ALGEBRA MID-TERM. 1 Suppose I is a principal ideal of the integral domain R. Prove that the R-module I R I has no non-zero torsion elements.
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1 ALGEBRA MID-TERM CLAY SHONKWILER 1 Suppose I s a prncpal deal of the ntegral doman R. Prove that the R-module I R I has no non-zero torson elements. Proof. Note, frst, that f I R I has no non-zero torson smple tensors, then t wll have no non-zero torson elements. Defne φ : I I I, gven by (m, n) mn. Ths s certanly well-defned. Furthermore, f m, n, m 1, m 2, n 1, n 2 I and r R, then and φ(m 1 + m 2, n) = (m 1 + m 2 )n = m 1 n + m 2 n = φ(m 1, n) + φ(m 2, n), φ(m, n 1 + n 2 ) = m(n 1 + n 2 ) = mn 1 + mn 2 = φ(m, n 1 ) + φ(m, n 2 ) φ(mr, n) = mrn = φ(m, rn), so φ s a blnear map. Hence, by the unversal property of the tensor product, φ nduces a homomorphsm Φ : I R I I such that m n = mn. Now, suppose there exsts m n I R I and non-zero r R such that r(m n) = 0. Now, snce m, n I and I s prncpal, then m = r 1 a and n = r 2 a for some r 1, r 2 R, where a s the generator of the prncpal deal I. If a = 0, then ths proposton s trvally true, snce ths would mply that I s the zero deal and so I I s just the zero module. Therefore, we can assume a 0. Snce Φ s a homomorphsm, ths mples 0 = Φ(r(m n)) = Φ(r(r 1 a r 2 a)) = r(r 1 r 2 a 2 ). Snce R s an ntegral doman, ths mples that ether r 1 = 0 or r 2 = 0. Hence, m n = r 1 a r 2 a = 0 r 2 a = 0 or m n = r 1 a r 2 a = r 1 a 0 = 0. Snce our choce of the smple tensor m n was arbtrary, we see that I R I contans no non-zero torson elements. 1
2 2 CLAY SHONKWILER Prove that Z[] Z R C as rngs. 2 Proof. We can thnk of elements of Z[] as beng of the form a + b for some a, b Z. Now, defne a map φ : Z[] R C gven by (a + b, r) r(a + b). Then, for a + b, c + d Z[], r, r 1, r 2 R and n Z, φ(a+b+c+d, r) = r(a+b+c+d) = r(a+b)+r(c+d) = φ(a+b, r)+φ(c+d, r), φ(a+b, r 1 +r 2 ) = (r 1 +r 2 )(a+b) = r 1 (a+b)+r 2 (a+b) = φ(a+b, r 1 )+φ(a+b, r 2 ) and φ((a + b)n, r) = r((a + b)n) = nr(a + b) = φ(a + b, nr), so φ s a blnear map and, hence, nduces a Z-module homomorphsm Φ : Z[] Z R C. Now, we defne the map Ψ : C Z[] Z R gven by z + w 1 z + w and we want to show that Φ and Ψ are nverses. Certanly f z 1 + w 1, z 2 + w 2 C and n Z, then Ψ(n(z 1 + w 1 ) + z 2 + w 2 ) = Ψ((nz 1 + z 2 ) + (nw 1 + w 2 )) = 1 (nz 1 + z 2 ) + (nw 1 + w 2 ) = n(1 z 1 + w 1 ) + 1 z 2 + w 2 = nφ(z 1 + w 1 ) + φ(z 2 + w 2 ) so Ψ s a Z-module homomorphsm. Now, f (a + b) r Z[] Z R, then (Ψ Φ)((a + b) r) = Ψ(Φ((a + b) r) = Ψ(r(a + b)) = 1 ra + rb = a r + b r = (a + b) r. Smlarly, f z + w C, then (Φ Ψ)(z + w) = Φ(1 z + w) = Φ(1 z) + Φ( w) = z + w. Hence, we see that Φ and Ψ are nverses, so Φ : Z[] Z R C s a Z-module somorphsm, whch s to say an somorphsm of abelan groups. To see that t s, n fact, an somorphsm of rngs, we need only show that Φ respects the multplcatve structure. To see ths, let (a+b) r 1, (c+d) r 2 Z[] Z R. Then Φ([(a + b) r 1 ][(c + d) r 2 ]) = Φ([(ac bd) + (ad + bc)] r 1 r 2 ) = r 1 r 2 ((ac bd) + (ad + bc)) = r 1 r 2 (a + b)(c + d) = Φ((a + b) r 1 )Φ((c + d) r 2 ). Therefore, we conclude that Z[] Z R C as rngs.
