerφ = 〈x2 + 1〉. By 1st isomorphism R[x]/〈x2 + 1〉 ∼= C.

There is a connection among irreducible polynomials, maximal ideas, and fields.

Theorem 12.1. Let F be a field and let p(x) ∈ F [x]. Then 〈p(x)〉 is a maximal ideal in F [x] if and only ifp(x) is irreducible over F .

Proof. (⇒) Suppose 〈p(x)〉 is a maximal ideal in F [x]. Then p(x) is neither the zero polynomial nor anunit in F [x]. If p(x) = g(x)h(x) (a factorization of lower degree). Then 〈p(x)〉⊂〈g(x)〉⊂F [x]. So either〈g(x)〉 = 〈p(x)〉 or 〈g(x)〉 = f [x]. If 〈g(x)〉 = 〈p(x)〉, the degree g(x) = deg p(x). If 〈g(x)〉 = F [x] thendeg g(x) = 0 and deg p(x) = deg h(x). So p(x) cannot be written as the product of lower degree.(⇐) Now, suppose p(x) is irreducible over F . Let I be an ideal of F [x] such that 〈p(x)〉⊂I⊂F [x]. Since F [x]is a principle ideal domain, I = 〈g(x)〉 for some g(x) ∈ F [x]. So p(x) ∈ 〈g(x)〉 ⇒ p(x) = g(x)h(x) whereh(x) ∈ F [x]. Since p(x) is irreducible either g(x) or h(x) is contant. If g(x) is constant, 〈g(x)〉 = I = F [x].If h(x) is constant, 〈p(x)〉 = 〈g(x)〉. So 〈p(x)〉 is maximal. �

Corollary 12.2. Let G be a field and p(x) an irreducible over F . Then F [x]/〈p(x)〉 is a field.

Definition. An integral domain D is a Unique Factorization Domain (UFD) if(1) Every nonzero element of D that is not a unit can be written as a product of irreducible of D, and(2) The factorization into irreducible is unique up to isomorphism up to associates and the order in

which the factors appear.

For F a field, F [x] is a UFD.

Definition. A field E is an extension field of a field F is F⊂E and the operations of F are those of Erestricted to F .

Theorem 12.3 (Fundamental Theorem of Field Theory (Kronecker’s Theorem)). Let F be a fieldand f(x) a nonconstant polynomial in F [x] Then there is an extension field E of F in which f(x) has a zero.

Proof. Since F [x] is a UFD, f(x) has an irreducible factor, say p(x). We can construct an extension fieldE of F in which p(x) has a zero. Consider F [x]/〈p(x)〉 = E. From our first theorem, F [x]/〈p(x)〉 is a field.The mapping φ : F → E given by φ(a) = a+ 〈p(x)〉. φ is a 1− 1 homomorphism (Think of E as containingF as a equal to a + 〈p(x)〉). Now, show p(x) has a zero in E. p(x) = anx

n + ... + a0. In E, x + 〈p(x)〉 is azero of p(x).

p(x+ 〈p(x)〉) = an(x+ 〈p(x)〉n + an−1(x+ 〈p(x)〉n−1 + ...+ a0

= an(xn + 〈p(x)〉) + ...+ a0

= anxn + an−1x

n−1 + ...+ a0 + 〈p(x)〉= p(x) + 〈p(x)〉= 〈p(x)〉= 0 + 〈p(x)〉

�

Examples. (1) Let f(x) = x2 + 1 ∈ Q[x] F = Q φ : F → E φ : a→ a+ 〈p(x)〉φ(x) = x+ 〈x2 + 1〉E = Q[x]/〉x2 + 1〉x → x+ 〈x2 + 1〉

f(x) → f(x+ 〈x2 + 1〉)(x+ 〈x2 + 1〉)2 + 1= x2 + 〈x2 + 1〉+ 1= x2 + 1 + 〈x2 + 1〉= 〈x2 + 1〉

(2) Let f(x) = x5+2x2+2x+2 ∈ Z3 = (x2+1)(x3+2x+2) E = Z[x]/〈x2+1〉 = {〈x2+1〉, 1+〈x2+1〉, 2+〈x2 +1〉, x+ 〈x2 +1〉, 2x+ 〈x2 +1〉, x+1+ 〈x2 +1〉, x+2+ 〈x2 +1〉, 2x+1+ 〈x2 +1〉, 2x+2+ 〈x2 +1〉}

Definition. Let E be an extension field of F and let f(x) ∈ F [x]. We say f(x) splits in E if f(x) can befactored as a product of linear factors in E[x]. We call E a splitting field for f(x) over F if f(x) splits in Ebut in no proper subfield of E.

