ALGEBRA HW 11 CLAY SHONKWILER

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1 ALGEBRA HW 11 CLAY SHONKWILER 1 Let V, W, Y be fiite dimesioal vector spaces over K. (a: Show that there are atural isomorphisms (V W V W Hom(V, W Hom(W, V. Proof. (V W V W : Defie the map φ : V W (V W by (f, g (v w f(vg(w. The, if f 1, f 2 V, g 1, g 2 W ad a, b K, the φ(af 1 + bf 2, g 1 (v w (af 1 + bf 2 (vg 1 (w ad v w (af 1 (v + bf 2 (vg 1 (w v w (af 1 (vg 1 (w + bf 2 (vg 1 (w (v w af 1 (vg 1 (w + (v w bf 2 (vg 1 (w a(v w f 1 (vg 1 (w + b(v w f 2 (vg 1 (w aφ(f 1, g 1 + bφ(f 2, g 1 φ(f 1, ag 1 + bg 2 v w f 1 (v(ag 1 + bg 2 (w v w f 1 (v(ag 1 (w + bg 2 (w v w (af 1 (vg 1 (w + bf 1 (vg 2 (w (v w af 1 (vg 1 (w + (v w bf 1 (vg 2 (w a(v w f 1 (vg 1 (w + b(v w f 1 (vg 2 (w aφ(f 1, g 1 + bφ(f 1, g 2, so φ is biliear. Therefore, by the uiversal property of the tesor product, φ iduces a uique liear map Φ : V W (V W. To show that Φ is a isomorphism, the, it suffices merely to show that Φ is ijective, sice V W ad (V W have the same dimesio (amely dim V dim W. Now, suppose f V ad g W such that Φ(f g 0. The 0 Φ(f g (v w f(vg(w, Suppose f 0. The f(v 0 0 for some v 0 V. The, sice 0 (v 0 w f(v 0 g(w 1

2 2 CLAY SHONKWILER for all w, we see that g(w 0 for all w W. Alteratively, if g 0 the f 0. Therefore, if f g ker Φ, the { } f 0 f g 0 0 g so ker Φ 0 ad so Φ is ijective ad, therefore, a isomorphism. V W Hom(V, W : Defie the map φ : V W Hom(V, W by φ : (x, y (v x(vy. If x 1, x 2 V, y 1, y 2 W ad a, b K, the ad φ(ax 1 + bx 2, y 1 v (ax 1 + bx 2 (vy 1 v (ax 1 (v + bx 2 (vy 1 v ax 1 (vy 1 + bx 2 (vy 1 (v ax 1 (vy 1 + (v bx 2 (vy 1 aφ(x 1, y 1 + bφ(x 2, y 1 φ(x 1, ay 1 + by 2 v (x 1 (v(ay 1 + by 2 v ax 1 (vy 1 + bx 1 (vy 2 (v ax 1 (vy 1 + (v bx 1 (vy 2 aφ(x 1, y 1 + bφ(x 1, y 2 so φ is biliear. Therefore, by the uiversal property of the tesor product, it iduces a uique liear map Φ : V W Hom(V, W. We wat to show that Φ is a isomorphism; to that ed, sice V W ad Hom(V, W have the same dimesio (amely dim(v dim(w, we eed oly show that Φ is ijective. Suppose Φ(f g 0 for f V ad g W. The 0 Φ(f g (v f(vg, which is to say that f(vg(w 0 for all v V ad w W. Clearly, this occurs oly if f 0 or g 0, so { } f 0 f g 0, 0 g so ker Φ 0 ad, thus, Φ is ijective ad, therefore, is a isomorphism. Hom(V, W Hom(W, V : Defie Φ : Hom(V, W Hom(W, V by (v g (w (v g(w ad Ψ : Hom(W, V Hom(V, W by (w f (v (w f(v.

