Let $R$ be a commutative local ring with maximal ideal $\mathfrak{m}$. Is it true in general that $\text{Hom}_R(\mathfrak{m},\mathfrak{m})\cong \text{Hom}_R(\mathfrak{m}, R)$? What if the Krull dimension of $R$ is equal to one?
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$\begingroup$ Not true. Take for $R$ a DVR with uniformizing parameter $\pi $, then $\times\, \pi ^{-1}:\mathfrak{m}\rightarrow R$ is surjective (and $\dim R=1$). $\endgroup$– abxCommented Aug 18, 2014 at 8:00
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$\begingroup$ Okey, I have replaced "=" by $\cong$. $\endgroup$– gamovCommented Aug 18, 2014 at 8:20
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Yes, this is always true.
Let us assume $\text{Hom}_R(\mathfrak m,\mathfrak m)\neq\text{Hom}_R(\mathfrak m,R)$, so we have a homomorphism $f:\mathfrak m\rightarrow R$ with image not contained in $\mathfrak m$: say $f(x)\in R^\times$ for some $x\in\mathfrak m$. We can w.l.o.g. assume that $f(x)=1$ (since $f(x/f(x))=1$). Now, we must have $y=yf(x)=f(xy)=xf(y)$ for all $y\in\mathfrak m$. Hence, $\mathfrak m$ is generated by $x$ and multiplication by $x$ is injective (since $xy\neq0$ for $0\neq y\in\mathfrak m$ follows from this equation and obviously $xy\neq0$ for $y\in R^\times$). But then we have $\mathfrak m\cong R$ as $R$-modules.
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$\begingroup$ This isn't a contradiction. In fact, $R \cong \mathfrak m$ precisely when $\mathfrak m$ is a principal ideal generated by a regular element. In particular, the requested isomorphism is obviously true in any such case. What you have shown is that this is precisely the condition needed to make the isomorphism true. If $R$ is a Noetherian domain, this is pretty close to saying you have a DVR; otherwise, there are lots of other valuation rings for which the isomorphism holds. $\endgroup$ Commented Aug 19, 2014 at 2:45
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$\begingroup$ @NeilEpstein: This isn't a proof by contradiction - what this shows is that if the canonical injection $\text{Hom}_R(\mathfrak{m},\mathfrak{m}) \hookrightarrow \text{Hom}_R(\mathfrak{m},R)$ is not surjective, then $\mathfrak{m}$ is principal and generated by a nonzerodivisor, and thus there always exists an isomorphism $\endgroup$– zcnCommented Aug 19, 2014 at 2:55
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$\begingroup$ @Neil Epstein : In fact, $R\cong \mathfrak{m}$ and $R$ noetherian already imply that $R$ is a DVR. $\endgroup$– abxCommented Aug 19, 2014 at 5:19