The (algebraic) dual of a completion

Question

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 $R$.

Answer 1

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.

Answer 2

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 $R \neq \hat{R}$.

Answer 3

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?

Answer 4

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 $0$.)