3 ALGEBRA HW 11 3 The, if (v g 1, (v g 2 Hom(V, W, the Φ((v g 1 + (v g 2 Φ(v (g 1 + g 2 w (v (g 1 + g 2 (w w (v (g 1 (w + g 2 (w w ((v g 1 (w + (v g 2 (w (w (v g 1 (w + (w (v g 2 (w Φ(v g 1 + Φ(v g 2, so Φ is a homomorphism. If (w f 1, (w f 2 Hom(W, V, the Ψ((w f 1 + (w f 2 Ψ(w (f 1 + f 2 v (w (f 1 + f 2 (v v (w (f 1 (v + f 2 (v v ((w f 1 (v + (w f 2 (v (v (w f 1 (v + (v (w f 2 (v Ψ(w f 1 + Ψ(w f 2, so Ψ is a homomorphism. Now, if (v g Hom(V, W ad we let f (v g(w, the Ψ Φ(v g Ψ(w (v g(w Ψ(w f v (w f(v v (w g(w v g. O the other had, if (w f Hom(W, V ad we let g (w f(v, the Φ Ψ(w f Φ(v (w f(v Therefore, we see that Φ(v g w (v g(w w (v f(v w f. Ψ Φ Id Hom(V,W Φ Ψ Id Hom(W,V, so, i fact, Φ ad Ψ are isomorphisms. (b: Show that there is a atural isomorphism Hom(V W, Y Hom(V, Hom(W, Y.

4 4 CLAY SHONKWILER Proof. Let π : V W V W be the stadard projectio ad suppose f Hom(V W, Y. The f π is a biliear map from V W to Y. Defie Φ : Hom(V W, Y Hom(V, Hom(W, Y by f (v (w f π(v, w. The, sice f π is biliear, w f π(v, w is a liear map. Now, if f 1, f 2 Hom(V W, Y ad a, b K, the Φ(af 1 + bf 2 v (w (af 1 + bf 2 π(v, w v (w (af 1 π(v, w + af 2 π(v, w v ((w af 1 π(v, w + (w bf 2 π(v, w v (a(w f 1 π(v, w + b(w f 2 π(v, w (v a(w f 1 π(v, w + (v b(w f 2 π(v, w a(v (w f 1 π(v, w + b(v (w f 2 π(v, w aφ(f 1 + bφ(f 2, so Φ is a homomorphism. Sice the dimesios of Hom(V W, Y ad Hom(V, Hom(W, Y are the same (amely the product of the dimesios of V, W ad Y, we eed oly show that Φ is ijective. Suppose, the, that Φ(f 0 for f Hom(V W, Y. The 0 Φ(f (v (w f π(v, w, so f π(v, w 0 for all v V ad w W. Thus, for v V ad w W, 0 f π(v, w f(v w, so f 0. Therefore, we see that ker Φ 0, so Φ is a isomorphism. (c: Show that Hom(V W, Y is aturally isomorphic to the vector space of biliear maps V W Y. Proof. Let B the space of biliear maps V W Y. By the uiversal property of tesor products, if f Hom(V W, Y, the f π is a biliear map from V W to Y (where π : V W V W is the stadard projectio. Therefore, defie Φ : Hom(V W, Y B by Φ(f f π. The, if f 1, f 2 Hom(V W, Y ad a, b K, the Φ(af 1 + bf 2 (af 1 + bf 2 π af 1 π + af 2 π aφ(f 1 + bφ(f 2, so Φ is a homomorphism. O the other had, if g B, the, by the uiversal property of the tesor product, g iduces a uique liear map g Hom(V W, Y such that g ḡ π. Defie Ψ : B Hom(V W, Y by Ψ(g ḡ. Now, suppose g 1, g 2 B, a, b K. The ag 1 + bg 2 Ψ(ag 1 + bg 2 π ag 1 + bg 2 π.