Examples. (1) f(x) = x2 + 1 ∈ Q[x] = (x+ i)(x− I) so f(x) splits in C. However Q[i] = {a+ bi|a, b ∈ Qis the splitting field of f over Q. f(x) = x2 + 1 ∈ R[x]. R[i] = {a+ bi|a, b,∈ R} is the splitting fieldof F over R.

(2) x2 − 2 = (x+√

2)(x−√

2) split in R but a splitting field for f(x) is Q[√

2] = {r + s√

2|r, s ∈ Q}.(3) f(x) = x4 − x2 − 2 = (x2 − 2)(x2 + 1) over Q. Zeros in C are ±

√2,±i. A splitting field of f(x) over

Q : Q(√

2, i) = Q(√

2)(i) = {α+ βi|α, β ∈ Q(√

2) = {(a+ b√

@) + (c+ d√

2)i}.

Notation. Let F be a field and let a1, a2, .., an be elements of some extension E of F . User F (a1, a2, ...an) todenote the smallest subfield of E that contains F and the set {a1, a2, ..., an}. Also, if f(x) ∈ F [x] and f(x)factors as f(x) = b(x − a1)(x − a2)...(x − an) over some extension E of F . Then F (a1, ...an) is a splittingfield over F in E.

Theorem 12.4. Let F be a field and p(x) ∈ F [x] an irreducible polynomial if a is root of p(x) then F (a) ∼=F [x]/〈p(x)〉.

(This theorem was not given in class but was given during the homework session.)

13. Algebraic Field Extensions (by Tim Ferdinands)

Theorem 13.1. Let F be a field and P (x) ∈ F [x] be an irreducible polynomial over F . If a is a zero of p(x)in some extension E of F , then F (a) is isomorphic to F [x]/〈p(x)〉.

Proof. Consider φ : F [x] to f(a) given by φ(f(x)) = f(a). We claim that kerφ = 〈p(x)〉.Since p(a) = 0, 〈p(x)〉 ⊆ kerφ. And since p(x) is irreducible, 〈p(x)〉 is maximal by a prior theorem. So if

〈p(x)〉 ⊆ kerφ ⊆ F [x],

then kerφ 6= F [x] because f(x) = 1 is not a kernel element. Thus 〈p(x)〉 = kerφ.When F is a field and p(x) is irreducible F [x]/〈p(x)〉 is a field. By the first isomorphism theorem

F [x]/〈p(x)〉 ∼= F (a) = φ(F [x]).

�

Definition. Let E be an extension field of a field F and let a ∈ E. We call a algebraic over F if a is thezero of a some nonzero polynomial in F [x].

If a is not algebraic over F it is called transcendental over F . An extension E of F is called an algebraicextension of F if every element of E is algebraic over F .

Example.√

2 is algebraic over Q. f(x) = x2 − 2. Q(√

2) = {a+ b√

2 : a, b ∈ Q} is an algebraic extension.

Definition.

F [x] ={f(x)g(x)

|f(x), g(x) ∈ F [x]}

is the field of quotients of F [x].

Example. Q(√

2) = {a+ b√

2 : a, b ∈ Q}.

a+ b√

2c+ d

√2

(c− d

√2

c− d√

2

)=ac− ad

√2 + bc

√2 + 2bd

c2 − 2d2=ac− 2bdc2 − 2d2

+bc− adc2 − 2d2

√2.

Example.

Q(π) ={anπ

n + an−1πn−1 + · · ·+ a1π + a0

bmπm + bm−1πm−1 + · · ·+ b1π + b0

}.

Theorem 13.2. Let E be an extension field of the field F with a ∈ E. If a is transcendental over F , then

F (a) ∼= F [x].

If a is algebraic over F thenF (a) ∼= F [x]/〈p(x)〉

where p(x) is a polynomial of minimum degree such that p(a) = 0. Moreover p(x) is irreducible over F .

Proof. Considerφ : F [x]→ F (a)

byφ(f(x)) = f(a).

If a is transcendental over F , then kerφ = {0} and we have an isomorphism

φ : F [x]→ f(a)

by

φ :f(x)g(x)

=f(a)g(a)

.

