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It is easy to show that $Hom(\mathbb{Z}_p,\mathbb{Z})=0$ (one just uses that each integer coprime to $p$ is a unit and therefore each image of such a homomorphism has infinitely many divisors). However, I wonder if this is true more generally:

Question: Given a commutative ring $R$ and a maximal ideal $m \subseteq R$, is it true that $Hom_R(\hat{R},R)=0$ where $\hat{R} = \varprojlim R/m^i$ ?

Of course, one has to require $R$ to be not complete with respect to $m$, since otherwise the Hom in question is simply the dual of $R$ and hence isomorphic to $R$.

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up vote 3 down vote accepted

Your example for $R=\mathbb{Z}$ (along with the proof) generalizes verbatim to the case $R$ a principal ideal domain and $m=(p)$ with a prime element $p$.

As another example let $R$ be a commutative Noetherian local domain that is not complete. Then also $Hom_R(\hat{R},R)=0$. A concrete example is $R=k[x_1,...,x_n]_{(x_1,...,x_n)}$ with $k$ a field.

This was shown by [Aldrich et. al: Derived Functors of Hom Relative to Flat Covers. Math. Nachr. 242(2002), 17-26], Lemma 3.3.

Added: The following is used in the proof of the cited result:

Lemma: If $R$ is a commutative Noetherian local ring then $\hat{R} \to Hom_R(\hat{R},\hat{R}),\;r \mapsto r \cdot id$ is an isomorphism of $R$-modules.

Proof: Let $E=E(R/m)$ be the injective hull of the residue field. By [Brodman, Sharp: Local Cohomology. Theorem 10.2.11] the map $\hat{R} \to Hom_R(E,E), \; r \mapsto r\cdot id_E$ is an isomorphism of $R$-modules (this is sometimes considered as part of Matlis duality). Hence we have $R$-module isomorphisms $$\begin{array}{lll} Hom_R(\hat{R},\hat{R}) & \cong & Hom_R(\hat{R},Hom_R(E,E))\cong Hom_R(\hat{R} \otimes_R E,E) \newline & \cong & Hom_R(E,E) \cong \hat{R} \end{array}$$ where the 3rd isomorphism uses $\hat{R} \otimes_R E \cong E$ (see [Matsumura: Comm. ring theory. Proof of 18.6 (iii)]. Composing the isomorphisms on element level now yields the map in the lemma.

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That are nice examples. Thanks. In the proof of the quoted lemma I don't understand the first sentence: "By Matlis duality any $R$-linear $\hat{R} \to \hat{R}$ is scalar multiplication by an $r \in R$." If I choose $r \in \hat{R}-R$ then scalar multiplication by $r$ is also $R$-linear and since $1 \mapsto r$ it doesn't coincide with scalar multiplication by some element from $R$. Am I missing something ? – tomasz Jul 15 '12 at 16:50
I think they mean that each $R$-linear map $f: \hat{R} \to \hat{R}$ is multiplication by some $r \in \hat{R}$ and in the case of interest, $f(\hat{R})\subseteq R$, so $r \in R$. – Ralph Jul 15 '12 at 18:01
OK that works. But I still don't understand why it follows from Matlis duality that each $R$-linear $\hat{R} \to \hat{R}$ is scalar multiplication by an $r \in \hat{R}$. To my knowledge, Matlis duality is an equivalence of functors $Hom_R(Hom_R(-,E),E)\cong id$ where $-$ is either an Artinian or a finitely generated Notherian $R$-module ? – tomasz Jul 15 '12 at 18:36
I added a proof of this fact. – Ralph Jul 15 '12 at 21:05
Thanks for clarifying this point. PS: Interesting lemma. – tomasz Jul 15 '12 at 22:08

Take $R = \mathbb{Z}_p \times \mathbb{F}_q$ for distinct primes $p,q$ and $\mathfrak{m} = qR$. Then $\hat{R} \cong 0 \times \mathbb{F}_q$ is isomorphic to an ideal in $R$, so $Hom_R(\hat{R}, R) \neq 0$.

This $R$ isn't complete with respect to $\mathfrak{m}$ since $R \neq \hat{R}$.

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Let me give another example that's finite type over a field. It uses a similar idea as Konstantin's but the associated primes are contained in one another (and so it is somehow more local). This should fix an error in an earlier version where I got $x$ and $y$ backwards.

Suppose $k$ is a field (for example $k = \mathbb{C}$). Consider $R = k[x,y]/\langle x^2, xy \rangle$ with maximal ideal $\mathfrak{m} = \langle x, y \rangle$. Then $\hat{R} = k[[x,y]]/\langle x^2, xy \rangle$.

However, there is a surjective map of $\hat{R}$-modules (and thus of $R$-modules) $\rho : \hat{R} \to \hat{R}/\langle y \rangle \cong k[x,y]/\langle x^2,y \rangle.$ Note $\rho(1) = 1$.

Consider the non-zero map of $R$-modules $\alpha: R \to R$ which sends $1$ to $x$. The kernel of $\alpha$ is $\langle x, y \rangle$. Thus we can factor alpha through a map $\phi : k[x,y]/\langle x^2,y \rangle \to R$ which sends $1$ to $x$.

It follows that $\phi \circ \rho \in \text{Hom}_R(\hat{R}, R)$ sends $1$ to $x$ and is thus non-zero.

EDIT: I wonder if the desired statement is true for $R$ a domain of positive dimension of finite type over a field?

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Please disregard this answer. It is not correct. I was thinking of ring homomorphisms the entire time.

There are truncation homomorphisms from a power series ring over a field $k[[X]]$ to the polynomial ring $k[X]$.

EDIT: Martin (below) is right. Truncation is not a homomorphism, however, $k[[X]]\rightarrow k[X]$ defined by $X\mapsto0$ still gives a nonzero homomorphism! ($\leftarrow$ truncation at degree $0$.)

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There is no homomorphism $k[[X]] \to k[X]$ (except for $X \mapsto 0$). – Martin Brandenburg Jul 14 '12 at 13:18
@Martin: You are right, but we still get a nonzero homomorphism! – Mahdi Majidi-Zolbanin Jul 14 '12 at 13:40
Mahdi, I'm confused, it's a homomorphism of Abelian groups, but why is it a homomorphism of modules? It needs to respect multiplication by elements of $k[x]$ of degree higher than the truncation. – Karl Schwede Jul 14 '12 at 13:47
@Karl: Yes, I realized that after Martin's comment. What I meant after, was we still get a nonzero homomorphism by sending $X$ to $0$. – Mahdi Majidi-Zolbanin Jul 14 '12 at 13:49
Mahdi, I don't think that's a module homomorphism (assuming you are sending $1$ to $1$). For example, say that map is called $\phi$. Then $0 = \phi(x) = x \cdot \phi(1) = x$. – Karl Schwede Jul 14 '12 at 14:00

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