5 Therefore, ALGEBRA HW 11 5 (aψ(g 1 + bψ(g 2 π (ag 1 + bg 2 π ag 1 π + bg 2 π ag 1 + bg 2 Ψ(ag 1 + bg 2 π; sice the iduced map is uique, we see that aψ(g 1 + bψ(g 2 Ψ(ag 1 + bg 2, so Ψ is a homomorphism. Now, if f Hom(V W, Y, the ad, if g B, the Ψ Φ(f Ψ(f π f Φ Ψ(g Ψ(g π g, so Ψ Φ Id Hom(V W,Y ad Φ Ψ Id B, so we see that Φ ad Ψ are isomorphisms. 2 Let R be the rig of polyomial fuctios o the uit sphere S 2 R 3. Thus this rig is give by R R[x, y, z]/(x 2 + y 2 + z 2 1. (a: Let P (0, 0, 1 S 2, ad let R P { f g f, g R; g(p 0}. Show directly that R P is a local rig, ad fid a set of geerators for I. Proof. Let U R P be the set of uits i R P. Let f g U. The g f R P, which meas that f(p 0. O the other had, if f g R P such that f(p 0, the g f R P ad so f g U. Therefore, { } f U f, g R, f(p 0, g(p 0. g Therefore, if J R P is a ideal, J R P U. However, R P U is itself a ideal: if f g R P U ad f g R P, the f g f g ff gg ad f(p f (P 0 f (P 0, so f g f g R P U; call this ideal I. The I is clearly the uique maximal ideal, sice we ve just see that all other proper ideals must be cotaied i I. Specifically, { } f I f, g R, f(p 0, g(p 0. g Now, certaily x, y, z 1 I; we claim that these three elemets geerate I. Now, if f(p 0, the f(x, y, z x α y β (z 1 γ h(x, y, z

6 6 CLAY SHONKWILER for α, β, γ 0 ad at least oe of the α, β, γ strictly positive ad h R P, so we see that the set {x, y, z 1} geerates I. (b: Show that I 2 I but that I 2 I. Let I/I 2 be the image of I uder the rig homomorphism R P R P /I 2. Show that I/I 2 is a 2-dimesioal vector space over R. z 1 Proof. Clearly, sice I is a ideal, I 2 I. Let f, g I. The, i R[x, y, z], deg(fg deg(f + deg(g. The oly relatio i R is x 2 + y 2 + z 2 1, which gives the relatios x 2 1 y 2 z 2, y 2 1 x 2 z 2 ad z 2 1 x 2 y 2. Note that oe of these relatios (except the origial oe are degree-reducig. Hece, viewed i R, deg(f g is equal to the degree of fg i R[x, y, z] uless fg is divisible by x 2 + y 2 + z 2. Sice x 2 +y 2 +z 2 is irreducible i R[x, y, z], fg caot be so divisible uless either f or g is divisible by x 2 +y 2 +z 2. However, sice x 2 +y 2 +z 2 1 i R, we ca make this substitutio prior to multiplyig ad so, if we take deg(f ad deg(g to be the miimal degrees of elemets of the equivalece classes of f ad g i R[x, y, z], we see that deg(fg deg(f + deg(g i R. Now, suppose x I 2. The there exist f, g I such that Hece, f(x, y, zg(x, y, z x. 1 deg(x deg(fg deg(f + deg(g, so either deg(f 0 or deg(g 0. This, however, implies that either f or g is costat i R; sice there are o costat terms i I, this is impossible. Therefore, we coclude that x / I 2 ad so I 2 I. Now, let f I/I 2. The, sice x, y, z 1 geerate I, f(x, y, z f 1 (x, y, zx α + f 2 (x, y, zy β + f 3 (x, y, z(z 1 γ for f 1, f 2, f 3 R P ad α, β, γ N {0}. Now, (z + 1(z 1 z + 1 sice x 2 + y 2 I 2, so, i fact, z2 1 z + 1 (x2 + y 2 1 z + 1 z + 1 (x2 + y 2 I 2, f(x, y, z f 1 (x, y, zx α + f 2 (x, y, zy β i I/I 2. Now, if α > 1, the x α x α 1 x I 2