Now if a is algebraic over F , then kerφ 6= 0 because f(a) = 0 for some nonzero f(x) and there is somepolynomial p(x) in F [x] such that kerφ = 〈p(x)〉. p(x) has minimal degree among elements of kerφ. By thefirst isomorphism theorem

F (a) ∼= F [x]/〈p(x)〉.�

Definition. Let E be an extension field of F . We say that E has degree n over F and write [E : F ] = n ifE has dimension n as a vector space over F . If [E : F ] is finite, E is called a finite extension of F .

Otherwise E is an infinite extension.

Example. C = {a+ bı|a, b ∈ R} has basis {1, ı} as a vector space over R. So [C : R] = 2.On the other hand, C has infinite degree over Q.

Example. Q(√

2) = {a+ b√

2 : a, b ∈ Q}. {1,√

2} is a basis for Q(√

2). Thus [Q(√

2) : Q] = 2.

Theorem 13.3. If E is a finite extension of F1, then E is an algebraic extension of F .

Proof. Suppose [E : F ] = n and a ∈ E. Then {1, a, . . . an} is linearly dependent over F . Thus there areelements c0, c1, . . . , cn ∈ F such that c0 + c1a+ c2a

2 + · · ·+ cnan = 0 with not all ci = 0.

So f(x) = c0 + c1 + · · · + cnxn is a nonzero polynomial that has a for a root. Thus a is algebraic over

F . �

Theorem 13.4. Let K be a finite extension field of E and let E be a finite extension field of F . Then K isa finite extension field of F and

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

Proof. Let X = {x1, x2 . . . xn} be a basis for K over E. So [K : E] = n, and let Y = {y1, y2, . . . yn} be abasis for E over F , so [E : F ] = m.We claim that

Y X = {yjxi|1 ≤ j ≤ m, 1 ≤ i ≤ n}is a basis for K over F .

Choose a ∈ K. Then there are elements b1, b2, . . . bn ∈ E such that

a = b1x1 + b2x2 · · ·+ bnxn.

For each i = 1, 2, . . . n there are ci1 , ci2 , . . . cin such that

bi = ci1y1 + ci2y2 + · · · cimym.

a =n∑i=1

bixi =n∑i=1

m∑j=1

cijyi

xi =∑i,j

cij (yjxi)

So Y X spans K over F (elements can be written as elements in F ).But is it linearly independent? ∑

i,j

cij (yjxi) =∑i

∑j

cijyj

xi = 0

Since X is a basis for K over E∑i,j cijyj = 0. And since Y is a basis for E over F ,

∑i,j cijyj = 0, and

cij ∈ F means that cij = 0. Thus we have linear independence. Therefore yjxi form a basis for K overF . �

14. Cyclotomic Polynomials and Group Rings (by Jayna Resman)

Cyclotomic Polynomials.• The complex zeros of xn − 1 are 1 and those of the form ω = cos( 2π

n ) + i sin( 2πn ).

•⟨ω⟩

forms a cyclic group of order n. The generators are of the form ωk for 1 ≤ k ≤ n andgcd(k, n) = 1. These are called the nth roots of unity.

• Define φ(n) = # of integers, k, less than or equal to n with gcd(n, k) = 1• Q(ω) = splitting field of xn − 1 called the nth cyclotomic extension of Q.• The irreducible factors of xn − 1 over Q are cyclotomic polynomials.

Definition. For any positive integer n, let ω1, ...ωφ(n) be the primitive roots of unity. The nth cyclotomicpolynomial over Q is the polynomial φn(x) = (x− ω1)(x− ω2)...(x− ωφ(n)) where ω1 = cos( 2π

n ) + i sin( 2πn ).

Examples. • For n = 1, ω = cos( 2π1 ) + sin( 2π

1 )i = 1. Therefore φ1(x) = (x− 1).• For n = 2, ω = cos( 2π

2 ) + sin( 2π2 )i = −1. Therefore φ2(x) = (x+ 1).

• For n = 3, ω1 = cos( 2π3 ) + sin( 2π

3 )i = −12 +

√3

2 i and

ω2 = ω21 =

(−12

+√

32i

)(−12

+√

32i

)=−12−√

32i.

Therefore, φ3(x) =(x−

(−12 +

√3

2 i))(

x−(−12 −

√3

2 i))

= x2 + x+ 1.

