1 Proofs to Know. 2 Intro to Groups. 2 INTRO TO GROUPS Algebra Study Guide

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1 2 INTRO TO GROUPS Algebra Study Guide 1 Proofs to Know Prop: Let R be commutative and let M be a left R-module. Suppose M has an R-basis B of cardinality n and an R-basis E of cardinality m. Then n = m. Proof. By the remark from class (which states that if M is a free R-module for a commutative ring R with a basis of n elements, M = R n ), R m = M = R n (can use the map ϕ : R n M given by ϕ(r 1,, r n ) = n 1 r is i, where {s i } n 1 is a basis for M it s surjective as it generates M and injective since ker ϕ = (0)). Let I be a maximal ideal of R, which exists as R is commutative. Then F = R/I is a field. By an exercise, R n /IR n = R/IR R/IR = F n }{{} = F m. By vector space theory, n = m. n times Prop: Assume the set A = {v 1,, v n } spans the vector space V but no proper subset of A spans V. Then A is a basis of V. Proof. Suffices to show v i are LI. If a 1 v a n v n = 0 and WLOG a 1 0, write v 1 as a linear combination of the others. So {v 2,, v n } spans V. ( ) Prop: Let F be a field and V an F -vector space. Let T S be subsets of V such that T is LI and S spans V. Then there exists a basis B such that T B S. Proof. Let Λ = {A : T A S, A LI}. Note Λ as T A. Λ is a poset under. Let C be a totally ordered subset of Λ. Let à = A. Show that à Λ. A C By Z-L there exists a maximal element of Λ, say B. Claim: B is a basis for V. Proof: It is enough to show that S span F B. If s S such that s / span F B, then B {s} is linearly independent. Thus B {s} Λ. Since B is maximal, B B {s} = B = B {s}. Thm: Let V be an F -vector space and let W be a subspace of V. Then V/W is a vector space with dim V = dim W + dim V/W (where if one side is infinite then both are). Proof. Let dim W = m and dim V = n and let w 1, w 2,, w m be a basis for W. These are LI and can be built up to a basis w 1, w 2,, w m, v m+1,, v n of V. The natural surjective projection of V into V/W sends all of the w i to 0 and none of the v i to 0 (as this would imply v i W ). Hence, V/W of the projection map is isomorphic to the subspace of V spanned by the v i, giving that dim V/W = n m. 2 Intro to Groups Prop: 1 (Z/nZ) = {a Z/nZ : gcd(a, n) = 1}. Proof. gcd(a, n) = 1 = x, y Z such that ax + ny = 1. Then n ax 1 = ax = 1 so a is a unit. Conversely, if ax = 1 in Z/nZ then n ax 1 = ax 1 = ny = gcd(a, n) = 1. Def: Let F be a field. Then the general linear group of F of degree n is defined by GL n (F ) := M n (F ) = {A M n (F ) : det A 0}. Def: Let G be a group and g G. Then the order of g is the least positive integer n (if it exists) s.t. g n = 1. If no such n exists, then g =. Def: Define the dihedral group of order 2n to be the group of rigid motions of the regular n-gon. Then D 2n = 2n. Let r be the counterclockwise rotation by 2π/n radians and s be the reflection about the x-axis. Note: r = n and s = 2. Then sr i sr j for 0 i < j n 1 (if sr i = sr j = r i = r j = 1 = r j i ). Then D 2n = r, s r n, s 2 = 1, rs = sr 1. Prop: 2 Let G be a group and x G. Let d = x <. Then for all n Z, x n = 1 d n. (For the forward direction, use the division algorithm.) Prop: Let G be a group, x G. Then x G. If G is a finite group, then every element of G has finite order. Proof. Nothing to show if G =. Assume m = G <. Consider { 1, x, x 2,, x m} G. Since G has only m distinct elements, there exists x i = x j where 0 i < j m Then x i j = 1 and 0 < j i m.

2 2 INTRO TO GROUPS Algebra Study Guide 2.1 Generating Groups Def: Let T be a set, and let S T = {f : T T : f is a bijection}. This is a group under composition. In the case T = {1, 2,, n}, denote this group S T by S n, and it s called the symmetric group of degree n. Note: S n = n!. Def: A k-cycle in S n is a permutation σ such that there exists {a 1, a 2,, a k } {1, 2,, n} such that σ(a 1 ) = a 2, σ(a 2 ) = a 3,, σ(a k ) = a 1 and σ(x) = x for all x {1,, n}\{a 1,, a k }. We write this k-cycle as (a 1 a 2 a k ). Rmrk: Any two disjoint cycles commute, and any permutation can be uniquely written as a product of disjoint cycles. Prop: Let a, b G and a = m < and b = n <. Suppose ab = ba and i, j Z, a i = b j a i = b j = 1. Then ab = lcm(m, n). Cor: Suppose σ, τ are disjoint cycles in S n. Then στ = lcm( σ, τ ). Thm: Let σ = c 1 c 2 c t, where this is a product of disjoint cycles. Let c i be an l i -cycle. Then σ = lcm(l 1, l 2,, l t ). Prop: 3 Let σ S n be a k-cycle. Then σ = k. Def: Let σ S n \ {1}. Suppose σ is a product of disjoint cycles σ = p 1 p l where p 1,, p l are of lengths 1 < k 1 k 2 k l. We say σ has cycle type {k 1, k 2,, k l }. Def: The quaternion group, Q 8, is defined by Q 8 = {1, 1, i, i, j, j, k, k}. Def: Let S G. The centralizer of S in G is C G (S) := {g S : gx = xg x S}. Def: Let H G. The normalizer of H in G is defined to be N G (H) := { g G : ghg 1 = H }. Rmrk: C G (H) N G (H). Prop: Let G be a group and {H α } α I be a family of subgroups of G. Then α I H α G. 2.1 Generating Groups Prop: Let S G with S. Then S = {s e1 1 sen n : s i S, e i = ±1 i}. If G is finite, S = {s 1, s n : s i S}. E.g.: 4 Q = {1/p m : p prime, m N}. Prop: S n = {(i j) : 1 i < j n} = {(1 2), (1 3),, (1 n)} = {(1 2), (2 3),, (n 1 n)} = {(1 2), (1 2 3 n)}. Def: If G = a, G is said to be a cyclic group (the cyclic group generated by a). Prop: a = a. If a = n, then a = { 1, a,, a n 1}. Prop: Any subgroup of a cyclic group is cyclic. Thm: 5 Let G = a be a cyclic group with G = n <. 1. For all i Z, a i = a d where d = gcd(i, n). Thus, { a d : d > 0, d n } is the set of all subgroups of G (these are all distinct since a d = a d = n/d). 2. a i = n/gcd(i, n). 3. Every subgroup of G has order dividing n. Cor: Let G = a be a cyclic group of order n. Then a i is a cyclic generator for G if and only if gcd(i, n) = 1. Cor: Let G = a, G = n, d a positive divisor of n. Then G has ϕ(d) elements of order d (where ϕ is Euler s ϕ function). Cor: Let n be a positive integer. Then d 1 d n ϕ(d) = n. Prop: Let G = a be a cyclic group. Then if G = n then G = Z/nZ. If G = then G = Z. Def: 6 The alternating group of order n, denoted by A n, is the subgroup of S n consisting of all even permutations. Prop: If n > 2, A n is generated by the set of all 3-cycles. Def: Let G be a group, H G. Then H is called normal, denoted H G if ghg 1 = H for all g G. Rmrk: Let G be a finite group and H G such that 2 H = G. Then H G. Proof. Let σ G \ H. Note H = σh and H σh =. Then H = G /2 = σh so H σh = G. Similarly, H Hσ = G and H Hσ =. Therefore G \ H = Hσ = σh, which implies σhσ 1 = H. Cor: A n S n. Prop: A 4 = 12. A 4 has no subgroup of order 6. Proof. The number of 3-cycles in A 4 (or S 4 ) is ( 4 3) 2 = 8. Suppose H A4 such that H = 6 = A 4 /2, which implies H A 4. So H contains some 3-cycle. WLOG, assume the 3-cycle (1 2 3) H. Then (1 3 2) H (as (1 3 2) 1 = (1 2 3). Since H is normal in A 4, setting σ = (1 2) implies σ(1 2 3)σ 1 = (1 4 2) H, so (1 2 4) H. Doing this again with, say, (1 3)(2 4) yields two more 3-cycles. Therefore H contains at least 6 3-cycles. ( )

