# Expressing, in terms of $I$ and $M$, the $R$-modules $\mathrm{Hom}_R(R/I,M)$, $\mathrm{Hom}_R(M,R/I)$, $\mathrm{Hom}_R(I,M)$, $\mathrm{Hom}_R(M,I)$

Let $R$ be a commutative unital ring, $I$ an ideal of $R$, and $M$ a $R$-module. It is known that $R/I \otimes_R M \cong M/IM$. Also, $\mathrm{Hom}_R(R,M)\cong M$.

Is there some similar formula for the $R$-module $$I\otimes_RM,$$ perhaps $I\otimes_RM\cong IM$? Are there any formulas that express in terms of $I$ and $M$ the $R$-modules $$\mathrm{Hom}_R(R/I,M),\;\;\; \mathrm{Hom}_R(M,R/I),\;\;\; \mathrm{Hom}_R(I,M),\;\;\; \mathrm{Hom}_R(M,I)?$$

How about the special case $\mathrm{Hom}_R(R/I,R/J)$? If not, how does one 'compute' such modules?

For example, can the modules $\mathrm{Hom}_\mathbb{Z}(\mathbb{Z}_m,\mathbb{Z}_n)$, $\mathrm{Hom}_\mathbb{Z}(\mathbb{Z}_m,\mathbb{Z})$, $\mathrm{Hom}_\mathbb{K[x]}(K[x]/(x^m),K[x]/(x^n))$ be expressed more nicely (Eisenbud, Commutative Algebra, p. 79, exc. 2.4)? Let me guess, the first one is $0$ if $m\!\neq\!n$, and $\mathbb{Z}_m$ if $m\!=\!n$; the second one is $0$. But I'd like to have a more general formula.

Update: $Hom(M,A/B) \cong Hom(M,A)/Hom(M,B)$?

Update: $Hom(M/A,N/B) \ncong \{f\in Hom(M,N); f(A)\subseteq B\}$ in general. For example, taking $A=M$ and $B=N$, the l.h.s. is $0$ and the r.h.s. is $Hom(M,N)$. The reason is that if $f(A)\subseteq B$ and $g(A)\subseteq B$ and $f|_A\neq g|_A$, then $f=g$ in the l.h.s. and $f\neq g$ in the r.h.s..

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It's unlikely you'll succeed in finding a nice formula for $\operatorname{Hom}_R(M, R/I)$. As seen in YACP's answer, $\operatorname{Hom}_R(R/I, M)$ is in natural bijection with the $I$-torsion elements of $M$. In other words, maps out of $R/I$ are easy to construct. This is usually an indication that, without some very specific restrictions on the objects in consideration, maps into $R/I$ are harder to construct. This is also precisely the reason that results like Serre duality in algebraic geometry are considered deep and powerful. – chthonian Jan 2 '13 at 6:52

We have $\operatorname{Hom}(R/I,M)\simeq (0:_MI)$ and this implies $\operatorname{Hom}(R/I,R/J)\simeq (J:I)/J$.

As a consequence we get $$\operatorname{Hom}(\mathbb Z_m,\mathbb Z_n)\simeq\mathbb Z_d,$$ where $d=\operatorname{gcd}(m,n)$

and $$\operatorname{Hom}_{K[X]}(K[X]/(X^m),K[X]/(X^n))\simeq (X^n):(X^m)/(X^n).$$ We have two cases: $m\ge n$, and in this case $(X^n):(X^m)/(X^n)=K[X]/(X^n)$ or $m<n$ and now we get $(X^n):(X^m)/(X^n)\simeq K[X]/(X^m)$.

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@LeonLampret I've edited my answer. – user26857 Jan 1 '13 at 22:30
Thanks! I'll accept this for now, but I'm still very much interested in general formulas regarding the matter. Is $I\otimes_R M \cong IM$? – Leon Jan 1 '13 at 22:46
This happens if $M$ is $R$-flat. If not, then there are counter-examples even on this site. – user26857 Jan 1 '13 at 22:57
Aha, here is the statement. One more question, is the following true: if $R$ is a (not necessarily commutative) unital ring, then a left $R$-module $M$ is flat iff for every right ideal $I$ of $R$ the map $I\otimes_R\longrightarrow IM$, $\sum_ki_k\otimes m_k\longmapsto\sum_ki_km_k$, is an isomorphism? – Leon Jan 1 '13 at 23:05
@LeonLampret As far as I know, yes. Furthermore, you can assume $I$ finitely generated when want to test the flatness. – user26857 Jan 1 '13 at 23:21

There is no general answer (as already the other ones mentioned), but let me say something about $\hom_R(I,M)$:

An important notion in homological algebra is that of injective modules. These have property that $\hom_R(-,M)$ maps monomorphisms to epimorphisms. The criterion of Baer states that it is enough to test this for inclusions of ideals, i.e. it is enough to check if $M \cong \hom_R(R,M) \to \hom_R(I,M)$ is surjective, i.e. if every homomorphism $I \to M$ is of the form $i \mapsto im$ for some $m \in M$.

It is a good exercise to show the injectivity of $\mathbb{Q}/\mathbb{Z}$ as a $\mathbb{Z}$-module with the help of this criterion. And this also shows you how hard it can be do describe $\hom_R(I,M)$ in general.

When $I$ is a principal ideal, say $I = (f)$, then $I \cong R/\mathrm{Ann}(f)$ as $R$-modules, so that there is an easy description. More generally, when $R$ is noetherian, a formula of Deligne states that for the associated quasi-coherent module $\tilde{M}$ on the affine scheme $\mathrm{Spec}(R)$ we have the formula $\Gamma(V(I)^c,\tilde{M}) = \mathrm{colim}_n \hom_R(I^n,M)$.

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The question asked in your update is false. For instance, consider the abelian groups $M = \mathbf{Z}/2\mathbf{Z}$, $A = \mathbf{Z}$ and $B = 2\mathbf{Z}$. Then $\operatorname{Hom}(M, A/B)$ has the nonzero identity map, but $\operatorname{Hom}(M, A) = \operatorname{Hom}(M, B) = 0$ since $A$ and $B$ both have no 2-torsion.

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