# Finitely generated projective resolution

Let $$K$$ be a field, $$A$$ be a finite dimensional $$K$$-algebra and $$M$$ be a finitely generated $$A$$-module. Is it true that $$M$$ admits a projective resolution by finitely generated projective $$A$$-modules?

As $$A$$ is a finite algebra over $$K$$ it is noetherian. As $$M$$ is finitely generated there is a surjection $$A^{\oplus n} \longrightarrow M$$. $$A^{\oplus n}$$ is noetherian as $$A$$ is. Let $$N$$ be the kernel of this map. By noetherianness, it is finitely generated, so there is a surjection $$A^{\oplus m} \longrightarrow N$$ and hence an exact sequence $$A^{\oplus m} \longrightarrow A^{\oplus n} \longrightarrow M \longrightarrow 0$$. Repeat this process to get a projective (in fact free) resolution by finitely generated modules. Note that we didn't need the full strength of the assumption that $$A$$ is finite over $$K$$ - only that it was noetherian.
• Is it easy to show that $A$ is noetherian? – Eduardo Longa Jun 10 '20 at 0:08
• How easy it is depends on your background knowledge. Here are some facts which will prove that $A$ is noetherian. 1. If $R$ is noetherian then the polynomial ring $R[x]$ is noetherian (this is called the Hilbert basis theorem and the proof is not that easy to come up with on your own). 2. If $R$ is noetherian and $I$ is an ideal, $R/I$ is noetherian. 3. Fields are noetherian. 4. A finite dimensional algebra $A$ over a field $K$ is finitely generated as an algebra, i.e. there is a surjection $k[x_1, \dots, x_n] \longrightarrow A$ for some $n$. – paul blart math cop Jun 10 '20 at 0:14