3 2 INTRO TO GROUPS Algebra Study Guide 2.2 Cosets 2.2 Cosets Def: Let H G, g G. Then the set gh = {gh : h H} is called a left coset of H. Prop: Let H G. (1) For all g 1, g 2 G, g 1 H = g 2 H. (2) For all g 1, g 2 G, g 1 H = g 2 H or g 1 H g 2 H =. (3) G = g G gh. Proof. (1) The maps f : g 1 H g 2 H defined by x g 2 g1 1 x and f : g 2 H g 1 H defined by y g 1 g2 1 y are mutual inverses. (2) Suppose g 1 H g 2 H. Then x g 1 H g 2 H, so x = g 1 h 1 = g 2 h 2. Let g 1 h 3 g 1 H. Then g 1 h 3 = g 1 h 1 h 1 1 h 3 = g 2 h 2 h 1 1 h 3 g 2 H. Def: If H G, let [G : H], called the index of H in G be defined to be the number of distinct left cosets of H in G. Thm: (Lagrange) Let G be a finite group, H G. Then G = H [G : H]. Proof. Let m = [G : H], say {g 1 H,, g m H} is the set of distinct left cosets of H. By (3) above, G = m i=1 g ih. By (2), g i H g j H = for i j. So G = g 1 H + + g m H = m H by (1). Cor: 7 Let G be a finite group. Then a G for all a G. 2.3 Group Actions Def: Let G be a group and A a set. An action of G on A is a function G A A ((g, a) g a) such that (1) for all a A, g 1, g 2 G, (g 1 g 2 )a = g 1 (g 2 a) and (2) for all a A, 1 a = a. Def: Let G be a group acting on a set A. For a A, the stabilizer G a of a is G a := {g G : ga = a}. Note that G a G. The kernel of the action is ker = a A G a. The action is called faithful if ker = {1}. The action is free if G a = {1} for all a A. For a A, the orbit O a of a is O a := {ga : g G}. The action is called transitive if O a = A for some a A. Lma: 8 Let ϕ : G G be a group homomorphism. Then ϕ is 1-1 iff ker ϕ = 1. Prop: Let A and B be sets such that A = B. Then S A = SB. Def: A homomorphism ϕ : G S A is called a permutation representation for ϕ. Rmrk: G acts on A iff for ϕ : G S A the kernel of the action = ker ϕ. Rmrk: G acts faithfully on A if and only if ϕ : G S A is injective. In this case G = ϕ(g) S A. Thm: (Cayley s Theorem) Let G be a group. Then G is isomorphic to a subgroup of S G. In particular, if G = n, then G = H S n. E.g.: Examples of Group Actions:

4 2 INTRO TO GROUPS Algebra Study Guide2.4 Quotient Groups and Homomorphisms Prop: Let H G, A = {xh : x G}. Then G acting on A by left multiplication gives a group homomorphism ϕ : G S A = Sn (where n = [G : H]) given by ϕ(g) = σ g : A A (where σ g (xh) = gxh) having kernel xhx 1. x G If x G xhx 1 = 1 then G is isomorphic to a subgroup of S n, where n = [G : H]. Prop: 9 Let G act on A and let a, b A. Then O a = O b or O a O b =. Thus, A is the disjoint union of the distinct orbits. If A < then A = n O ai, where O a1,, O an are the distinct orbits of A. i=1 Cor: If G acts transitively on A, then O a = A for all a A. Therefore, G acts transitively on A iff a, b A there exists g G such that ga = b. Prop: Let G act on A. Let a A. Then there exists a bijection ϕ from {gg a : g G} O a given by gg a ga. Hence O a = [G : G a ]. Cor: G acts on A, where A <. Let O a1,, O an be the distinct orbits. Then A = n [G : G ai ]. E.g.: (The Class Equation) Let G be a finite group. Let G act on itself by conjugation, i.e., G = A, g a = gag 1. Let O a1,, O an be the distinct orbits of G under this action. O a1 = { ga 1 g 1 : g G } is called the conjugacy class of a 1. Then G a = { g G : gag 1 = a } = C G (a). Therefore, G = n [G : C G (a i )]. Notice that [G : C G (a)] = 1 if and only if C G (a) = G if and only if a Z(G). Therefore, i=1 G = Z(G) + [G : C G (a)], where a runs through the distinct conjugacy classes and a / Z(G). Prop: Let G be a group with G = p n, where p is prime. Then Z(G) 1 (follows from the fact that [G : C G (a)] = p l, l 1). 2.4 Quotient Groups and Homomorphisms Def: A subgroup H of G is normal in G if xhx 1 = H for all x G. Rmrk: H is normal in G iff xhx 1 H for all x G. Rmrk: Suppose H = S. Then H G iff gsg 1 H for all s S and for all g G. Rmrk: If H Z(G) then H G. Prop: Let ϕ : G 1 G 2 be a group homomorphism. Then ker ϕ G 1. Rmrk: Let H G, A = {gh : g G}. Let G act on A as usual. Recall, G xh = xhx 1. Therefore x G xhx 1 G (this is the largest normal subgroup of G contained in H). In particular, if [G : H] = n then there exists a group homomorphism ϕ : G S n = SA where kernel = x G xhx 1. Prop: 10 Let ϕ : G 1 G 2 be a group homomorphism, let K = ker ϕ. Then there exists a bijection {xk : x G 1 } imϕ (xk ϕ(x)). In particular, if G 2 <, [G : K] = imϕ divides G 2. Prop: Let G be a finite group and p the smallest prime divisor of G. Suppose there exists a subgroup H of G such that [G : H] = p. Then H G. STUDY THIS PROOF Cor: If [G : H] = 2. Then H G. Cor: If G is odd and [G : H] = 3 then H G. Def: A group G 1 is called simple if it has no nontrivial normal subgroups. Prop: Let G be a group and H G. Define an operation on the left cosets {xh : x G} by (xh) (yh) := (xy)h. This operation is well-defined if and only if H G. Prop: Suppose H G. Then the set of cosets of H is a group under the operation defined above. This group is denoted G/H and is referred to as a quotient group. Rmrk: If G < then G/H = G / H (Lagrange). Prop: 11 Let ϕ: G A be a surjective group homomorphisms. Then there exists a bijective correspondence between {subgroups of G containing ker ϕ} {subgroups of A} given by K ϕ(k). Furthermore, K G ϕ(k) A. Prop: Let G be a group and H G. Consider the function ϕ : G G/H given by ϕ(g) = gh. Then ϕ is a surjective group homomorphism with ker ϕ = H. Cor: Let H G, ϕ : G G/H be the canonical map. Then there exists a bijection between {K : K G, H K} and {A : A G/H} given by K K/H. Furthermore, if H K G, K G K/H G/H. i=1