3 Prove that C R C C C as rngs. ALGEBRA MID-TERM 3 3 Proof. Let φ : C C C C be the dentty map. Then φ s certanly R-blnear, so, by the unversal property of the tensor product, t nduces an R-module homomorphsm Φ : C R C C C. If we defne the map Ψ : C C C R C gven by (z, w) z w. Then we want to show that Φ Ψ = Id and Ψ Φ = Id, whch wll mean that Ψ s the nverse of Φ and so Φ s an somorphsm. If z w C R C, then (Ψ Φ)(z w) = Ψ(Φ(z w)) = Ψ(z, w) = z w and f (z, w) C C, then (Φ Ψ)(z, w) = Φ(Ψ(z, w)) = Φ(z w) = (z, w). Hence, we see that the R-module homomorphsm Φ s nvertble and so s an somorphsm. To see that t s a rng somorphsm, we must show that t s a rng homomorphsm and the the bjectvty wll follow from what we ve just shown. To that end, let z 1 w 1, z 2 w 2 C R C. Then Φ((z 1 w 1 )(z 2 w 2 )) = Φ(z 1 z 2 w 1 w 2 ) = (z 1 z 2, w 1 w 2 ) = (z 1, w 1 )(z 2, w 2 ) = Φ(z 1 w 1 )Φ(z 2 w 2 ). Hence, we conclude that C R C C C. Calculate the character table of S 3 S 3. Answer: See attached sheet. 4 If ρ s the character of V, show that Ind G H(ρ)(g) = 5 k ρ( gg ) where g 1,..., g k are representatves of the left cosets of H n G, and ρ( gg ) s defned to be 0 f gg s not n H. Proof. Let ψ be the representaton of H wth character ρ. We know that G = k g H.
4 4 CLAY SHONKWILER Hence, Therefore, [ k ] k C[G] = C g H = g C[H]. ( k ) Ind G H(V ) = C[G] C[H] V = g C[H] C[H] V = k (g C[H] C[H] V ) = k (g C[H] V ). Now, for each g G, gg = g j h for a sngle j = 1,..., k and some h H. Ths mples that gj 1 gg H for ths value of j and for no other. Snce gg = g j h, Ind G H(V )(g)(g C[H] V ) = gg C[H] V = g j h C[H] V = g j C[H] ψ(h)v = g j C[H] V. Hence, f j, whch s the same thng as sayng that g 1 gg / H, then there are no egenvalues of Ind H G (V )(g) n g C[H] V, meanng ths pece contrbutes nothng to Ind G H (ρ). On the other hand, f = j, then we see that and so gg H, Ind G H(V )(g)(g C[H] V ) = gg C[H] V = g gg C[H] V = g C[H] ψ( gg )V. Hence, any egenvalues n g V of Ind G H (V ) are precsely those gven by the acton of gg on V. Therefore, we see that whch we can rewrte as Ind G H(ρ) = gg H ρ( gg ), Ind G H(ρ) = k ρ( gg ) where we follow the conventon that ρ(g 1 gg ) = 0 f gg / H.
5 ALGEBRA MID-TERM 5 6 Let H G as above, and let ρ be a character of H, and χ a character of G. Prove the followng formula, called the Frobenus recprocty formula: χ H, ρ H = χ, Ind G H(ρ) G where χ H denotes the restrcton of χ to H. Proof. Recall, frst of all, that Ind G H(ρ)(g) = k ρ( gg ) as we showed n problem 5 above. Thus, χ, Ind G H(ρ) G = 1 χ(g)ind G H (ρ)(g) = 1 k χ(g) ρ( gg ). Now, snce χ s a class functon, we can re-arrange the summatons lke so: χ, Ind G H(ρ) G = 1 k χ(g 1 gg )ρ( gg ). Snce ρ( gg ) 0 only when g 1 gg H, ths s the same as χ, Ind G H(ρ) G = 1 χ(g 1 gg )ρ( gg ). gg H Now, for each g, there are exactly H elements of g H, so there are H elements g G such that gg g H. Furthermore, f g g, then gg g g, whch we can see smply by multplyng both sdes of ths expresson by. Hence, f we denote g g = g h then the h run over all of H. Hence, for each h H, we see that h = g 1 gg exactly once for each g, so Therefore, 1 gg H χ, Ind G H (ρ) G = 1 = 1 χ(g 1 gg )ρ( gg ) = k χ(h)ρ(h). h H = k = 1 H gvng us the desred result. gg H χ(g 1 gg )ρ( gg ) k h H χ(h)ρ(h) h H χ(h)ρ(h) h H χ(h)ρ(h) = χ H, ρ H,
