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Let's say I have a ring $R$ and a subring $I$. Let $M$ be a $R$-module. Is it true that, as $R$-modules (or as groups):

$$\text{Hom}_R (M, R \otimes_I M) \simeq \text{Hom}_I(M, M)$$

If so, why?

I am aware of a particular case where:

$$\text{Ext}_R (M, R \otimes_I M) \simeq \text{Ext}_I(M, M)$$

And I am wondering if it is true in general, for I think of $\text{Ext}$ as $H^*(\text{Hom})$.

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    What makes you think this holds? – 2017-02-07
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    It's the other way around, actually! Any $R$ linear map $R\otimes_I M \longrightarrow M$ identifies bijectively with an $I$ linear endomorphism on $M$. – 2017-02-07

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Actually, I think this is false, a counter example takes $R = M = \mathbb{Z}[x]$ and $I = \mathbb{Z}$.

$\operatorname{Hom}_{\mathbb{Z}[x]}(\mathbb{Z}[x], \mathbb{Z}[x] \otimes_{\mathbb{Z}} \mathbb{Z}[x]) \simeq \operatorname{Hom}_{\mathbb{Z}[x]}(\mathbb{Z}[x], \mathbb{Z}[x, y]) \simeq \oplus \mathbb{Z}[y]$

However, $\operatorname{Hom}_{\mathbb{Z}}(\mathbb{Z}[x], \mathbb{Z}[x]) \simeq \prod \mathbb{Z}[y]$, because as a $\mathbb{Z}$-module, $\mathbb{Z} = \oplus \mathbb{Z}$, and since the maps are only $\mathbb{Z}$-linear each map is uniquely specified by where we send $(x, x^2, x^3, \ldots)$, which is an infinite number of polynomials (no need for it to have only finitely many nonzero).