5 2 INTRO TO GROUPS Algebra Study Guide 2.5 Direct Products Thm: Let ϕ : G A be a group homomorphism and let H G with H ker ϕ. Then there exists a group homomorphism ϕ : G/H A given by gh ϕ(g). Furthermore, imϕ = imϕ and ker ϕ = ker ϕ/h. Cor: (1st Isomorphism Theorem) Let ϕ : G A be a group homomorphism. Then G/ ker ϕ = imϕ. Cor: Let H K G and H G, K G. We already know that K/H G/H. Then (G/H)/(K/H) = G/K via the map (gh)/(k/h) gk. (We have ker ϕ = K/H in the induced map.) Rmrk: 12 Let ϕ : G A be a group homomorphism. Let x G with x <. Then ϕ(x) x. Thm: (Cauchy s Theorem) Let G be a finite group and p a prime such that p G. Then G has an element of order p. Proof. Case 1. G is abelian. Use induction on G = n. If n = p, then G is cyclic of order p. If n > p, choose x G, x 1. Let H = x. If p H, p x, say x = ps. Then x s = p. If p H, then p [G : H] = G / H. Since G is abelian, H G. Therefore, p G/H = [G : H]. As H 1, G/H < G. By induction, G/H has an element of order p, say yh. Therefore p y, and hence y = tp, then y t = p. Case 2. G arbitrary. If Z(G) = G, then G is abelian and we re done. So assume Z(G) G. If p Z(G), then Z(G) has an element of order p (as Z(G) is abelian). If p Z(G), use the class formula: G = Z(G) + [G : C G (x)]. x / Z(G) Since p G, p Z(G), there exists x / Z(G) such that p [G : C G (x)] = G / C G (x). Therefore, p C G (x). As x / Z(G), C G (x) < G. By induction on G, C G (x) has an element of order p. Hence, G does. Lma: Let a, b G, a <, b <. Suppose (1) a b = {1} and (2) ab = ba. Then ab = lcm( a, b ). Lma: Let H 1, H 2 G. Suppose H 1 H 2 = 1. Then ab = ba for all a H 1, b H 2. Cor: Suppose G = pq and let a be an element of order p, b be an element of order q, p < q, p, q primes. Suppose a G and b G. Then G is cyclic. Proof. a b a, b and a b = 1 by Lagrange. So ab = ba and hence ab = a b = pq. E.g.: Let G be a group, G = 6. Then G = Z/6Z or G = S 3. Proof. Suppose G is not cyclic. By Cauchy there exists a, b G such that a = 2, b = 3. Note if ab = ba then ab = 6 which implies G is cyclic ( ). Let H = a, K = b. As [G : K] = 2, K is normal. Also, H K = 1 (as gcd( H, K ) = 1). By the lemma, A = {xh : x G}. A = [G : H] = 3. Let G act on S A as usual. This gives a group homomorphism ϕ : G S 3 and ker ϕ is the largest normal subgroup of G contained in H. As H = 2 is not normal, ker ϕ = 1. Therefore ϕ is 1-1. As G = S 3 = 6, ϕ is onto. 2.5 Direct Products Def: 13 Let G be a group, A, B G (nonempty). Define AB := {ab : a A, b B}. Prop: Let H, K G. Then HK H K = H K. Prop: H, K G. Then HK G HK = KH. Cor: If H, K G and either H or K is normal in G, then HK G. Def: Let G be a group, H 1,, H t G. Suppose (1) H i G for all i, (2) G = H 1 H 2 H t = {h 1 h 2 h t : h i H i }, and (3) H i H 1 Ĥi H t = 1 for all i. We say that G is the (internal) direct product of H 1,, H t. Prop: Suppose G is the internal direct product of H 1,, H t. Then G = H 1 H t (externally). Prop: Let G be a group of order p 2, where p is a prime. Then G = C p 2 or G = C p C p. (For proof, suppose not cyclic and find two distinct subgroups of order p.) Thm: 14 Let H, K G, K G. Then (1) H K H and (2) HK/K = H/H K. 2.6 Fund. Thm. of FG Abelian Groups Thm: 15 Let G be a finitely-generated abelian group. Then there exists unique integers r 0 and p 1,, p k primes, m 1,, m k > 0 such that p 1 p k and G = C r C m p 1 C m 1 p k, so G is finite iff r = 0. If G < then k G = p m1 1 p m k k. Special Case: Suppose G = pn (p a prime) is abelian. Then G = C p m 1 C p m k, where mi = n.

6 2 INTRO TO GROUPS Algebra Study Guide 2.7 Automorphisms 2.7 Automorphisms Def: Let G be a group. An isomorphism ϕ : G G is called an automorphism of G. Let Aut(G) denote the set of automorphisms, which is a group under function composition. Def: Let G be abelian, m Z. Then the map ϕ m : G G given by ϕ m (g) = g m is a group homomorphism. Note that ϕ m ϕ l = ϕ ml. Prop: Let G = C n, n 1. Then Aut(C n ) = {ϕ m : C n C n : gcd(m, n) = 1} = (Z/nZ). Prop: 16 C m C n = Cmn gcd(m, n) = 1. Cor: If m 1,, m k are pairwise relatively prime positive integers then C m1 C mk = C m1 m k. Def: Let G be a group. Define the exponent of G by e(g) = inf { d 1 : g d = 1 g G }. If G <, then by Lagrange, e(g) G. Notation: Let G be a group, x G. Define ψ x : G G by ψ x (g) = xgx 1. Def: The map ψ x is a homomorphism for all x G, and is called a inner automorphism of G. Define the group of inner automorphisms to be Inn(G) = {ψ x : x G}. Exercise: Inn(G) Aut(G). Note: Aut(G)/ Inn(G) is called the group of automorphisms. Exercise: Inn(G) = G/Z(G) (isomorphism given by ψ x xz(g)). Def: Let H G. H is called a characteristic subgroup of G, denoted H char G if ϕ(h) = H for all ϕ Aut(G). Rmrk: H char G ϕ(h) H for all ϕ Aut(G). Rmrk: H char G = H G. Rmrk: If H is the unique subgroup of order H then H char G. Rmrk: Z(G) char G. Rmrk: In general, K H and H G = K G. Rmrk: K char H and H char G implies K char G. Rmrk: K char H and H G = K G. Def: Let G be a group, x, y G. The element denoted [x, y] := xyx 1 y 1 is called the commutator of x and y. The subgroup of G generated by all commutators is called the commutator subgroup (sometimes denoted G or [G, G]). So G = xyx 1 y 1 x, y G. Rmrk: G char G. Rmrk: 17 Let G be a group, H G. (1) If H G then H G and G/H is abelian. (2) The converse is also true. If H G and G/H is abelian then H G. 2.8 Sylow s Theorems Def: Let G be a finite group, p a prime such that p G. Suppose p n G but p n+1 G. A p-subgroup of G is a subgroup of G of order p α for some α 0. Let Syl p (G) denote the set of Sylow p-subgroups of G. Let n p = Syl p (G). Thm: (Sylow I) If p α G then G has a subgroup of order p α. Proof. Induction on G = n. If n = 1, the statement is vacuously true. Then the class equation gives G = Z(G) + l [G : C G (g i )] where C G (g i ) G. i=1 Case 1. p Z(G). By Cauchy s Theorem there exists a Z(G) where a = p. Let K = a G (as a Z(G)). Consider G = G/K. Then G /p < G. Also, p α 1 G. By the induction hypothesis G has a subgroup of order p α 1. Such a subgroup has the form H/K, where K H G. Then H/K = H / K = p α 1, which implies K = p = H = p α. Case 2. p Z(G). Then the class equation implies there exists some i for which p [G : C G (g i )]. Then p G / C G (g i ) implies p α C G (g i ). But C G (g i ) < G. Since [G : C G (g i )] > 1, C G (g i ) has a subgroup of order p α (by induction). Thm: (Sylow II) If G is a finite group, p n G, p n+1 G. Let P Syl p (G), n p = Syl p (G). Then 1. n p = [G : N G (P )] for all P Syl p (G). In particular, Syl p (G) = {P } if and only if P G (i.e., G = N G (P )). 2. n p 1 mod p 3. Any two Sylow p-subgroups of G are conjugate. 4. Any p-subgroup of G is contained in a Sylow p-subgroup.

