Math 101 Fall 2013 Homework #7 Due Friday, November 15, 2013

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1 Math 101 Fall 2013 Homework #7 Due Frday, November 15, Let R be a untal subrng of E. Show that E R R s somorphc to E. ANS: The map (s,r) sr s a R-balanced map of E R to E. Hence there s a group homomorphsm φ : E R R E satsfyng s r sr. Snce these rngs are untal, φ s clearly surjectve. But f t = n s r s n the kernel of φ, then n s r = 0. But then t = s r = s r r 1 = ( n ) s r 1 = 0. Hence φ s a group somorphsm. It then follows that the two are somorphc as left E-modules, rght R-modules or even (E, R)-bmodules. Remark: If E s commutatve or f R s n the center of E, then both E and R are R-algebras. In partcular, E R R has a rng structure and φ s easly seen to be a rng somorphsm. 2. Show that Q Z Q and Q Q Q are somorphc. (Show both are vector spaces over Q of dmenson one.) ANS: Both are left Q-modules and hence vector spaces over Q. Hence t suffces to see that they have the same dmenson. But Q Q Q = Q, so t clearly has dmenson 1. But f a b c d Q ZQ, then a b c d bd( = a b c ) = 1 ac (a c) = d bd bd (1 1). Hence {1 1} s a spannng set for Q Z Q. Ths suffces. 3. Show that as left R-modules, C R C and C C C are not somorphc. ANS: C C C = C, whch s a two-dmensonal real vector space. On the other hand, as a real vector space, C = R 2. Hence C R C = R 2 R R 2 whch s somorphc to R 4, whch s a 4-dmensonal real vector space. 4. Recall that for R-modules, we wrte I M n place of the coproduct I M. Then show that tensor products commute wth drect sums. That s, show that N R M = N R M, I I 1

2 and that an somorphsm s gven by n (m ) (n m ). (I suggest usng the unversal property of the tensor product. What assumptons are you makng about R, N and the M?) ANS: To begn wth, we assume that N s a rght N-module and that each M s a left R-module. Let : N M N R M be the unversal R-balanced map (n,(m )) = n (m ). I clam j : N M N RM s also unversal, where j(n,(m )) = (n m ). Let f : N M A be an R-balanced map nto an abelan group A. Let k : M k M be the natural njecton. Then j k s an R-balanced map of N M k nto A. Hence there s a unque group homomorphsm j k : N M k A such that N M k 1 k N M f N R M k commutes. Then we can defne f : N RM A by f((n m )) = j (n m ). (Ths s well defned only because only fntely many terms are nonzero.) But snce f s R-balanced, and hence addtve n ts second varable, and snce (m ) = m ǫ, we see that f(n,(m )) = f(n,m ǫ ) = j (n,m ). (Wth the last equalty due to the commutatvty of the above dagram. That s, the dagram N R M j f N M A f commutes. Snce both and j are unversal, the result follows easly: we get unque group homomorphsms ī and j such that N R M j N M ī j A j k N R M commutes. Unqueness forces ī and j to be nverses of one another. In partcular, ī(n (m )) = j(n,(m )) = (n m ) as requred. Note that f N s a (E,R)-bmodule, then ī s easly seen to be an E-module map. Or f each M s a (R,P)-bmodule, then ī s a rght P-module map. 2

