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Here's the Exercise 3.30 from the textbook of Görtz and Wedhorn:

Let $k$ be a field, and let $A$ be a local $k$-algebra of finite type. Prove that $\operatorname{Spec} A$ consists of a single point, and that $A$ is finite-dimensional as a $k$-vector space. In particular $A$ is a local Artin ring (why?), and $\kappa (A)/k$ is a finite field extension.

I see how this follows from some well-known results:

  1. Every finitely generated algebra over a Jacobson ring is Jacobson, so that $A$ is Jacobson. As $A$ is local, this amounts to saying that its maximal ideal is the only prime ideal.

  2. In general, for every maximal ideal $\mathfrak{m}$ in a finitely generated $k$-algebra $A$, the extension $\kappa (\mathfrak{m})/k$ is finite (which follows again from the general results about Jacobson rings).

  3. A ring is Artinian iff it is Noetherian and every prime ideal is maximal, which is the case here.

  4. A finitely generated $k$-algebra is Artinian iff it is finite (Atiyah-Macdonald, Exercise 8.3).

However, all this seems to be an overkill, and I think the authors had in mind some direct argument for the very special case when $A$ is local. Do you see one?

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  • $\begingroup$ Just an opinion, but this doesn't seem like overkill to me. $\endgroup$ – RghtHndSd Jan 5 at 22:29
  • $\begingroup$ I think there is something wrong with the finite type and I think you need finite(i.e. integrality condition.) For finite type, I assume you mean finitely generated over $k$. However, consider $k[x,y]$ and localize away from $(x,y)=(0,0)$ points. This gives a local ring of dimension 2 and this is finite type over $k$. The spectrum cannot be made of single point. If it is made of 1 point, one would expect this had better be a finite field extension as indicated in the conclusion. $\endgroup$ – user45765 Jan 10 at 16:07
  • $\begingroup$ @user45765 Every algebra of finite type over a field is a Jacobson ring, and a Jacobson ring is local when its maximal ideal is the only prime ideal. (See e.g. stacks.math.columbia.edu/tag/00FZ for the statements and proofs.) So a local ring of dimension $> 0$ can't be of finite type over a field. $\endgroup$ – Sr. Tacuacín Jan 10 at 22:25

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