7 ALGEBRA HW 11 7 ad, similarly, if β > 1, y β I 2, so we see that α β 1. Sice f 1, f 2 R P, f 1 g 1 h 1 ad f 2 g 2 h 2 for g 1, g 2, h 1, h 2 R ad h 1 (P 0, h 2 (P 0. Hece, Let f(x, y, z g 1(x, y, z h 1 (x, y, z x + g 2(x, y, z h 2 (x, y, z y. h i (x, y, z h i (x, y, z h i (0, 0, 1. The h i (P 0 ad so h i I 2. Thus, i I/I 2, h 1 h2 0 ad f h i 0, so h 2 (x, y, zg 1 (x, y, zx + h 1 (x, y, zg 2 (x, y, zy f(x, y, z(h 1 (x, y, zh 2 (x, y, z f(x, y, z((h 1 (x, y, z h 1 (0, 0, 1 + h 1 (0, 0, 1(h 2 (x, y, z h 2 (0, 0, 1 + h 2 (0, 0, 1 f(x, y, z(( h 1 (x, y, z + h 1 (0, 0, 1( h 2 (x, y, z + h 2 (x, y, z f(x, y, z( h 1 (x, y, z h 2 (x, y, z + h 1 (x, y, zh 2 (0, 0, 1 + h 2 (x, y, zh 1 (0, 0, 1 + h 1 (0, 0, 1h 2 (0, 0, 1 f(x, y, zh 1 (0, 0, 1h 2 (0, 0, 1. Therefore, sice g 1 h 2 g 1 h 1 (0,0,1h 2 (0,0,1 ad g 2 h 1 g 2 h 1 (0,0,1h 2 (0,0,1 R, f(x, y, z g 1(x, y, zx + g 2(x, y, zy for g 1, g 2 R. Let g i (x, y, z g i(x, y, z g i(0, 0, 1. The g i (P 0, so g i I ad hece g i (x, y, zx I2. Thus, g 1(x, y, zx (g 1(x, y, z g 1(0, 0, 1+g 1(0, 0, 1x g 1 (x, y, zx+g 1(0, 0, 1x, which is equal to simply g 1 (0, 0, 1x i I/I2. A similar argumet for g 2 ad y yields the result f(x, y, z g 1(0, 0, 1x + g 2(0, 1, 0y. Therefore, sice g 1 (0, 0, 1 ad g 2 (0, 1, 0 are simply scalars, we see that I/I 2 is a 2-dimesioal vector space over R with basis {x, y}. 3 I the situatio of problem 2: (a: Let T R 3 be the taget plae to S 2 at P. Thus, T {(x, y, 1 x, y R}. Show that T is a 2-dimesioal vector space over R, uder the additio (x, y, 1 + (x, y, 1 (x + x, y + y, 1 ad scalar multiplicatio c(x, y, 1 (cx, cy, 1. What is the 0-vector of T?

8 8 CLAY SHONKWILER Proof. Let (x, y, 1, (x, y, 1 T. The ad, if c R, (x, y, 1 + (x, y, 1 (x + x, y + y, 1 T c(x, y, 1 (cx, cy, 1 T, so T is closed uder additio ad scalar multiplicatio. Also, c((x, y, 1 + (x, y, 1 c(x + x, y + y, 1 (c(x + x, c(y + y, 1 ad, for d R, (cx + cx, cy + cy, 1 c(x, y, 1 + c(x, y, 1 (c+d(x, y, 1 ((c+dx, (c+dy, 1 (cx+dx, cy+dy, 1 c(x, y, 1+d(x, y, 1, so the distributive laws hold ad so T is a vector space. Note that (0, 0, 1 is the 0 vector of T. If (x, y, 1 T, the ad, if c, d R such that the (x, y, 1 x(1, 0, 1 + y(0, 1, 1 c(1, 0, 1 + d(0, 1, 1 0, 0 c(1, 0, 1 + d(0, 1, 1 (c, d, 1, so c d 0. Therefore, we see that {(1, 0, 1, (0, 1, 1} is a basis for T, ad so we coclude that T is two-dimesioal. (b: Let f T, the dual space of T. Show that f : T R exteds to a uique liear fuctioal f : R 3 R. Let f : S 2 R be the restrictio of f to S 2. Show that f R, ad moreover f I. Proof. Let f T. The defie f : R 3 R by f (x, y, z f(x, y, 1. Let (x, y, z, (x, y, z R 3 ad let a, b R. The f (a(x, y, z + b(x, y, z f (ax + bx, ay + by, az + bz f(ax + bx, ay + by, 1 af(x, y, 1 + bf(x, y, 1 af (x, y, z + bf (x, y, z, so we see that f is liear. Furthermore, if there exists liear g : R 3 R such that g T f, the, for c R, g(0, 0, c g(c(0, 0, 1 cg(0, 0, 1 c 0 0.