3 5. LetAbeafnteabelangroupoforderp α mwthp m. ProvethatZ p α Z Assomorphc to the p-sylow subgroup of A. ANS: Snce all the Sylow subgroups of an abelan group are normal and have trval ntersecton wth one another, A s the nternal drect sum A(p ) where A(p ) s the p -Sylow subgroup. We can assume p 1 = p. Recall that Z n Z Z m = {0} f (n,m) = 1. Hence Z p α Z A = Z p α A(p). On the other hand, A(p) = k Z p n k wth n k = α. In partcular, p α p n k for all k. But then Z p α Z Zp n k = Z p n k = A(p). 6. Recall that f S s a multplcatve subset of a commutatve rng R and M s an R- module, thenwecanformthefractonmodules 1 M. Becausebothsharethesameunversal property, we also observed that m s 1 1 m nduces an somorphsm of S 1 M onto S 1 R R M. Except for part (a), we ll take R = Z and S 1 R = Q n ths problem. But you mght want to thnk about generalzatons of parts (b) and (c). (a) Show that 1 s m s zero n S 1 M R M f and only f there s a s S such that s m = 0. ( Use the somorphsm Luke. ) (b) Let A be an abelan group. Show that Q Z A = {0} f and only f A s torson. (c) Recall that f φ : M M s an R-module map, then we get a homomorphsm 1 φ : N R M N R M for any rght R-module N characterzed by φ(n m ) = n φ(m ). Show that f 1 A B j C 1 s a short exact sequence of abelan groups, then 1 Q Z A 1 Q Z B 1 j Q Z C 1 s a short exact sequence of vector spaces. (One says that Q Z don t forget...) s exact. And Luke, (d) Is t always true that f M s a submodule of M then N R M s a submodule of N R M. (That s, f ι : M M s the ncluson, s 1 ι necessarly njectve?) ANS: (a) In vew of the somorphsm above, 1 s m = 0 f and only f s m = 0 n S 1 M. But that means s m = 0 1. But that happens only when there s a s S such that s m = 0. (b) If a A s not a torson element, then n a 0 for all n Z \ {0}. Hence 1 a 0 by part (a). Hence f A s not all torson, Q Z A {0}. On the other hand, f A s torson, then gven a A, there s a n Z such that n a = 0. Then q a = 1 n q n a = 0. That s Q Z A = {0}. 3

4 (c) Note that f t = q a Q A, then there are ntegers m and n such that t = k m n a = k 1 n m a = 1 ( n m a ). But ths just says every element n Q Z A can be wrtten n the form 1 s a for some nteger s and some a A. But f 1 s a ker1, then 1 s (a) = 0 n Q Z B. But then part (a) mples that (a) = 0. But njectve forces a = 0. Thus 1 s njectve. Snce 1 j s clearly surjectve, we only have to worry about exactness n the mddle. Snce (1 j) (1 ) s clearly zero, t s enough to see that the kernel of 1 j s n the mage of 1. But f t ker1 j, then by the above we can assume t = 1 s b wth j(b) = 0. But the there s an a such that (a) = b, and (1 )( 1 s a) = 1 s b. (d) See Lecture notes. 7. In queston 4, we observed the tensor products commute wth drect sums. Do they commute wth drect products? Let P be the set of prmes n Z and consder M = p P Z p and our old frend Q Z. ANS: We just need to observe that Q Z M {0}. Note that the element x := ( 1, 1, 1,...) has nfnte order n M. Hence t generates a subgroup x somorphc to Z. Snce, as we now say, Q s flat, we have an njecton of Q = Q Z Z nto Q Z M. In partcular, M {0}. 8. If R s a untal subrng of E, then we formally have two meanngs for E R M for an R-module M: we frst extended the scalars from R to E, but we could also consder E as a (E, R)-bmodule and form the (general) tensor product. Explan why these are the same thng. ANS: Orgnally, we defned E R M so that ī : M E R M was unversal as a R-module map of M nto an E-module L. Let s show that the full-fledged tensor product E R M has the rght unversal property. But f f : M L s an R-module map nto an E-module L, then g(s,m) = sf(m) s an R-balanced map nto L. Hence there s a E-module map ḡ : E R M to L such that ḡ(s m) = sf(m). But then f(m) := ḡ(1,m) does the old job. Hence unversal nonsense says they are the same. 9. Let F(S) be a free R-module wth bass S and let M be an R-module. Show that every element t of F(S) R M has a unque representaton n the form t = s S s m s, ( ) 4

5 where only fntely many m s are nonzero. In partcular, f t = 0, then all the m s are zero. (If we specalze to the case where S s fnte and F(S) = R n, every element of R n R M has a unque representatve of the form e m where e s the usual bass vector and m M. We also note that t s crtcal that S be a bass. The set S = {s 1,s 2 } = {(2,0),(0,2)} s Z-lnearly ndependent n Z 2, but s 1 1+s 2 1 = 0 n Z 2 Z Z 2.) ANS: Note that every element x n F(S) can be wrtten unquely as s S s r s where all by fntely many r s are zero. Thus x m = s r s m. It now follows that every element t can be wrtten n the form ( ). The map f s0 : F(S) R gven by sendng s S s r s to r s0 s an R-module map. Smlary the map (x,m) f s0 (x) m s an R-balanced map from F(S) M to M. Hence we get a map S s0 : F(S) R M M that takes t above to m s0. Ths mples unqueness. Thus f t = 0, then m s = 0 for all s. 5