# If $M, N$ are finitely generated, then $\operatorname{Ext}_R^1(M,N)$ has finite dimension

Problem

Let $$K$$ be a field, and fix $$R$$ a finite dimensional algebra over $$K$$. If $$M,N$$ are finitely generated $$R$$-modules, then $$\operatorname{Ext}_R^1(M,N)$$ is a finite-dimensional $$K$$-vector space.

Attempt

Since $$M,N$$ are finitely generated, we have that $$M = \langle x_1, \ldots, x_m\rangle$$ and $$N = \langle y_1, \ldots , y_n \rangle$$, for some $$x_i, y_j$$. In this case, writing $$M = \bigoplus_{i = 1}^m x_i R$$ and $$N = \bigoplus_{i = 1}^n y_i R$$ yields (after some computations) $$\operatorname{Ext}_R^1(M,N) = \prod_{j = 1}^n(\sum_{i = 1}^m \operatorname{Ext}^1_R(x_iR,y_jR)).$$ So, to solve the problem, I would need to verify that $$\operatorname{Ext}^1_R(x_iR,y_jR)$$ is a finite-dimensional vector space.

To do so, I've considered the short exact sequences $$0 \rightarrow \operatorname{Ker}(p_i) \hookrightarrow R \xrightarrow{p_i} x_i R \rightarrow 0$$ in which $$p_i(a) = x_i a$$, and $$0 \rightarrow \operatorname{Ker}(q_j) \hookrightarrow R \xrightarrow{q_j} y_j R \rightarrow 0$$ in which $$q_j(a) = y_j a$$. From the long cohomology exact sequence of those, we can determine that $$\operatorname{Ext}_R^2(x_i R, \operatorname{Ker}(q_j)) \cong \operatorname{Ext}_R^1(x_iR, y_jR) \cong \operatorname{Ext}_R^2(\operatorname{Ker}(p_i), y_jR).$$ At which point I am stuck.

Questions

Am I on the right track? If yes, how do I end the proof? If not, could anyone point out a better approach?

Also, if I change $$\operatorname{Ext}$$ by $$\operatorname{Tor}$$, does this result still holds?

• You are claiming that a f.g. module is a direct sum of cyclic modules, which is not true (you should be able to find a counterexample from linear algebra, say over $\mathbb R$). You also say the Ext must have finite dimension, but you have not introduced any field.
– Pedro
Jul 8, 2021 at 18:22
• You are right, @PedroTamaroff. My claim is very wrong. As for the field, I will edit the post. Jul 8, 2021 at 18:27
• It is not true that if $R$ is an algebra over a field $k$, then $\mathrm{Ext}^1_R(M,N)$ is finite dimensional over $k$ for all finitely generated $R$-modules $M,N$. For example, take $V$ to be any $k$-vector space, and $R=\begin{pmatrix}k&0\\V&k\end{pmatrix}$. Then there are two simples, both one dimensional over $k$, and their ext group (one way round) is isomorphic to the dual of $V$. Jul 8, 2021 at 18:27
• I've fixed the question. It was wrong - there is a hypothesis on $R$ being finite-dimensional. Sorry. Jul 8, 2021 at 18:30
• I would suggest trying to instead work directly with a free resolution of $M$. You don't need to compute it explicitly; you just need to know that certain things are finite-dimensional. Jul 8, 2021 at 23:46

$$\rightarrow^{d_3} P_2 \rightarrow^{d_2} P_1 \rightarrow^{d_1} P_0 \rightarrow^{d_0} M \rightarrow 0$$
be a projective resolution of $$M$$. Since $$M$$ is a finitely generated as $$R$$-module and since $$R$$ is finite dimensional you may choose $$P_i:=R^{n_i}$$ for some integer $$n_i\geq 1$$ for all $$i$$. Hence $$Hom_R(P_i,N)$$ is finite dimensional over $$K$$ for all $$i$$. It follows
$$Ext^i_R(M,N) :=ker(d_i)/ker(d_{i-1})$$
is finite dimensional over $$K$$ for all $$i$$.