7 2 INTRO TO GROUPS Algebra Study Guide 2.9 Solvable Groups Applications Prop: 18 Let G = pq where p < q primes and p q 1. Then G is cyclic. Proof. Let n p G /p = q implies n p = 1 or q and n q p implies n q = 1 or p. And n p 1 mod p and n q 1 mod q. As p < q, p 1 mod q. Therefore n q = 1. As q 1 mod p, n p = 1. Therefore both the Sylow p-subgroup P and P Q the Sylow q-subgroup Q are normal. Therefore P Q = = pq. So G = P Q, P Q = 1, P G, Q G,and P Q hence G = P Q = C p C q = Cpq. Exercise: Suppose G = p 2 q where p, q are primes. Prove G is not simple. E.g.: 19 No group of order 144 is simple. Prop: Let G be a finite group. G is cyclic if and only if G has a unique subgroup of order d for every positive divisor d of G. Study Proof! 2.9 Solvable Groups Def: Let G be a group. A normal series for G is a sequence of subgroups 1 = H n H n 1 H 0 = G. Note: G, nontrivial, is simple iff its only normal series is the trivial normal series. The groups H i /H i 1 are called the factor groups of the series. The length of the series is the number of nontrivial factor groups. Def: A composition series for G is a normal series such that all the factor groups are simple. E.g.: Easy composition series: {1} A 5. E.g.: {1} {1, (1 2)(3 4)} V 4 A 4 S 4 (A 4 /V 4 = C3, S 4 /A 4 = C2, V 4 /H = C 2 ). Def: 20 A subgroup H of G is called a maximal normal subgroup if H G, H G and there are no normal subgroups properly between H and G (equivalently, H G and G/H is simple). Prop: Any finite group has a composition series. Thm: (Jordan-Hölder Theorem) Suppose G has a composition series. Then any two composition series for G have the same length, and the set of factor groups are the same up to isomorphism. Def: A refinement of a normal series for G is just another normal series for G obtained by inserting more subgroups between the terms of the original. Rmrk: If G has a composition series, any normal series has a refinement which is a composition series. Note: A finite abelian group is simple iff G has prime order. Note: The factor groups for any composition series for an abelian group have prime order. Def: A solvable series for a group G is a normal series in which the factor groups are all abelian, i.e., 1 = H n H 0 = G such that H i /H i+1 is abelian. A group is solvable if it has a solvable series. Prop: Suppose G <. Then G is solvable iff all the factor groups in the composition series have prime order. Exercise: Prove this! Def: 21 Define G (n) := G (n 1) where G (0) = G (recall G is the commutator subgroup of G). Recall that G char G. Since char is transitive, G (n) char G for all n 0. In particular, G (n 1) /G (n) is abelian. The series G (2) G (1) G (0) = G is called the derived series. Prop: G is solvable iff G (n) = 1 for some n 1 (i.e., the derived series is normal).

8 3 RING THEORY Algebra Study Guide Lma: Let ϕ : G A be a surjective group homomorphism. Then ϕ(g (n) ) = A (n) for all n 0. Cor: Suppose H G. Then (G/H) (n) = G (n) H/H. Thm: Let G be a group, H G. (1) If G is solvable, so is H. (2) If H G then G is solvable iff H and G/H are solvable. Prop: Any p-group is solvable. (For proof, use induction and note that Z(G) and G/Z(G) is solvable.) 3 Ring Theory 3.1 Intro to Ring Theory Def: 22 Let R be a ring. Let a R. Then a is called a zero-divisor if there exists nonzero b R such that either ab = 0 or ba = 0. If R is commutative and has no nonzero zero divisors (nonzero-divisors or NZDs) then R is called an integral domain (or just a domain). Def: Let R be a ring with 1. Let a R. An element b R is called a left-inverse for a if ba = 1 and b is a right-inverse for a if ab = 1. Rmrk: Let a R. Suppose b is a left-inverse for a and c is a right-inverse. Then b = c. E.g.: Let (G, +) be an abelian group. Let Hom(G, G) = {f : G G : f a group homom.}(= End(G)). Define + and on Hom(G, G) as follows: given f, g Hom(G, G), (f + g)(x) = f(x) + g(x) and (fg)(x) = (f g)(x) for all x G. Then Hom(G, G) is a ring with 1. E.g.: Let G = Z, and R = Hom(G, G). Let f, g R be given by f(a 1, a 2, a 3, ) = (0, a 1, a 2, ) and g(a 1, a 2, a 3, ) = (a 2, a 3, ). Then gf = 1 but fg((1, 0, 0, )) = (0, 0, ) so fg 1. Thus, f is left-invertible but not right-invertible and vice versa for g. (And actually, if α n R is given by α n ((a 0, a 1, )) = (na 0, 0, 0, ), (g + α n )f = 1, so f has infinitely many left inverses.) Rmrk: 23 A unit in a ring cannot be a zero-divisor. Def: A division ring is a ring R {0} with 1 such that every nonzero element is a unit. A field is a commutative division ring. Prop: Let n 2. Then n is prime iff Z/nZ is a field iff Z/nZ is a domain. 3.2 Polynomial Rings Rmrk: If R is a domain, then R[x] is a domain. Thm: 24 (Division Theorem) Let R be commutative with 1, and x a variable. Let f(x), d(x) R[x], where the leading coefficient of d(x) is a unit in R. Then there exist unique polynomials q(x), r(x) R[x] such that f(x) = d(x)q(x) + r(x) and deg r < deg d. Def: A polynomial is called monic if its leading coefficient is a unit. So the division theorem works whenever you divide by a monic polynomial. Cor: Let f(x) R[x], r R. Then f(r) = 0 iff (x r) f(x). 3.3 Ideals and Ring Homomorphisms Def: A map ϕ : R S (R, S rings) is called a ring homomorphism if ϕ(a+b) = ϕ(a)+ϕ(b) and ϕ(ab) = ϕ(a)ϕ(b) for all a, b R. E.g.: 25 Suppose ϕ : R S is a nonzero ring homomorphism. Suppose that R, S have identity and S is a domain. Then ϕ(1 R ) = 1 S. Def: 26 Let R be commutative with 1. A nonempty subset I of R is an ideal if (I, +) (R, +) and for all r R and i I, ri I. Def: Let R be commutative with 1. Let A R. Define the ideal of R generated by A by (A) := I. Exercise: (A) = {r 1 a r n a n : r i R, a i A, n N}. Exercise: Consider the ideal (2, x) Z[x] is not principal. Def: Let R be commutative with identity, I R an ideal. Let R/I denote the set of left cosets of (I, +) in (R, +). Define +, on R/I by (1) (a + I) + (b + I) = (a + b) + I and (2) (a + I) (b + I) = (a b) + I. Prop: 27 Let ϕ : R S be a homomorphism of commutative rings. Then R/ ker ϕ = imϕ (as rings). Prop: Suppose I J are ideals of R. Then J/I is an ideal of R/I. I ideal A R

9 3 RING THEORY Algebra Study Guide 3.4 Operations on Ideals Prop: (R/I)/(J/I) = R/J. Prop: Let I be an ideal of R. Then there exists a bijection, inclusion-preserving correspondence between {J : J an ideal of R} and {ideals of R/I} given by J J/I = ϕ(j) and K ϕ 1 (K), where ϕ : R R/I is the canonical surjective homomorphism. 3.4 Operations on Ideals Def: Let I, J be ideals of R. Then we define I +J := {a + b : a I, { b J}. This is the smallest ideal of R containing n } I J (which is not an ideal). Further, IJ = {ij} i I, jıj = i k j k : i k I, j k J, which is the smallest k=1 ideal containing {ij} i I, j J. Thm: (Chinese Remainder Theorem) Let R be commutative with identity, I 1,, I n ideals of R such that I i + I j = R for all i j. Then (1) I 1 I n = I 1 I 2 I n and (2) there is an isomorphism of rings ϕ : R/(I 1 I n ) R/I 1 R/I n given by r + I 1 I n (r + I 1,, r + I n ). Thm: 28 Let d 1,, d n be pairwise relatively prime positive integers and a 1,, a n any integers. Then there exists x Z such that x a i mod d i for i = 1,, n. Rmrk: If R, S are rings with 1, then (R S) = R S as groups. Def: Let R be commutative with 1. An ideal m of R is called maximal if m R and the only ideals containing m are m and R. Prop: An ideal I of R is maximal iff R/I is a field. Rmrk: Let ϕ : R F is a surjective ring homomorphism such that ϕ 0, F a field. Then ker ϕ is a maximal ideal, as R/ ker ϕ is isomorphic to a field. Rmrk: Let m, n be distinct maximal ideals of a commutative ring R. Then m + n = R. Def: An element r R is called nilpotent if r n = 0 for some n N. R is called reduced if there are no nonzero nilpotents. 3.5 Zorn s Lemma Def: 29 Let A be a set. A partial order on A is a binary relation with the following properties: for all a, b, c A, (1) a a; (2) a b and b a iff a = b; (3) a b, b c implies a c. Additionally, is a total order on A if for all a, b A, a b or b a. Def: Let (A, ) be a poset, and B a subset of A. An element x A is an upper bound for B if b x for all b B. An element b B is maximal if for every y B with y b implies y = b. Thm: (Zorn s Lemma) Let (A, ) be a nonempty poset and suppose every totally ordered subset of (A, ) has an upper bound in A. Then A is a maximal element. Prop: Let R 0 be commutative with 1 and I R an ideal. Then there exists a maximal ideal m of R containing I. Def: Let R be commutative. An ideal P of R is called prime if P R and whenever ab P, then either a P or b P. Rmrk: P is prime iff R/P 0 is a domain. Rmrk: Let ϕ : R S be a commutative ring homomorphism. Suppose ϕ 0 and S is a domain. Then ker ϕ is a prime ideal of R. Rmrk: R is a domain iff {0} is prime. R is a field iff {0} is maximal. Rmrk: If P is prime and P IJ then P I or P J. 3.6 Quotient Fields Def: Given a domain R, the smallest field containing R is called the quotient field (or field of fractions) of R, denoted by Q(R). Prop: Let R be a domain and suppose R F, F a field. Then there exists an injective field homomorphism f : Q(R) F that fixes R. 3.7 Euclidean Domains Def: 30 Let R be a domain. R is called a Euclidean domain if there exists a function δ : R\{0} N 0 = {0, 1, 2, } such that for all a, b R with b 0 there exists q, r R such that a = bq + r with r = 0 or δ(r) < δ(b).

10 3 RING THEORY Algebra Study Guide 3.8 Factorization Conditions Thm: Z[i] is a Euclidean domain with δ : Z[i] N 0 by a + bi a 2 + b 2. Thm: Z[i] = {a + bi : a, b Z} is a Euclidean Domain with δ : Z[i] N given by a + bi a 2 + b 2 = (a + bi)(a bi). Proof. Note that, for all α, β Z[i], δ(αβ) = δ(α)δ(β). Suppose β 0. Then α β = αβ = x + yi, for x, y Q. ββ Let a, b Z and u, v Q such that x = a + u and y = b + v for u, v 1/2. Then observe α = β(x + yi) = β(a + bi) + β(u + vi). Let γ = β(u + vi) and note γ = α β(a + bi) implies γ Z[i]. Claim: If we show δ(γ) < δ(β), we re done: α = β(a + bi) + γ. Notice that δ can be extended to a function on Q[i] Q 0 by x+yi x 2 +y 2. Then δ(γ) = δ(β(u+vi)) = δ(β)δ(u+vi), and δ(u + vi) = u 2 + v = 1 2. Thus, δ(γ) 1 2δ(β) < δ(β). Def: An integral domain in which every ideal is principal is a PID. Thm: Euclidean Domains are PIDs. Proof. Let R be a ED under δ and let I R be an ideal. If I = {0}, we re done. Assume I {0}. Choose d I \{0} such that δ(d) δ(a) for all a I \ {0}. Claim: I = (d). Pick a I. Then a = dq + r for some q, r R with r = 0 or δ(r) < δ(d). Since r = a dq I, we may not have δ(r) < δ(d), as δ(d) δ(b) for all b I. Thus, r = 0, and a (d). Rmrk: The domain R = Z[ 5] is not a PID (the ideal (3, 2 + ] 5) is not principal). Rmrk: 31 The domain R = Z is a PID that is not a Euclidean Domain. [ Prop: Let R be a PID and I = (p) 0 a prime ideal. Then I is maximal. 3.8 Factorization Conditions Def: Let R be a domain. An element r R is called irreducible if r is not a unit and whenever r = ab, then one of a or b is a unit. An element p R \ {0} is called prime if (p) is a prime ideal (i.e. if p ab then p a and p b). Prop: Let R be a domain. Then any prime element is irreducible. Rmrk: 2 R = Z[ 5] is irreducible but not prime. Def: 32 Let R be a domain. Then R is a UFD if (1) every nonunit of R is a product of irreducibles and (2) if a 1 a 2 a m = b 1 b n, where a i, b j is irreducible, then m = n and after reindexing a i = u i b i for u i R. Def: Let R be a commutative ring. R is said to satisfy the ascending chain condition (ACC) on principal ideals if given any ascending chain of principal ideals (a 1 ) (a 2 ) there exists n 1 such that for all k n, (a k ) = (a k+1 ) i.e., (a 1 ) (a 2 ) (a n ) = (a n+1 ) = (a n+2 ) =. Rmrk: Let R be a commutative ring. Then R satisfies ACC on principal ideals iff every nonempty set Λ of principal ideals of R has a maximal element in Λ. Thm: Let R be a domain. Then R is a UFD if and only if R satisfies ACC on principal ideals and every irreducible element is prime. Prop: 33 Let R be a PID. Then R is a UFD. Def: Let R be a domain, a, b R. An element d R \ {0} is a greatest common divisor (gcd) of a, b if (1) d a, d b and (2) if e a, e b, then e d. Rmrk: If d 1, d 2 are gcds of a and b then (d 1 ) = (d 2 ). Prop: Let R be a UFD. Then gcd(a, b) exists for all a, b R. Def: Let R be a domain, a, b R both nonzero. The least common multiple (lcm) of a and b is an element m R \ {0} such that (1) a m, b m, and if a m and b n, then m n. Rmrk: Any two lcms of a and b are associate. Exercise: Let R be a UFD, a, b R \ {0}. Then (a) (b) = (lcm(a, b)). Exercise: Find an example of a domain R and nonzero elements a, b such that gcd(a, b) does not exist. (R = Z[ 5], a = 2(1 + 5), b = 6.)