9 Hece, for ay (x, y, z R 3, ALGEBRA HW 11 9 f (x, y, z g(x, y, z f(x, y, 1 g((x, y, 1 + (0, 0, z 1 f(x, y, 1 (g(x, y, 1 + g(0, 0, z 1 f(x, y, 1 f(x, y, 1 g(0, 0, z 1 g(0, 0, z 1 0; sice our choice of (x, y, z was arbitrary, we see that g f, so f is the uique liear extesio of f to R 3. Let f : S 2 R be the restrictio of f to S 2. Now, let c 1 f(1, 0, 0, c 2 f(0, 1, 0 ad c 3 f(0, 0, 1. The c 3 0 sice (0, 0, 1 is the 0 elemet of T ad, for (x, y, z R 3, f(x, y, z f((x, 0, 0 + (0, y, 0 + (0, 0, z f(x, 0, 0 + f(0, y, 0 + f(0, 0, z x f(1, 0, 0 + y f(0, 1, 0 + z f(0, 0, 1 c 1 x + c 2 y + c 3 z c 1 x + c 2 y, so f R. Furthermore, f(p f(0, 0, 1 0, so f I. (c: If f T, let φ(f I/I 2 be the image of f I uder I I/I 2. Show that φ : T I/I 2 is a isomorphism of vector spaces. Proof. Note, first, that if f T, φ(f f where f is as defied i (b above. Let e 1 (1, 0, 1 ad e 2 (0, 1, 1. The e 1, e 2 defies a basis o T. Let f 1, f 2 be the correspodig dual basis for T. Now, sice f 1 (x, y, z x exteds f 1 ad f 2 (x, y, z y exteds f 2 ad we saw i (b that such extesios are uique, we see that f 1 ad f 2 are the uique liear extesios of f 1 ad f 2 to maps R 3 R. Let f 1 ad f 2 be their restrictios to S2. The φ(f 1 f 1 x ad φ(f 2 f 2 y, the basis o I/I 2. Now, let f T. The f af 1 + bf 2 for a, b R. Furthermore, if f(x, y, z ax + by, the f(x, y, 1 ax + by af 1 (x, y, 1 + bf(x, y, 1 f(x, y, 1,

10 10 CLAY SHONKWILER so f is the uique extesio of f to a liear fuctioal o R 3. Hece, if f is the restrictio of f to S 2, the φ(af 1 +bf 2 (x, y, z φ(f(x, y, z f (x, y, z ax+by aφ(f 1 (x, y, z+bφ(f 2 (x, y, z, so we see that φ is liear o the basis elemets of T. Thus, φ is liear; sice φ maps the basis of T bijectively oto the basis of I/I 2, we see that φ is, i fact, a isomorphism. (d: Coclude that T is isomorphic to (I/I 2 via φ. Proof. Sice φ : T I/I 2 is a isomorphism, the followig sequece is exact: 0 T φ I/I 2 0. I PS10#4, we showed that iduced maps o duals preserve exact sequeces, so 0 T φ (I/I 2 0 is exact, ad so φ : (I/I 2 T is a isomorphism. Now, T is aturally isomorphic to T by the map ψ : T T defied i PS10#5, so we see that the compositio is a isomorphism. ψ φ : (I/I 2 T 4 Let V be a K-vector space ad let 0 W W W 0 be a exact sequece of K-vector spaces. Show that the iduced sequeces 0 V W V W V W 0; 0 Hom(V, W Hom(V, W Hom(V, W 0; ad 0 Hom(W, V Hom(W, V Hom(W, V 0 are also exact. Proof. Let us deote the maps as follows: (1 0 W G W F W 0 (2 0 V W G V W F V W 0 (3 0 Hom(V, W G Hom(V, W F Hom(V, W 0 G (4 0 Hom(W, V Hom(W, V Hom(W, V 0 Let e 1,..., e be a basis for V. F

