Let $A$ and $B$ be a commutative ring and $B$ an $A$-module, and suppose that $B$ is finitely generated and free as an $A$-module. Is $Hom_{A}(B,A)$ free over $A$ with the same rank as B?
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Is that hom as A-modules or A-algebras? – Daniel McLaury Sep 08 '13 at 02:31
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@DanielMcLaury $\text{Hom}{A-alg}$ isn't an A-module--adding ring maps often isn't a ring map. For a concrete example, it's a common fact that if $L/K$ is separable of degree $n$, then $\text{Hom}{K-alg}(L,\overline{K})$ is of size $n$--and so clearly not a $K$-module for any infinite field $K$. – Alex Youcis Sep 08 '13 at 03:56
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Yeah, I didn't even think about that. I was just wondering why he gave $B$ a ring structure since it doesn't seem to come up in the problem. – Daniel McLaury Sep 08 '13 at 07:34
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Yes.
Hint: What if you considered the map $\text{Hom}_A(A^n,A)\cong A^n$ by the map $f\mapsto (f(e_1),\ldots,f(e_n))$ where $\{e_1,\ldots,e_n\}$ is a basis for $A^n$.
Also useful is the fact that the same statement is true if all instances of "free" are replaced with "projective". Then you merely note that if $P\oplus C=A^n$ then by the above, and the fact that $\text{Hom}_A$ is additive, you get that $\text{Hom}_A(P,A)\oplus\text{Hom}_A(C,A)\cong A^n$. Thus, $\text{Hom}_A(P,A)$ is also f.g. projective.
Alex Youcis
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I thought that $(f_{i}){i=1,2,...,n}$ such that $f{i}(a_{1}e_{1}+\dots +a_{n}e_{n})=a_{i}$ is a basis for $Hom_{A}(A^n,A)$ but the unit element is not preserved by $f_{i}$ unless $1=e_{1}+\dots +e_{n}$. Is there always exits such a basis? – Sep 08 '13 at 09:50
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@user53216 Did you see the above discussion in the comments? Did you mean A-algebra homomorphisms? – Alex Youcis Sep 08 '13 at 09:51
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@user53216 Then it certainly means A-module homomorphisms. In that case, module maps needn't preserve unit. – Alex Youcis Sep 08 '13 at 10:00
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