Theorem 14.1. For every n ∈ N, xn − 1 =∏d|n φd(x) where the product runs over all possible divisors, d,

of n.

Example. x3 − 1 = φ1(x)φ3(x) = (x− 1)(x2 + x+ 1).

Theorem 14.2. The cyclotomic polynomials φn(x) are irreducible over Z.

Group Rings.• Fix a commutative ring R with unity 1, and let G = {g1, g2, ..., gn} be any finite group with multipli-

cation. The group ring R[G] with coefficients in R is the set of formal sums: a1g1 + a2g2 + ...+ angnwith ai ∈ R.

• If g1 is the multiplicative identity in G, then aig1 = ai and 1 · gj = gj for all gj ∈ G.• Addition is component-wise.

(a1g1 + a2g2 + ...+ angn) + (b1g1 + b2g2 + ...+ bngn) = (a1 + b1)g1 + (a2 + b2)g2 + ...+ (an + bn)gn• Multiplication

(a1g1 + a2g2 + ...+ angn)(b1g1 + b2g2 + ...+ bngn) =∑

aibjgk where (gk = gigj).

Example. Recall the Quaternion group Q8 = {1,−1, i,−i, j,−j, k,−k}. Then 2i− 3j and 4 + 2k− j are bothelements in Z[Q8].

Theorem 14.3. If R is a subring of S, then R[G] is a subring of S[G] for any group G.

Proof. First note that R[G] is non-empty since R is a ring (and thus contains an additive identity) and Gis a group (and thus contains its own identity). Let α, β ∈ R[G] and α = (a1g1 + a2g2 + ... + angn), β =(b1g1 + b2g2 + ...+ bngn) such that ai, bi ∈ R and ai ∈ S. Therefore, (a1g1 + a2g2 + ...+ angn) ∈ S[G], too.

α− β = (a1 − b1)g1 + (a2 − b2) + ...+ (an − bn)gnBut, (ai − bi) ∈ R, thus α− β ∈ R[G].

αβ =(∑

aigi

)(∑bjgj

)=∑

aibjgk,

and ibj ∈ R, thus αβ ∈ R[G]. Thus, R[G] is non-empty and closed under subtraction and multiplication, soR[G] is a subring of S[G]. �

Definition. A ring R with unity is called reversible if ab = 0 implies ba = 0 for all a, b ∈ R.

Definition. A ring R with unity is called symmetric if abc = acb for all a, b, c ∈ R.

Appendix: Notes from David Jorgensen (by Becky Egg)

Two ways to study rings• inside:

Does the ring have– zero divisors?– idempotents?

What is the– radical of an ideal?– the Jacobson radical (the intersection of all maximal left ideals)?– the nilradical (the set of all nilpotent elements)?– ideals?

• outside:– modules– free modules– projective modules– injective modules– flat modules– extensions– functors– derived functors

an analogy for modules:

k a field R a ringV a vector space M a modulelinear transformations R-module homomorphisms

Example. Rn = {(a1, a2, · · · , an)|ai ∈ R} is a vector space over R.

Definition. A left R-module M is an abelian group under + along with ring multiplication r · x such thatthe following axioms hold: For all r, s ∈ R and all x, y ∈M

(1) r · x ∈M (closure)

(2)r · (x+ y) = r · x+ r · y(r + s) · x = r · x+ s · x

}(distributive properties)

(3) r · (s · x) = (rs) · x (associativity)In addition, if 1R ∈ R, then

(4) 1R · x = x.

If (4) holds, then M is called a unitary module over R.

Example. (1) R is itself an R-module.(2) I an ideal of R is an R-module.(3) R/I is an R-module.(4) Rn = R×R× · · · ×R︸ ︷︷ ︸

n copies

= {(a1, . . . , an)|ai ∈ R} is an R module, where the operations are defined

componentwise. In fact, Rn is a free module.

Definition. A subset {xλ}λ∈Λ of an R-module M is a generating set for M if

M = {r1xλ1 + r2xλ2 + · · ·+ rnxλn|ri ∈ R},

i.e., M is the set of finite R-linear combinations of elements of {xλ}λ∈Λ.If there exists a generating set {xλ}λ∈Λ such that |Λ| < ∞, then M is called a finitely generated module.[Note that in our analogy between modules and vector spaces, a generating set corresponds to a spanningset.]