11 (2: To see that (2 is exact, we eed to show: ALGEBRA HW ker G 0 im G ker F im F V W. To see that ker G 0, suppose v w ker G. The 0 G (v w v G(w, so either v 0 or G(w 0. Sice G is ijective, this implies that either v 0 or w 0. Sice 0 w 0 v 0, we see that v w ker G implies v w 0. Hece, G is ijective. To see im F V W, let k v i w i V W. The, sice F is surjective, there exist w 1,..., w k W such that F (w i w i. Therefore, F ( v i w i F (v i w i v i F (w i v i w i, so we see that F is surjective. Fially, to see that im G ker F, first let k v i w i im G. The there exist w i W such that G(w i w i for all i 1,..., k. Now, F ( v i w i F (v i w i v i F (w i v i F (G(w i v i 0 0,

12 12 CLAY SHONKWILER sice im G ker F, so we see that im G ker F. O the other had, if k v i w i ker F, the, for each i 1,..., k, v i j1 a i j e j ad 0 F ( v i w i v i F (w i a ij e j F (w i j1 ( e j a ij F (w i j1 ( e j F (a ij w i j1 ( e j F a ij w i ; j1 sice e i w is liearly idepedet of e j w for ay w, w W ad i j, we see that this implies that ( F a ij w i 0 for all j 1,...,. Therefore, sice (1 is exact, there exists w j W such that G(w j k a i j w i for each j 1,...,.

13 Now, G e j w j j1 ALGEBRA HW G (e j w j j1 e j G(w j j1 ( e j a ij w i j1 j1 j1 j1 e j a ij w i a ij e j w i a ij e j w i a ij e j w i j1 v i w i, so ker F im G. Havig proved cotaimet both ways, we coclude that ker F im G, ad so coclude that (2 is exact. (3: To see that (3 is exact, we must show that ker G 0 im G ker F im F Hom(V, W. Note that for φ Hom(V, U ad H : U U, H Hom(V, U is defied by for ay vector space U. Now, let φ ker G. The H (φ(v (H φ(v 0 G (φ(v (G φ(v G(φ(v : Hom(V, U for all v V. Sice G is ijective, this i tur meas that φ(v 0 for all v V, so φ 0. Therefore, we see that ker G 0. Now, let φ Hom(V, W. Let v V. The there exists w W such that F (w φ(v, sice F is surjective. Therefore, defie ψ : V W by ψ(v w.

14 14 CLAY SHONKWILER To see that ψ is liear, let v, v V ad a, b K. The ψ(v w ad ψ(v w for w, w W. The F (w φ(v ad F (w φ(v ad, sice F is liear, F (aw + bw af (w + bf (w. Hece, ψ(av + bv aw + bw aψ(v + bψ(v, so ψ Hom(V, W. Furthermore, for all v V, there exists w W such that F (w φ(v ad F (ψ(v (F ψ(v F (ψ(v F (w φ(v, so we see that F (ψ φ, so F is surjective. Now, to see that im G ker F, let φ im G. The there exists ψ Hom(V, W such that G (ψ φ. Hece, for v V, F (φ(v (F φ(v F (φ(v F (G (ψ(v F (G(ψ(v (F G(ψ(v 0, sice F G 0. Sice our choice of v was arbitrary, we see that F (φ 0, so im G ker F. O the other had, if φ ker F, the 0 F (φ(v (F φ(v F (φ(v for all v V. Sice (1 is exact, there exists w W such that G(w φ(v for all v V. Hece, defie ψ : V W by ψ(v w. The, if v, v V ad a, b K, the ψ(v w ad ψ(v w for some w, w W. Furthermore, G(w φ(v ad G(w φ(v ad, sice G is liear, G(aw + bw ag(w + bg(w. Hece, ψ(av + bv aw + bw aψ(v + bψ(v, so ψ Hom(V, W. Now, for v V, ψ(v w for some w W ad G (ψ(v (G ψ(v G(ψ(v G(w φ(v, so G (ψ φ ad so we see that ker F im G. Therefore, sice cotaimet has bee show both ways, we see that im G ker F ad so (3 is exact. (4: To see that (4 is exact, we must show that ker F 0 im F ker G im G Hom(W, V. Note that for φ Hom(U, V ad H : U U, H : Hom(U, V Hom(U, V is defied by H (φ(v (φ H(u for ay vector space U. Now, let φ ker F. The, for all w W, 0 F (φ(w (φ F (w φ(f (w. Sice F is surjective, this implies that φ(w 0 for all w W, so φ 0. Hece, ker F 0. Let φ Hom(W, V. π G : W W be the orthogoal projectio oto the image of G i W. Defie ψ : W V by ψ(w (φ G 1 π G (w.

