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I have a question . Can anyone give me examples for $Hom(B,\oplus A_j)$ not isomorphic to $\oplus Hom(B,A_j)$ or $\prod Hom(B,A_j)$ as abelian groups? Here $A_j$ and B are both modules. I have read $Hom(B,\prod A_j)$ is isomorphic to $\prod Hom(B,A_j)$. Thanks for any hint!

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  • $\begingroup$ think about how you could encode the universal property of direct products into $hom$. And then note that direct sum/product only coincide for finitely many objects. $\endgroup$ – Daniel Valenzuela Mar 6 '15 at 21:22
  • $\begingroup$ Sorry I still don't know how to deal with it... $\endgroup$ – user146507 Mar 7 '15 at 2:47
  • $\begingroup$ There is a canonical homomorphism $\oplus Hom(B,A_j) \to Hom(B,\oplus A_j)$. Are you asking for an example where this canonical hom. isn't an isomorphism or are you asking for an example where no isomorphism $\oplus Hom(B,A_j) \cong Hom(B,\oplus A_j)$ exists at all ? $\endgroup$ – tj_ Mar 8 '15 at 0:32
  • $\begingroup$ I am looking for an example where no isomorphism exists. Thanks! $\endgroup$ – user146507 Mar 8 '15 at 1:44
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There is one example in Rotman's book.

Let $p$ be a prime and $A_n$ a cyclic group of order $p^n$, let $$ B=\oplus_{n=1}^{\infty}A_n. $$ Then $\text{Hom}(B,\oplus_{n=1}^{\infty}A_n)$ is not isomorphic to $\oplus_{n=1}^{\infty}\text{Hom}(B,A_n)$. (There is an element of infinite order in $\text{Hom}(B,B)$, but every element in $\oplus_{n=1}^{\infty}\text{Hom}(B,A_n)$ has finite order.)

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  • $\begingroup$ Would you mind explaining why Hom(B,B) is infinite and $\oplus_{n=1}^{\infty} Hom(B,A_n)$ is finite? Thanks! $\endgroup$ – user146507 Mar 18 '15 at 3:06
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    $\begingroup$ @user12345 : That's not what he said. He said there is an element of infinite order in one group, and such element does not exist in the other. You should try to find that element! $\endgroup$ – Patrick Da Silva Mar 18 '15 at 23:06
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To be able to state generally that $\mathrm{Hom}(M, \oplus N_i) \simeq \oplus \mathrm{Hom}(M, N_i)$, you want the canonical homomorphism $\oplus \mathrm{Hom}(M,N_i) \to \mathrm{Hom}(M,\oplus N_i)$ (which essentially just sums your maps) to be an isomorphism, otherwise any map you can think of which is an isomorphism would rely on the context and is not general. The canonical map above is not always an isomorphism ; consider for instance a module $M$ which possesses a strictly increasing chain of proper submodules, say $$ M_0 \subsetneq M_1 \subsetneq \cdots \subsetneq M_n \subsetneq \cdots \subseteq M. $$ Then the map $\oplus \mathrm{Hom}(M, M/M_i) \to \mathrm{Hom}(M, \oplus M/M_i)$ is not surjective since there is a canonical morphism $M \to \oplus M/M_i$ (sending $a$ to $(a+M_0) \oplus (a+M_1) \oplus \cdots \oplus (a+M_n) \oplus \cdots$) which is not the direct sum of finitely many maps $M \to M/M_i$. Of course, this doesn't prove that $\oplus \mathrm{Hom}(M,M/M_i)$ is not isomorphic to $\mathrm{Hom}(M,\oplus M/M_i)$, but an isomorphism between them has to be "unnatural".

Hope that helps,

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  • $\begingroup$ I guess I want to find a map (which may not be general) such that $Hom(M,\oplus N_i)$ is not homeomorphic to $\oplus Hom(M,N_i)$. But your argument about not surjectivity makes sense to me. Thanks! $\endgroup$ – user146507 Mar 18 '15 at 3:14
  • $\begingroup$ Can you re-read your comment? It just sounds weird. I'd be happy to answer then. (A map which is not homeomorphic? We have no topologies here, and even if I wrote isomorphic, it still sounds weird.) $\endgroup$ – Patrick Da Silva Mar 18 '15 at 4:41
  • $\begingroup$ Oh, I am sorry. It should be isomorphic. I am sorry for the confusion. Here I mean I want to find an example of M and $N_i$ such that $ Hom(M,\oplus N_i)$ is not isomorphic to $\oplus Hom(M, N_i)$ in whatever map, including the canonical homomorphism. It might be some very weird example and cannot be applied to general cases. Thanks for your comments and help:) $\endgroup$ – user146507 Mar 18 '15 at 6:10

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