Definition. The set {xλ}λ∈Λ is linearly independent if for all xλ1 , . . . , xλn,

r1xλ1 + · · ·+ rnxλn= 0

implies r1 = · · · = rn = 0.

Definition. An R-module M is called free if it has a linearly independent generating set, called a basis.

Example. Consider the following Z-modules:(1) For n ∈ N, Zn is a free module.(2) (2) is generated by 2. Note that r ∈ Z with r2 = 0 implies that r = 0, so {2} is a basis, and hence

(2) is a free module.(3) (2, 3) = Z is also a free module, with basis {1}.(4) M = Z/(2) = {0+(2), 1+(2)} is a Z-module; its only generating set is {1+(2)}. Note however that

2(1 + (2)) = 2 + (2) = (2) = 0Z/(2),

so Z/(2) is finitely generated, but not a free module.

Example. Let k be a field, and consider the following k[x, y]-modules:(1) M = (x) is a module with generating set {x}. Since r(x, y)x = 0 implies that r(x, y) = 0, M is a

free module.(2) N = (x, y) is generated by {x, y}. Note however that (−y)x+ (x)y = 0, so N is not a free module.

If turns out that every R-module M “looks like” (i.e., is isomorphic to) F/S, where F is a free module,and S is a submodule of F .

Exercises: Let k be a field and G = 〈g〉 be a cyclic group of order 2. Let R = k[G].(1) Let

M ={[

αβ

] ∣∣∣∣α, β ∈ k} .Determine which of the following define R-modules; justify your answers.(a)

e

[αβ

]=[α0

], g

[αβ

]=[

0β

](b)

e

[αβ

]=[αβ

], g

[αβ

]=[α+ 1β + 1

](c)

e

[αβ

]=[αβ

], g

[αβ

]=[α0

]

(d)

e

[αβ

]=[αβ

], g

[αβ

]=[βα

](e)

e

[αβ

]=[αβ

], g

[αβ

]=[αβ

]Now let

M =

αβγδ

∣∣∣∣∣α, β, γ, δ ∈ k

,

and determine which of the following define R-modules:(f)

e

αβγδ

=

αβγδ

, g

αβγδ

=

βαδγ

(g)

e

αβγδ

=

αβγδ

, g

αβγδ

=

δαβγ

(2) Of the modules given in (1), determine which are free modules.(3) A representation of a group G is a group homomorphism ρ : G → GL(V ), where V is a finite di-

mensional vector space over a field k and GL(V ) is the set of invertible linear transformations fromV to V . If dimV = n, then ρ is an n-dimensional representation.Define a bijective correspondence between n-dimensional representations of G and n-dimensionalk[G]-modules.

Let R be an associative ring with 1.

Recall the correspondence between a vector space over a field and a module of a ring. Free modules, i.e.,those which have a basis, have the clearest correspondence to vector spaces over a field.

Definition. A subset S of an R-module M is called a submodule of M if it is an R-module whose ringmultiplication is compatible with that of M .

Definition. A function f : M → N between R-modules M and N is called an R-module homomorphism iffor all x, y ∈M and all r ∈ R

(1) f(x+ y) = f(x) + f(y)(2) f(rx) = rf(x).

Exercises:(1) Let f : M → N be an R-module homomorphism. Show that ker f and im(f) are R-submodules of

M and N , respectively.(2) If N is a submodule of the R-module M , then M/N is also an R-module.

Theorem 14.4. Every R-module is the homomorphic image of a free module.

Proof. Let M be an R-module and {xλ}λ∈Λ be a generating set for M . Let F be the free R-module on thebasis {eλ}λ∈Λ. Define

f : F →M

eλ 7→ xλ,

and extend by linearity (i.e., f(rλ1eλ1 + · · · + rλneλn

) = rλ1xλ1 + · · · + rλnxλn

). By construction, f is asurjective homomorphism. �

Using the first isomorphism theorem, we have the following:

Corollary 14.5. M ∼= F/S, where S = ker f , with f given above.

This result is the first step in the idea of a ‘free resolution’, the ultimate determination in the failure of amodule to be free.

Definition. A sequence

Lf // M

g // N

of R-module homomorphisms is called exact if ker g = im f .

Example.

