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Let $M$ and $N$ be $A$-modules, $\operatorname{Hom}_A(M,N)$ the set of all $A$-module homomorphisms $M\rightarrow N$. $\operatorname{Hom}_A(M,N)$ can be viewed as a subset of the cartesian product $\prod_{m\in M}N$. But doesn't it require axiom of choice for $\prod_{m\in M}N$ to be a set? Is there a chance that $\operatorname{Hom}_A(M,N)$ may be a set without axiom of choice?

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up vote 13 down vote accepted

The fact that $\prod_{m\in M} N$ is a set holds regardless of the Axiom of Choice. A product of a family of sets is always a set in ZF, you don't need AC for that. What the Axiom of Choice might play a role in doing is telling you whether the product is empty or not.

However, in the case of modules, you can prove the product is not empty without the Axiom of Choice. This, because modules have a distinguished element (the zero element), so if $\{N_i\}_{i\in I}$ is a nonempty family of modules, then $\prod_{i\in I}N_i$ is necessarily nonempty: it contains the function $f\colon I\to \cup N_i$ with $f(i)=0$ for all $i$.

Moreover, even for sets, one can show that an arbitrary power of a nonempty set is nonempty without the Axiom of Choice. Meaning, if $A$ is a nonempty set and $I$ is a nonempty set, then $\prod_{i\in I}A\neq\emptyset$ holds in ZF, even without choice: since $A$ is nonempty, there exists $a\in A$; define $f\colon I\to A$ by $f(i)=a$ for all $i$, and this shows the product is nonempty.

And so $\mathrm{Hom}_A(M,N)$ is a set, regardless of anything else; and moreover, it is provably nonempty, since the function $f\colon M\to N$ given by $f(m)=0$ for all $m$ lies in the set.

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I almost got ahead of you this time :-) –  Asaf Karagila Sep 20 '11 at 13:55
    
@Asar: Had the question been posted five minutes later, you would have. I'm off to teach for two hours now... (-: –  Arturo Magidin Sep 20 '11 at 13:57
    
Shame. Oh well, enjoy teaching. :-) –  Asaf Karagila Sep 20 '11 at 13:58
    
Thanks for clarifying that. My brain was a little short of oxygen when I was writing that question. –  ashpool Sep 20 '11 at 14:03
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Assuming the set theory of ZF, it is a set.

Since $M$ is a set and $N$ is a set, we have that $f\colon M\to N$ is a subset of $\mathcal P(M\times N)$. The latter, is a set by the axiom of the power set.

Now we can define (using $A$ as a parameter, of course) all the maps which has a certain property, and by the axiom schema of replacement (or subset, if you want) it is indeed a set.

Without the axiom of choice, however, the product may be empty. The empty set, however, is still a set. Although in algebraic structures this is often not the case, since $m\mapsto 0_N$ is a homomorphism, and it certainly exists.

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