11 4 MODULES Algebra Study Guide Def: 34 Let R be a UFD, x a variable. Let f(x) = a n x n + + a 0 R[x] \ {0}. The content of f(x) is defined by c(f) = gcd(a n,, a 1, a 0 ). f(x) is called primitive if c(f) = 1. Rmrk: c = c(f) implies f = cf 1, where f 1 is primitive. Rmrk: If d R, d c(f), then f/d R[x] and c(f/d) = c(f)/d. Rmrk: If f = cg where c R, then c c(f). Rmrk: Suppose f = gh, f, g, h R[x] then c(g)c(h) c(f). Thm: (Gauss Lemma) Let R be a UFD, f = gh R[x] \ {0}. Then c(f) = c(g)c(h). Cor: 35 The product of primitive polynomials is primitive. Cor: Let R be a UFD, F = Q(R). Let f(x) R[x], g(x) R[x] primitive. Suppose g f in F [x] (i.e. f = gh for h F [x]). Then g f in R[x]. Cor: Let R be a UFD and F = Q(R), f(x) R[x]. Suppose f(x) = g(x)h(x), where g, h F [x]. Then there exist a, b F such that ag, bh R[x] and f = (ag)(bh). Cor: Let R be a UFD and F = Q(R). Let f(x) R[x] be primitive. Then f(x) is irreducible in R[x] if and only if f(x) is irreducible in F [x]. Thm: Let R be a UFD. Then R[x] is a UFD. Cor: If k is a field (or even just a UFD), k[x 1,, x n ] is a UFD for all n. Rmrk: 36 If deg f(x) > 1 and f(x) is irreducible then f(x) has no roots in F. Rmrk: (x 2 + 1) 2 Q[x] is reducible but has no roots. Rmrk: If deg f = 2 or 3 then f is irreducible if and only if f has no roots in F. Thm: (Rational Root Theorem) Let R be a UFD and f(x) = a n x n + a 0 R[x]. Suppose α F = Q(R) is a root of f. Then α = c/d where c a 0 and d a n, for c, d R. Thm: (Eisenstein s Criterion) Let R be a UFD and f(x) = a n x n + +a 0 R[x] and suppose there exists an irreducible element π R such that (1) π a n ; (2) π a i for 0 i n 1, and (3) π 2 a 0. Then f(x) is irreducible in F [x], where F = Q(R). 4 Modules 4.1 Intro to Module Theory Def: 37 Let R be a ring. A left R-module is an abelian group (M, +) together with an R-action, i.e., a map from R M M ((r, m) rm) with the following properties: For all r, s R and m, n M, r(m + n) = rm + rn (r + s)m = rm + sm (rs)m = r(sm) 1m = m. E.g.: Let R be a ring {0}, R Let R be commutative. Any ideal I of R is a left (or right) R-module. R a ring, n N. Define the set R n := {(x 1,, x n ) : x i R}. R n is a left R-module, and is called a free left R-module of rank n. Let R = Z. Any abelian group (G, +) is a Z-module via the action of adding elements up m or m times. Let R be a ring and n N. Let S = M n (R). Then S acts on R n via matrix multiplication. Let ϕ : R S be a ring homomorphism such that ϕ(1 R ) = 1 S. Then S is a left R module via the action r s = ϕ(r)s. If R S and 1 R = 1 S, then S is a left R-module via r s = rs. Let R be commutative and I and ideal of R. There is a canonical surjective ring homomorphism ϕ : R R/I given by ϕ(r) = r + I. Thus, R/I is an R-module.

12 4 MODULES Algebra Study Guide 4.1 Intro to Module Theory Def: Let F be a field. An F -vector space is a left F -module. Def: Let R be a ring and let M be a left R-module. A left R-submodule of M is a subgroup N of M such that N is a left R-module via the action of R on M restricted to N. Prop: Let M be a left R-module and N a nonempty subset of M. Then N is a left R-submodule of M if for all r R and for all x, y N, x + ry N. E.g.: Let R be a ring. Then N = { (r 1, r 2, r 3 ) R 3 : r 1 + 2r 2 r 3 = 0 } is a left R-submodule of R 3. E.g.: Let R be a commutative ring and M a left R-module. Let r R and N = {m M : rm = 0}. Then N is a left R-module. Rmrk: 38 Let ϕ : R S be a ring homomorphism with ϕ(1 R ) = 1 S. Then any left S-module M is also a left R-module in a natural way: r R, m M, r m = ϕ(r) = m. E.g.: ϕ : Z Q. Then Q 2 is a Z-module in a natural way. E.g.: Let E be a field and F a subfield. Then ϕ : F E is a field homomorphism via the inclusion map. Then E is an F -vector space and F is an F -subspace of E. Def: Let M, N be R-modules. A function f : M N is an R-module homomorphism (or an R-linear map) if f(x + y) = f(x) + f(y) for all x, y M. f(rx) = rf(x) for all r R, x M. Rmrk: Given an R-linear map f from M to N, ker(f) := {m M : f(m) = 0} and im(f) := {f(m) : m M}. ker(f) and im(f) are submodules of M and N, respectively. Def: Let M be an R-module and {M i } i I a collection of submodules of M. Define M i := { m i1 + + m in : n N, m ij M ij, for j = 1,, n }. i I Then M i is the smallest submodule of M containing M i. In the case I = {1,, n}, we write M 1 + M 2 + i I i I + M n = i I M i. Def: If S is a nonempty subset of M, then RS := Rs. Def: Let M be an R-module. The annihilator of M is defined by s S Ann R (M) := {r R : rm = 0 m M}. Note: Ann R (M) is a two-sided ideal of R. Rmrk: Let R be commutative and M an R-module and I an ideal of R with I Ann R (M). Then M is an R/I module. Def: 39 Let {M 1,, M n } be a collection of R-modules. The external direct product is defined to be M 1 M 2 M n := {(m 1,, m n ) : m i M i }. The general case: Let {M i } i I be a collection of R-modules. Then M i := {(m i ) i I : m i M i, m i = 0 for all but finitely many i}. i I Def: Let {M i } i I be a set of R-modules. We define the external direct product to be M i := {(m i ) : m i M i i}. i I Thus, M i M i, with equality if I <. i I i I Prop: Let M be an R-module and {M i } i I a collection of submodules of M. Then M = M i if and only if (1) i I ( M = ) M i and (2) for all i I, M i M j = (0). i I j i Prop: Let {M i } i I be submodules of M. If M = i I M i (internal) then M = i I M i (external). Def: Let M be an R-module. A subset S of M is called an R-basis for M if