6 6 CLAY SHONKWILER 7 Suppose that G = H K. Let V be a representaton of H. Prove that Ind G H (V ) = V R K, where R K denotes the regular representaton of K. Proof. Frst, note that, by hypothess, C[G] = C[H K]. Now, we showed n class (February 11) that F [G 1 G 2 ] F [G 1 ] F F [G 2 ] for any feld F and groups G 1 and G 2. Hence, C[G] = C[H K] C[H] C C[K] whch s, of course, somorphc to C[K] C C[H]. Therefore, we see that Ind G H(V ) = C[G] C[H] V (C[K] C C[H]) C[H] V. Snce C[H] s a (C, C[H])-bmodule, we know that ths expresson s assocatve; that s (C[K] C C[H]) C[H] V C[K] C (C[H] C[H] V ). In turn, t s certanly true that C[H] C[H] V V, so we see that, when we combne all of the above Ind G H(V ) = C[G] C[H] C[K] C (C[H] C[H] V ) C[K] C V. Snce C[K] s R K, we conclude that Ind G H(V ) R K C V V C R K. 8 (a) Let F p denote the fnte feld wth p elements. Suppose that [K : F p ] = n, prove that K has p n elements. Proof. Let a 1,..., a n K be such that K = F p (a 1,..., a n ) and {a 1,..., a n } forms a bass for K over F p. Then each element β K can be wrtten as β = β 1 a β n a n where β k F p. Snce there are p possble choces for each β k, we see that K can contan at most p n dstnct elements. Now, suppose there are fewer than p n dstnct elements n K. Ths means that, for some β 1,..., β n, γ 1,..., γ n F p, β 1 a β n a n = γ 1 a γ n a n, or (β 1 γ 1 )a (β n γ n )a n = 0. Now, snce the a k form a bass for K as a vector space over F p, ths mples that β k γ k = 0
7 ALGEBRA MID-TERM 7 for all k = 1,..., n. Hence, β k = γ k and we see that dstnct lnear combnatons of the bass elements correspond to dstnct elements of K, meanng that K contans at least p n elements. Snce we ve shown that K p n and K p n, we conclude that K contans exactly p n elements. (b) Prove that f(x) = x 5 ax 1 Z[x] s rreducble unless a = 0, 2, 1. Proof. If f s not rreducble, then there are two cases to consder: t has a lnear factor, or t s the product of an rreducble cubc polynomal wth an rreducble quadratc. If f has a lnear factor, then ths mples that f has a root r Z,.e., r 5 ar 1 = 0. Ths mples that 1 = r 5 ar = r(r 4 a), so t must be true that r dvdes 1; that s, r = ±1. If r = 1 s a root of f, then 0 = f(r) = f(1) = 1 5 a(1) 1 = 1 a 1 = a, so t must be the case that a = 0. On the other hand, f r = 1 s a root of f, then 0 = f(r) = f( 1) = ( 1) 5 a( 1) 1 = 1 + a 1 = a 2, so t must be the case that a = 2. Snce these are the only possble nteger roots of f, we conclude that f s rreducble or only reduces nto factors wth degree greater than 1 unless a = 0, 2. Now, suppose x 5 ax 1 = f(x) = (hx 2 + bx + c)(jx 3 + dx 2 + ex + g). Then ether h = j = 1 or h = j = 1; wthout loss of generalty, we may assume that h = j = 1, so x 5 ax 1 = (x 2 + bx + c)(x 3 + dx 2 + ex + g) = x 5 + (b + d)x 4 + (c + bd + e)x 3 + (cd + be + g)x 2 + (ce + bg)x + cg. Ths gves us the followng system of equatons: b + d = 0 c + bd + e = 0 cd + be + g = 0 ce + bg = a cg = 1. Hence, we can conclude mmedately that d = b and c = ±1. Now, suppose c = 1. Then g = 1 and the system above reduces to: 1 b 2 + e = 0 b + be 1 = 0 e b = a.
8 8 CLAY SHONKWILER Hence, by the second lne, b(e 1) = 1 b = 1 e 1. Snce b must be an nteger, ths mples e = 0, 2. If e = 2, then b = 1 and so, by the frst equaton, 0 = 1 b 2 + e = = 2 whch s certanly not true. If e = 0, then b = 1 and so, by the thrd equaton n our revsed system, a = 1. On the other hand, suppose c = 1. Then g = 1 and the orgnal system of equatons reduces to: Hence, by the second lne, 1 b 2 + e = 0 b + be + 1 = 0 b e = a b(e + 1) = 1 b = 1 e + 1. Snce b s an nteger, ths mples that b = ±1, whch means that b 2 = 1. By the frst equaton, then 0 = 1 b 2 + e = e = e 2 e = 2. However, f e = 2, then b = 1 e + 1 = = 1 / Z. 3 Hence, we conclude that c 1. Therefore, we see that f can only be factored nto the product of a cubc and a quadratc f a = 1. Combned wth our other two possble values of a that make f reducble, 0 and 2, we conclude that f s rreducble unless a = 0, 2, 1. DRL 3E3A, Unversty of Pennsylvana E-mal address: shonkwl@math.upenn.edu
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