0
$\begingroup$

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?

$\endgroup$
7
  • 5
    $\begingroup$ 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. $\endgroup$
    – Pedro
    Jul 8, 2021 at 18:22
  • $\begingroup$ You are right, @PedroTamaroff. My claim is very wrong. As for the field, I will edit the post. $\endgroup$
    – Brass One
    Jul 8, 2021 at 18:27
  • 2
    $\begingroup$ 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 $V$. $\endgroup$ Jul 8, 2021 at 18:27
  • $\begingroup$ I've fixed the question. It was wrong - there is a hypothesis on $R$ being finite-dimensional. Sorry. $\endgroup$
    – Brass One
    Jul 8, 2021 at 18:30
  • 5
    $\begingroup$ 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. $\endgroup$ Jul 8, 2021 at 23:46

1 Answer 1

1
$\begingroup$

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

Answer: Let

$$\rightarrow^{d_3} P_2 \rightarrow^{d_2} P_1 \rightarrow^{d_1} P_0 \rightarrow^{d_0} M \rightarrow 0$$

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 $i$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .