# Exact sequences and Hom

I have a exact sequence $$1 \longrightarrow A \overset{\phi}\longrightarrow B \overset{\psi}\longrightarrow C \longrightarrow 1$$ of commutative rings $A,B,C$ and the the exact sequence is such that we only consider the multiplicative structure then if I take a free abelian group $X$, how would I form $$0 \longrightarrow Hom(X,A) \overset{\phi'}\longrightarrow Hom(X,B) \overset{\psi'}\longrightarrow Hom(X,C) \longrightarrow 0$$ since in the second sequence the groups are under addition. I have seen this done many times when both sequences are of additive groups and you just define $\phi'\circ \alpha=\phi \circ \alpha$ and this works as both $\alpha,\phi$ are additive. But I cant see how to do this when $\phi$ is multiplicative and $\phi'$ has to be additive.

Or more generaly if $A,B,C,X$ where all $G$-modules for some finite group $G$

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## 2 Answers

If I understand your question correctly, then the answer is: it doesn't matter. These are groups, not rings: multiplication and addition mean the same thing, i.e. an application of the group operation. And everything's abelian, so talking about addition is fine.

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What if $A,B,C$ where rings but in the first exact sequence we had them as groups under multiplication? Ill fix the question as this is what I mean. Sorry –  Chris Birkbeck Nov 26 '12 at 17:43
A ring is a group under multiplication if and only if it is the zero ring. You will have to be much more precise about what you're doing here. –  Zhen Lin Nov 26 '12 at 18:52

Another thing to add to Clive's answer is that, in case $X$ is indeed a free Abelian group on $\kappa$ generators, then we have $\hom(X,A)\cong \oplus^\kappa A$. That may make your second exact sequence clearer.

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