13 5 VECTOR SPACES Algebra Study Guide 4.2 Module Isomorphism Theorems 1. M = RS (i.e., S generates M) 2. S is R-linear independent (i.e., if r 1 s r n s n = 0 (with r i R, s i S and distinct) then r i = 0 for all i). If M has an R-basis, then M is called a free R-module. Rmrk: If M has a basis of n elements, then M = R n. Rmrk: If R is not commutative, then a free module may have bases of different cardinalities. If R is commutative, then any two bases for a given free module have the same cardinality. This cardinality is called the rank of the free module. Rmrk: Suppose R is commutative. Let I be an ideal. Then I is free if and only if I = (0) or I = Rs = (s), where s is a non-zero divisor. Rmrk: Let R be commutative and I an ideal of R. Then R/I is free if and only if I = R or I = (0). 4.2 Module Isomorphism Theorems 1. Let ϕ : M N be an R-module homomorphism. Then M/ ker ϕ = ϕ(m). 2. Let A, B be submodules of M. Then (A + B)/B = A/(A B). 3. Let L M N be R-modules. Then (N/L)/(M/L) = N/M. 4. Let N M be R-modules. Then there exists a one-to-one correspondence between the set of R-submodules of M containing N and the R-submodules of M/N (i.e., Q Q/N). 5 Vector Spaces Let R = F a field. Then a set V is an F -module if and only if V is an F -vector space 40. Def: If V is an F -vector space and S V, then F S = span F (S). If span F (S) = V, we say S spans V. Prop: Let F be a field and V and F -vector space. Let T S be subsets of V such that (1) T is linearly independent and (2) S spans V. Then there exists a basis β with T β S. Cor: Let V be an F -vector space for a field F. Then 1. V has a basis. 2. Every linearly independent subset of V is contained in a basis. 3. Every spanning set for V contains a basis. Thm: (Replacement Theorem) Let V be an F -vector space. Let S be a spanning set for V and T a linearly independent subset of V. For every (distinct) t 1,, t n T there exists s 1,, s n S (distinct) such that (S \ {s 1,, s n }) {t 1,, t n } spans V. Cor: Let V be a finitely-generated F -vector space. Let S be a finite spanning set for V and T a linearly independent subset of V. Then T S. Cor: Let V be a finitely-generated F -vector space. Then (1) V has a finite basis and (2) any two bases have the same number of elements. Def: Let V be an F -vector space and dim V = β for any basis β of V (finite or infinite). Cor: Let V be a finite-dimensional vector space and W a subspace of V. Then dim W dim V and W = V if and only if dim W = dim V. Prop: 41 Let F be a finite field. Then F = p n for p Z a prime and n 1. Prop: Let ϕ : V W be a linear transformation of F -vector spaces. Suppose dim V = n <. Let K = ker ϕ and C = imϕ. Then dim F V = dim F K + dim F C. Def: The rank of ϕ is rank(ϕ) := dim C and the nullity of ϕ is nullity(ϕ) := dim K. Cor: Let ϕ : V W be an F -linear transformation. Suppose dim V = dim W <. Then the following are equivalent: 1. ϕ is an isomorphism. 2. ϕ is injective.

14 6 MATRIX ALGEBRA Algebra Study Guide 3. ϕ is surjective. 4. ϕ takes a basis of V to a basis of W. The proof follows from the rank + nullity theorem. 6 Matrix Algebra 6.1 Intro to Matrices Def: Let R be a commutative ring and m, n N. An m n matrix over R is an array (a ij ) with m rows, n columns, and mn th entry a mn. Let M m n (R) denote the set of m n matrices over R. We add them in the usual way. Then (M m n (R), +) is an abelian group. Define an R-action on M m n (R) by r (a ij ) = (ra ij ) for all r R. Thus, M m n (R) is an R-module. So for 1 i m and 1 j n let E ij denote the m n matrix with 1 = a ij and 0 elsewhere. Then {E ij : 1 i m, 1 j n} is an R-basis for M m n (R). Then M m n (R) = R mn. Let A = (a ij ) be an m n matrix and b = (b j1 ) be an n 1 matrix. Define where c i = n a ij b j1. j=1 A b = c 1., c m Prop: Let R be a commutative ring, and M an R-module with B = {e 1,, c n } M. The following are equivalent: 1. B is a basis for M. 2. Given any R-module N and x 1,, x n N, there exists a unique R-module homomorphism ϕ : M N such that ϕ(e i ) = x i for all i. 6.2 Transition Matrices Def: Let R be a commutative ring, and let M and N be R-modules. Define Hom R (M, N) = {f : M N : f R-linear}. Rmrk: Hom R (M, N) is an R-module with R-action: for r R, f : M N, define r f : M N by m rf(m). Rmrk: If M = N, then Hom R (M, M) is a ring with identity (and multiplication given by composition). Hom R (M, M) is usually denoted End R (M) (the endomorphism ring). Prop: Let R be a commutative ring, M an R-module, β = {e 1,, e n } M. The following are equivalent: 1. β is a basis for M 2. Given any R-module N and x 1,, x n N there exists a unique R-homomorphism ϕ : M N such that ϕ(e i ) = x i for all i. Def: Let R be commutative, M, N free R-modules of ranks m and n, respectively. Let β = {β 1,, β m } and E = {E 1,, E n } be bases for M, N, respectively. Given an R-homomorphism ϕ : M N we have for 1 j m, there exist unique elements a ij R such that ϕ(β j ) = m a ij E i. The n m matrix (a ij ) is called the matrix of ϕ with i=1 respect to β and E and denoted by Mβ E(ϕ). Prop: Let R be commutative, and M and N free R-modules of ranks m, n respectively. Fix bases B, E for M, N. Then the map from Hom R (M, N) M n m (R) taking ϕ MB E (ϕ) is an R-module homomorphism. Prop: Let ϕ : M N, ψ : L M be R-module homomorphisms. Let δ, B, E be bases for L, M, and N respectively. Then Mδ E(ϕ ψ) = M B E(ϕ)M δ B(ψ). Cor: If M is free of rank n, then the map End R (M) M n n (R) given by ϕ MB B (ϕ) is a ring isomorphism. Cor: Let F be a field and A M n (F ). TFAE:

15 6 MATRIX ALGEBRA Algebra Study Guide 6.3 Hom and Duality 1. A is invertible. 2. Ax = 0 if and only if x = 0 for all x M n 1 (F ). 3. For all y M n 1 (F ) there exists x M n 1 (F ) such that Ax = y. Def: 42 Let A, B M n n (R). Then A and B are called similar if there exists an invertible n n matrix P such that A = P 1 BP. E.g.: Let F be a field and let V = {f F [x] : deg f 2}. Let B = { 1, x, x 2} and E = { 1, 1 + x, 1 + x + x 2}. Then MB(I) E = and ME B (I) = Hom and Duality Def: Let R be commutative, M an R-module. The dual module (or dual space) of M is M := Hom R (M, R). In the case that R is a field, the elements of M are called linear functionals. Prop: Let F be a free R-module of rank n. Let {e 1,, e n } be a basis for F. For each i {1,, n}, define maps e i := F R by e i (e j) = δ ij (the Dirac delta function). Then {e 1,, e n} is a basis for F called the dual basis for {e 1,, e n }. In particular, F is free of rank n. Def: Let M be an R-module, m M. Define a map ev m : M R by f f(m). Note: ev m is R-linear. Therefore, ev m Hom R (M, R) = M. This gives a natural homomorphism from M to M, i.e., ϕ : M M given by ϕ(m) = ev m. Rmrk: For finite-dimensional vector spaces, V = V = V. 6.4 Determinants Def: 43 Let R be a commutative ring and M an R-module with n N. Then an n-multilinear form on M is a function f : M n R such that for all i and for all m 1,, ˆm i,, m n M (fixed) the function ϕ i : M R given by x f(m 1,, m i 1, x, m i+1,, m n ) is an R-module homomorphism. We say f is alternating if for all m 1,, m n M there exists i such that m i = m i+1 im plies f(m 1,, m n ) = 0. Prop: Let f be an alternating n-multilinear form on M. Let m 1,, m n M. Then 1. f(m 1,, m i 1, m i+1, m i, m i+2,, m n ) = f(m 1,, m n ) (i.e., the value of ϕ on an n-tuple is negated if two adjacent components are interchanged). 2. σ S n, f(m σ(1),, m σ(n) ) = sign(σ)f(m 1,, m n ). 3. If m i = m j for some i j, f(m 1,, m n ) = α R and i j, f(m 1,, m i + αm j,, m n ) = f(m 1,, m n ) (where m i + αm j is in the ith position). Proof. 1. Let ψ : M M R be given by ψ(x, y) = f(m 1,, m i 1, x, y, m i+2,, m n ). then ψ(x+y, x+y) = 0 as f is alternating. By the fact that ψ is an R-linear map, ψ(x + y, x + y) = ψ(x, x + y) + ψ(y, x + y) = ψ(x, x) + ψ(x, y) + ψ(y, x) + ψ(y, y) = 0 = ψ(x, y) + ψ(y, x).