15 ALGEBRA HW The ψ is well-defied, sice G 1 : im G W is well-defied (because G is ijective. Note that, if w 1, w 2 im G ad a, b K, the G(w 1 w 1 ad G(w 2 w 2 for some w 1, w 2 W ad G(aw 1 + bw 2 ag(w 1 + bg(w 2 aw 1 + bw 2, so G 1 (aw 1 + bw 2 aw 1 + bw 2. Therefore, if w, w W ad a, b K, the ψ(aw + bw (φ G 1 π G (aw + bw φ(g 1 (π G (aw + bw φ(g 1 (aπ G (w + bπ G (w φ(a(g 1 π G (w + b(g 1 π G (w a(φ G 1 π G (w + b(φ G 1 π G (w aψ(w + bψ(w sice φ ad π G are liear. Therefore, we see that ψ Hom(W, V. Furthermore, for all w W, (G ψ(w G (ψ(w (ψ G(w ψ(g(w (φ G 1 π G (G(w φ(g 1 (π G (G(w φ(g 1 (G(w φ(w, so G ψ φ ad thus G is surjective. To see that im F ker G, let φ im F. The φ F ψ for some ψ Hom(W, V. Therefore, for w W, (G φ(w (φ G(w ((F ψ(g(w (ψ F (G(w ψ(f (G(w ψ(0 0, sice F G 0. Therefore, sice our choice of w was arbitrary, we see that G φ 0, so im F ker G. O the other had, suppose φ ker G. The, for all w W, φ(g(w (φ G(w G (φ(w 0. Defie ψ : W V by ψ(w φ(w, where F (w w. Sice F is surjective, this ψ is defied for all w W. The, if w 1, w 2 W ad a, b K, F (w 1 w 1 ad F (w 2 w 2 for some w 1, w 2 W. Furthermore, so F (aw 1 + bw 2 af (w 1 + bf (w 2 aw 1 + bw 2 ψ(aw 1 + bw 2 φ(aw 1 + bw 2 aφ(w 1 + bφ(w 2 aψ(w 1 + bψ(w 2; hece, ψ Hom(W, V. Now, if w W, the ψ(f (w φ(w 0 for some w 0 W such that F (w F (w 0. Hece, 0 F (w F (w 0 F (w w 0,

