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For a commutative ring $R$, an $R$-module $M$ and an ideal $I\trianglelefteq R$, I want to show that $\text{Hom}_R(I, \text{Hom}_R(I,M)) = \text{Hom}_R(I^2, M)$.

Is this true in general?

Suppose $I$ is generated by two elements $I = (x,y)$.

Then if $\eta : I^2 \rightarrow M$, then $\eta$ defines a map $\eta': I\rightarrow \text{Hom}(I,M)$ by sending $x\mapsto (x\mapsto \eta(x^2), y\mapsto \eta(xy))$ and similarly for $y$.

However if $\varepsilon : I\rightarrow \text{Hom}(I,M)$, I want to define $\varepsilon'$ by sending $x^2 \mapsto \varepsilon(x)(x)$, $xy\mapsto\varepsilon(x)(y)$ and $y^2 \mapsto \varepsilon(y)(y)$. This is an issue for $xy$, since I could also have defined it to send $xy \mapsto \varepsilon(y)(x)$, so there isn't a single choice for the map $\varepsilon'$.

I don't know how I could prove that $\varepsilon(y)(x) = \varepsilon(x)(y)$ in general. Am I approaching this problem incorrectly?

IAlreadyHaveAKey
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1 Answers1

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$$\newcommand{\Hom}{\text{Hom}}\Hom_R(I,\Hom_R(I,M))$$ is the set of $R$-bilinear maps from $I\times I$ to $M$. These are the same as the linear maps from $I\otimes_R I$ to $M$, so $$\newcommand{\Hom}{\text{Hom}}\Hom_R(I,\Hom_R(I,M)) \cong\Hom_R(I\otimes_R I,M).$$ This is an isomorphism of functors on $M$. From the usual abstract nonsense we find that $$\newcommand{\Hom}{\text{Hom}}\Hom_R(I,\Hom_R(I,M)) \cong\Hom_R(I^2,M)$$ if and only if $I^2\cong I\otimes_R I$ as $R$-modules.

In general, $I\otimes_R I$ is not isomorphic to $I^2$ as $R$-modules. A classic example is $I=(X,Y)$ in a polynomial ring $R=k[X,Y]$ over a field. Then $I^2$ is torsion-free, but $I\otimes_R I$ is not.

Angina Seng
  • 158,341
  • Does the answer change if we assume that $R$ is graded and so is $M$, and $I$ is a finitely generated homogeneous ideal, which turns the Hom-sets themselves into graded modules? – IAlreadyHaveAKey Sep 28 '17 at 08:46