# Is the dual of a finitely generated module finitely generated?

I recently thought of this and have no idea whether over a general commutative unitary ring the dual of a finitely generated module is finitely generated. This must be known.

• It is true for Noetherian rings: take a surjection $R^n\rightarrow M$ and apply $\mathrm{Hom}_R(-,R)$ to get an injection $\mathrm{Hom}_R(M,R)\rightarrow\mathrm{Hom}_R(R^n,R)=R^n$. Because $R$ is Noetherian, submodules of finite $R$-modules are finite. I'm not sure about the general case. – Keenan Kidwell May 15 '13 at 18:10
• @carmelodevisu You can find an example here. – user26857 May 15 '13 at 21:12

## 2 Answers

If $I$ is an ideal of $R$, the dual of $R/I$ is isomorphic to $\mathrm{Ann}(I) = \{r \in R : rI = 0\}$, and this doesn't have to be finitely generated. Take for instance $R = k[y,x_1,x_2,\dotsc]/(y x_i : i \geq 1)$ and $I=(y)$.

A more natural question would be: How can we characterize commutative rings with the property that duals of f.g. modules over that ring are f.g.? Even more natural: How can we characterize commutative rings with the property that hom modules between f.g. modules are f.g.? For example, noetherian commutative rings satisfy this property (see the comment by Keenan Kidwell). But I think that there are more examples (perhaps coherent rings?).

• Nice answer @Martin! – Keenan Kidwell May 15 '13 at 21:27

Given a finitely generated module $B$, then $B^* = \text{Hom}_R(B,R)$ is a finitely generated module.

Indeed, take a module $B'$, we have a $n$ and $\varphi:B^n\rightarrow B'$ a surjective morphism.

Now for $B^*$. Note first that

$$(B^*)^n = \text{Hom}_R(B,R)^n = \text{Hom}_R(B^n,R) = (B^n)^*$$

To construct our surjective morphism, we take $\{e_i\}_{i\in I}$ a basis of $B^n$. This gives us $\{e_i^*\}_{i\in I}$ where $e_i^*(e_j) = \delta_{ij}$. This set can be completed into a basis denoted $\{e_j^*\}_{j\in J}$ since it is a linearily independent set. We then define our surjective morhism as follows \begin{align*} \varphi' &:(B^*)^n = (B^n)^*\rightarrow B' \\ &:\sum_{j\in J} e_j^*r_j\mapsto \sum_{i\in I} \varphi(e_i^*)r_i \end{align*}

i.e. $\varphi'$ is a composition of a projection on the submodule generated by the dual elements of the original base of $B^n$ and of $\varphi$

NB: I think this is correct but not sure

• If the question has a negative answer, in general, then how could be your answer correct? (-1) – user26857 Mar 23 '18 at 9:46
• ok. I didn't pay attention sorry. Any chance of saying where the mistake is then? – tomak Mar 23 '18 at 9:52