16 6 MATRIX ALGEBRA Algebra Study Guide 6.4 Determinants 2. Recall: every σ S n can be written as a product of transpositions of the form (i i + 1). We do induction on the number of transpositions in such a decomposition. If σ = (i i + 1), we re done by part 1. Now let σ = τ(i i + 1), where τ is a product of n 1 such transpositions. Then f(m σ(1),, m σ(n) ) = f(m τ(1),, m τ(i 1), m τ(i+1), m τ(i),, m τ(n) ) = f(m τ(1),, m τ(n) ) (by 1) = sign(τ)f(m 1,, m n ) = sign(σ)f(m 1,, m n ). 3. Let σ be any permutation which fixes i and sends j to i+1. Then f(m 1,, m n ) = sign(σ)f(m σ(1),, m σ(n) ) = 0, as m j is now in the i+1 st component, giving that f equals 0 by the definition of an alternating n-multilinear form. 4. f(m 1,, m i + αm j,, m n ) = f(m 1,, m n ) + αf(m 1,, m j, m n ) = f(m 1,, m n ) + 0(by 3, as the ith and jth components are equal in the second summand). Prop: Let f be an alternating n-multilinear form on M. Suppose for j = 1,, n and w j = n α ij v i, where w i M, v i M, α ij R. Then f(w 1,, w n ) = ( σ S n sign(σ)α σ(1)1 α σ(n)n ) f(v 1,, v n ). Def: Let R be a commutative ring and n N. A determinant on R is a function det : M n n (R) R such that det is an alternating n-multilinear function on the columns of R (M = R n ) and such that det(i n n ) = 1. Notation: If A M n n (R), let A 1,, A n denote the columns of A, i.e., det(a 1,, A n ) is alternating multilinear. Thm: There exists a unique determinant function on R. If A = [a ij ] then det(a) = σ S n sign(σ)a σ(1)1 a σ(n)n. Prop: 44 Let A, B M n n (R). Then det(ab) = det(a) det(b). Def: Let A M m n (R). Say A = [a ij ]. The transpose of A, denoted A t, is the n m matrix whose ij th entry is a ji. Prop: Let A M n n (R). Then det(a) = det(a t ). Thm: (Cramer s Rule) Let A M n n (R) with columns A 1,, A n. Let [x 1,, x n ] M n 1 (R). Let [y 1,, y n ] = x 1 A x n A n (so AX = Y ). Then x i det A = det(a 1,, A i 1, Y, A i+1,, A n ). Def: Let A = [a ij ] M n n (R). Then the ij th cofactor of A is ( 1) i+j det(a ij ), where A ij is the n 1 n 1 submatrix of A obtained by deleting the ith row and jth column. Prop: (Cofactor Expansion) Let A = [a ij ]. Fix i {1,, n}. Then det A = n ( 1) i+j det(a ij ). Def: Let A M n n (R). Define the adjoint of A, denoted adj A, to be the n n matrix whose ijth entry is ( 1) i+j det(a ji ). Thm: Let A M n n (R). Then adj(a) A = A adj(a) = det(a)i. Cor: 45 Let A M n n (R). Then A is invertible IFF det(a) R. Def: Let R be a domain, A M n n (R). Then the column rank of A is the maximum number of linearly independent columns of A, (and similarly for row rank). Prop: Let R be a domain, A M n n (R). TFAE: 1. col rank(a) = n 2. row rank(a) = n 3. det(a) 0. Def: Let R be commutative, A M m n (R). Let 1 r min {m, n}. An r r minor of A is the determinant of an r r submatrix of A obtained by deleting any m r rows and n = r columns. Def: Define the rth fitting ideal of A, denoted I r (A), to be the ideal of R generated by all the r r minors of A. Rmrk: i=1 i=1

17 7 MODULES OVER A PID Algebra Study Guide 1. I 0 (A) = R and I r (A) = 0 for r min {m, n} by convention. 2. I r (A) = I r (A t ) for all r I r+1 (A) I r (A) for all r For any A, B, I r (AB) I r (A) I r (B). Thm: Let R be a domain and A an arbitrary m n matrix over R. Then col rank(a) = sup {r : I r (A) (0)}. 7 Modules over a PID 7.1 Basic Theory Def: 46 Let M be an R-module. M is called Noetherian if M satisfies the ascending chain condition on submodules, i.e., given any ascending chain N 0 N 1 N 2 of submodules of M there exists t such that for all n 0, N t = N t+n, N t = N t+1 = N t+2 =. We say R is Notherian as a ring if it is Noetherian as an R-modules (i.e., R satisfies ACC on ideals). Prop: Let M be an R-module. The following are equivalent: 1. M is Noetherian 2. Every submodule of M is finitely generated. 3. Every nonempty set of submodules of M has a maximal element. Cor: R is a Noetherian ring if and only if for all ideals I of R, I is finitely generated. Prop: Let M N be R-modules. Then N is Noetherian if and only if M and N/M are Noetherian. Cor: Let M 1,, M n be R-modules. Then M 1 M 2 M n is Noetherian if and only if each M i is Noetherian. Proof. By induction on n (M 1 M 2 )/M 1 = M2. Prop: Let R be a Noetherian ring. Then every finitely-generated R-module is Noetherian. Prop: Let R be a domain and M a free R-module of rank n. Then every set of n+1 elements of M is linearly dependent. Def: Let R be a domain and M an R-module. A set x 1,, x n M is said to be a maximal linearly independent set if {x 1,, x n } is linearly independent and {x 1,, x n, z} is linearly dependent for all z M. The rank of M is defined to be rank M = sup {n : x 1,, x n M forming a maximal LI set}. Prop: 47 Let R be a domain and let L M be R-modules. Then Cor: rank M 1 rank M n = n rank M i. i=1 Proof. Induction and (M 1 M 2 )/M 1 = M2. rank M = rank L + rank M/L. Def: Let R be a domain and M an R-module. Define the torsion submodule of M to be Tor R (M) := {m M : rm = 0 for some r R \ {0}}. If M = Tor R (M), M is called torsion. If Tor R (M) = 0, M is said to be torsion-free. Rmrk: Let R be a domain. 1. If M is free, M is torsion-free. 2. Submodules of torsion-free modules are torsion-free.