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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?

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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.

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  • $\begingroup$ Is it easy to show that $A$ is noetherian? $\endgroup$ – Eduardo Longa Jun 10 '20 at 0:08
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    $\begingroup$ 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 $n$. $\endgroup$ – paul blart math cop Jun 10 '20 at 0:14

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