16 16 CLAY SHONKWILER so w w 0 ker F im G. Hece, w w 0 G(w for some w W. Therefore, F (ψ(w (ψ F (w ψ(f (w φ(w 0 φ(w + (w 0 w φ(w + φ(w 0 w φ(w + φ(g(w φ(w, sice φ(g(w 0 for all w W. Sice our choice of w was arbitrary, we see that F ψ φ, so ker G im F. Havig show cotaimet both ways, we coclude that im F ker G ad thus that (4 is exact. 5 Call T Ed V ilpotet if T m 0 for some m > 0. Show that if T is ilpotet, the T has o o-zero eigevalues. Proof. Suppose a K is a eigevalue of T. The, for some o-zero v V, Sice T m 0, T v av. 0 T m v T m 1 (T v T m 1 av at m 1 v... a m 1 T v a m v. Sice v 0, this implies that a m 0, so a is a zero divisor. However, sice o field cotais ay o-zero zero divisors, this i tur implies that a 0. Therefore, sice our choice of eigevalue a was arbitrary, we coclude that T has o o-zero eigevalues. 6 Let V be a vector space. If T Ed V ad W V, call W a T -irreducible subspace if W is T -ivariat ad the oly T -ivariat subspaces of W are 0 ad W. (a: Suppose that V is a fiite-dimesioal C-vector space, that T Ed V ad its powers form a group of order uder compositio, ad that W is a T -ivariat subspace. Show that W has a T - ivariat complemet W. Proof. Let W be a arbitrary complemet of W. The V W W ; that is, if v V, the v ca be uiquely decomposed as v w + w for w W ad w W. Defie P : V W by P (v P (w + w w.

17 ALGEBRA HW The, if v 1, v 2 V ad a, b K, the v 1 w 1 +w 1 ad v 2 w 2 +w 2 for w 1, w 2 W ad w 1, w 2 W ad thus P (v 1 + v 2 P (a(w 1 + w 1 + b(w 2 + w 2 P ((aw 1 + bw 2 + (aw 1 + bw 2 aw 1 + bw 2 ap (v 1 + bp (v 2, so P is liear. Now, defie S : W W by S(v 1 1 T i P T i (v. i0 The, if v 1, v 2 V ad a, b K, S(av 1 + bv T i P T i (av 1 + bv 2 i0 1 1 T i P (at i (v 1 + bt i (v 2 i0 1 1 T i (ap T i (v 1 + bp T i (v 2 i0 1 1 (at i P T i (v 1 + bt i P T i (v 2 i0 1 a 1 T i P T i (v 1 + b 1 1 T i P T i (v 2 i0 as(v 1 + bs(v 2, so S is liear. Now, sice P (v W for all v V, we see that P (T i (v W for all v ad i ad, sice W is T -ivariat, T i P T i (v W for all v V ad i 1,...,. Hece, S(v W for all v V ad so im S W. O the other had, sice P is the idetity o W, we see that for all w W, so T i P T i (w T i T i (w w i0 S(w 1 1 T i P T i (w 1 1 w w i0 Therefore, W im S. Havig show cotaimet both ways, we coclude that im S W. Furthermore, sice we just showed that S W is the idetity; sice im S W, this implies that S 2 S. Thus, i0

18 18 CLAY SHONKWILER by our work i PS10#7, we kow that V W W where W ker S, so W is a complemet of W. Now, let w W. The S(T (w 1 1 T i P T i (T (w i0 1 1 T i P T (i 1 (w i0 1 1 T T i 1 P T (i 1 (w i0 ( 1 1 T T i 1 P T (i 1 (w T i0 ( 1 1 T i 1 P T (i 1 (w i0 T (S(w T (0 0, so T (w W ad, hece, W is T -ivariat. (b: Uder the hypotheses of (a, show that V ca be writte as the direct product of T -irreducible subspaces. Proof. By iductio o dim V. If dim V 0, the we re doe. Similarly, if dim V 1, the V V is already the direct product of T -irreducible subspaces. Now, suppose all vector spaces of dimesio k ca be writte as a direct product of T -irreducible subspaces. Let V be a vector space of dimesio k + 1. The, by our result i (a, we kow that V W W where W ad W are T -ivariat. Furthermore, dim W k ad dim W k so, by the iductio hypothesis, W ad W m W i W i

19 ALGEBRA HW for T -irreducible subspaces W 1,..., W, W 1,..., W m. The ( m V W i W 1... W W 1... W m, j1 W j so V is the product of T -irreducible subspaces. Havig show the base case ad the iductive step, we coclude, by iductio, that for ay fiite-dimesioal vector space V, V ca be writte as the direct product of T -irreducible subspaces. DRL 3E3A, Uiversity of Pesylvaia address: shokwil@math.upe.edu