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Let $R$ be a ring, and $M,N$ be left $R$-modules. Then is it not true that $\operatorname{Hom}_R(M,N)$ has the structure of an $R$-module?

I was reading the preface of the Homological Algebra book by Rotman and was quite surprised to learn that this is not the case. I think all the axioms for being a module are satisfied by $\operatorname{Hom}_R(M,N)$, but Rotman is very unlikely to make a mistake. What is it that I am missing?

Under what circumstances is this true?

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    $\begingroup$ You say «you think the axioms for being a module are satisfied», but have you even tried to define an action of $R$ on the $\hom$? $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Mar 18, 2015 at 20:45
  • $\begingroup$ Inother words, do you have any candidate for a function $R\times\hom_R(M,N)\to\hom_R(M,N)$? $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Mar 18, 2015 at 20:46
  • $\begingroup$ @MarianoSuárez-Alvarez I should have mentioned my attempt at the definition of multiplication in the question, I apologize. I asked this long back, but I guess what I had in mind could have been $(r f)(m)=r \times f(m)$. I will try to be less lazy in future and to write down details of my thought process as well. $\endgroup$
    – user90041
    Commented Mar 19, 2015 at 16:11
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    $\begingroup$ The point of my question was not to complain about your laziness, if there was any, but to suggest that you actually try to define an action, so that you see what goes wrong. For example, that action you describe in your comment does not work: in general, the map $rf$ that you defined is not a homomorphism. $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Mar 19, 2015 at 18:37

4 Answers 4

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This is true if $R$ is commutative.

Otherwise, say that you are dealing with left $R$-modules, for instance. If you attempt to define multiplication by $r$ by $(rh)(m) = rh(m)$ for any homomorphism $h \colon M \to N$, then you run into the problem that the mapping $rh$ may not be $R$-linear.

For example, let $h \colon R \to R$ be the identity map. Then $k(s) = (rh)(s)$ would be $rs$. But in general, $k(s1) = rs \neq sr = sk(1)$.

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  • $\begingroup$ However, this does not show that there cannot be any $R$-module structure on $\operatorname{Hom}_R(M, N)$. $\endgroup$
    – Bubaya
    Commented May 9, 2019 at 19:48
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    $\begingroup$ Of course there is a structure of r module: the trivial one, or any other one on the underlying set. You probably want to impose some compatibility condition, and the most natural one I see is that the map $Hom_R(M,N) \otimes M \to N$ of evaluation should be a map of R modules. This mean exaclty that the action should be $(rh)(m) = r(h(m))$. Of course then it depends on what compatibility you search for. $\endgroup$
    – Andrea Marino
    Commented Dec 14, 2019 at 20:36
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Presumably, you'll want to define $\psi=r.\phi$ by $$\psi(m)=r.\phi(m)$$ (where $r\in R, \phi\in\mathrm{Hom}_R(M,N)$ and $m\in M$.) However, this map, which is a morphism of abelian groups, need not be $R$-linear when $R$ isn't commutative : $R$-linearity would imply that for all $r'\in R$ (and all $m\in M$) $\psi(r'.m)=r'.\psi(m)$, i.e., by $R$-linearity of $\phi$, $$rr'.\phi(m)=r'r.\phi(m)$$ Since $rr'\neq r'r$ in general, there is no reason the above identity should hold.

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Any natural definition of $R$-action works only, when $R$ is commutative. For example, if you try to define $$ (rf)(m)=r(f(m)) $$ for all $r\in R$, $f\in Hom_R(M,N)$, $m\in M$, then the mapping $rf$ fails to be homomorphism of $R$-modules in general. If $s\in R$ is such that $sr\neq rs$, then $$ (rf)(sm)=r(f(sm))=r(sf(m))=(rs)(f(m))\neq s((rf)(m)) $$ in general.

OTOH, if one of the modules, $M$ or $N$, is an $(R,R)$-bimodule, then you do get a module structure.

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    $\begingroup$ An important case of a bimodule is $N=R$. $\endgroup$
    – Jyrki Lahtonen
    Commented Jan 14, 2014 at 10:07
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What action of the ring $R$ do you expect on $\operatorname{Hom}_R(M, N)$? The left action of $R$ on $M$ and $N$ is "used up" when you consider $R$-linear homomorphisms $$ \operatorname{Hom}_R(M, N) \subseteq \operatorname{Hom}_{\Bbb{Z}}(M, N). $$

If $M$ is in fact an $(R, R)$-bimodule (this is automatic if $R$ is commutative), then the right action of $R$ on $M$ yields a left action of $R$ on $\operatorname{Hom}_R(M, N)$ via $$ (rf)(m) = f(mr). $$

On the other hand, if $N$ is an $(R, R)$-bimodule, then the right action of $R$ on $N$ yields a right action of $R$ on $\operatorname{Hom}_R(M, N)$ via $$ (fr)(m) = f(m)r. $$

Of course, if both $M$ and $N$ are bimodules, then $\operatorname{Hom}_R(M, N) $ is a bimodule, as well.

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