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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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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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    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.2012-11-26
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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.