0 // S1⊆ // F0

f // M // 0

is a short exact sequence. Note that since the image of the injection from S1 to F0 is all of S1, we have thatker f = S1.We can repeat this process on S1:

0 // S2// F1

// S // 0 ,

and then splice or compose these two sequences to get

F1d1 //

AAA

AAAA

A F0π // M // 0

S1

>>}}}}}}}

The above sequence is exact, and called a free presentation of M . If F1, F0 are finitely generated free mod-ules, then M is called finitely presented.

Suppose that M is finitely presented. Then F1, F0 are finitely generated free modules; let F1 be of rankn, and F0 be of rank m. Then we can represent the R-module homomorphism d1 by an m× n matrix, withentries in R

d1 =

r11 r12 . . . r1n

r21 r22 . . . r2n

......

. . ....

rm1 rm2 . . . rmn

,where this matrix is given with respect to the bases {e1, . . . , en} of F1 and {e′1, . . . , e′m} of F0. Then we have

π(d1(e1)) = π(r11e′1 + r21e

′2 + · · ·+ rm1e

′m)

= r11x1 + r21x2 + · · ·+ rm1xm

= 0.

This non-trivial 0-linear combination shows that x1, . . . , xm are not linearly independent, i.e., M is not afree module.

In fact, the column space of d1 generates all the nontrivial linear relations of the generators of M . “All”the information about M is contained in the presentation matrix d1.

We can continue this process:

0 // S2// F1

d1 // F0// M // 0,

0 // S3// F2

// S2// 0,

which we compose to get

F2d2 // F1

d1 // F0// M // 0.

What we’ve just constructed is called a free resolution.

Definition. A free resolution of an R-module M is an exact sequence

· · · // Fndn // · · · // F2

d2 // F1d1 // F0

// M // 0,

where each Fi above is a free R-module.

Example. (1) R = f [x, y],M = 〈x, y〉. We have

S1// R2 // M // 0,

where the map from R2 to M is given by

e1 7→ x

e2 7→ y,

and e1, e2 are the standard basis vectors for R2. A vector(ab

)maps to ax+ by, so for

(ab

)in

the kernel of this map, we have a = −a′y, b = −a′x, i.e.,(ab

)= a′

(−yx

).

Thus(−yx

)generates the kernel of this map. Thus S1 =

(−yx

). Since this is a one-dimensional

module, we have

Rd2 // S1

// 0,

given by e1 7→(−yx

). Composing these sequences gives

R // R2 // M // 0.

Note that if a ∈ ker d2, then

a

(−yx

)=(

00

)implies that a = 0. So ker d2 = {0}, and hence

0 // R // R2 // M // 0

is a free resolution of M .(2) R = k[x, y],M = R/(x, y), which is generated by 1 + (x, y). We have

(x, y) // R // M // 0,

where the map from R to M is given by e1 7→ 1. Since we previously found the free resolution of(x, y), we have

0 // R // R2 // R // M // 0.

(3) Resolutions need not be finite: R = k[X,Y ]/(X,Y ), M = (x) and x = X + (XY ). We have

0 // (y) // R // M // 0,

where the map from R to M is given by e1 7→ x. This resolution is infinite:

· · · // R[y] // R

[x] // R[y] // R // M // 0.

The modules Si given in a free resoultion are called Syzygy modules of M .

Definition. A complete resolution of an R-module M is an exact sequence

F : · · · // F2d2 // F1

d1 // F0d0 // F−1

d−1 // F−2) // · · · ,

where the Fi’s are finitely generated, M = im(d0), and homR(F,R) is exact.

Example. The previous example

· · · // R[y] // R

[x] // R[y] // R // M // 0.

is a complete resolution.

Exercises:(1) Compute the free resoultions of all the modules from yesterday’s problem 1.(2) Let k be a field.

(a) Let R = k[G] where G = 〈g〉 is a cyclic group of order 2. Let M = 〈e− g〉. Find the completeresolution of this ideal.

(b) Let R = k[G], where G = 〈g1〉 × 〈g2〉 and g1, g2 are of order 2. Compute the free resolution ofM = 〈e1 − g1, e2 − g2〉. Hint: consider example 2.

(3) Two resolutions ρ1 : G → GL(V ) and ρ2 : G → GL(W ) are called equivalent if there exists andisomorphism α : V →W such that for all g ∈ G,

α ◦ ρ1(g) ◦ α−1 = ρ2(g).

Show that the equivalence classes of n-dimensional representations are in bijective correspondenceto the isomorphism classes of n-dimensional k[G]-modules.