Consider an algebra $A$ over a field and suppose that $A$ is zerodimensional as a ring. It is wellknown that if, in addition, $A$ is finitely generated, it has a finite vector space dimension. Assume now that $A$ is not necessarily finitely generated, but let $A$ be noetherian (as a ring). Does it follow that the vector space dimension of $A$ is finite? What if $A$ is assumed to be noetherian, and also local and Gorenstein? Are there any infinitedimensional examples then?

$\begingroup$ What do you mean by the dimension of a ring? Obviously not Noetherian dimension... $\endgroup$ – Dylan Thurston Apr 6 '11 at 2:21

2$\begingroup$ I mean the Krull dimension of course. $\endgroup$ – Alexander Isaev Apr 6 '11 at 2:22

2$\begingroup$ Take $A$ to be a field of infinite degree over the base field $k$. $\endgroup$ – Anatoly Preygel Apr 6 '11 at 2:46

$\begingroup$ Such a field is not a local algebra over the field $k$ though. I think if the algebra is local, one can argue as follows. Since $A$ is Noetherian and zerodimensional, it is Artinian (in the sense of having the Descending Chain Condition). But then its composition series is finite. Since $A$ is also local, it follows that it is finitedimensional as a vector space. Being Gorenstein plays no role here. $\endgroup$ – Alexander Isaev Apr 6 '11 at 3:10
As Toly commented, an infinite field extension of $k$ gives a counterexample. However, this is the only way to get a counterexample: if you assume all the residue fields of $A$ are finite over $k$, then $A$ must be finitedimensional. Indeed, 0dimensionality implies that $\text{spec }A$ is Hausdorff, and any Noetherian Hausdorff space is automatically finite and discrete. So $A$ is a finite product of Artinian local rings, each of which is finitedimensional over its residue field.
(Proof that Noetherian and Hausdorff implies finite and discrete: fix a point $x$; by Hausdorffness, for any $y\neq x$, there is a closed neighborhood $C_y$ of $x$ that does not contain $y$. By Noetherianness, the intersection of all the $C_y$'s can be obtained by intersecting only finitely many of them, and is hence still a neighborhood of $x$. But this intersection is just $\{x\}$. This shows the space is discrete; finiteness then follows by Noetherianness.)

$\begingroup$ Thank you for that. I think the proof I gave above for the local case is OK too. $\endgroup$ – Alexander Isaev Apr 6 '11 at 3:41

$\begingroup$ @Alexander Isaev: As Eric remarks, your argument works only if the residue field is finite over $k$. $\endgroup$ – Daniel Litt Apr 6 '11 at 3:56

$\begingroup$ (Indeed, your argument is the proof of the last line of the first paragraph of Eric's answer.) $\endgroup$ – Daniel Litt Apr 6 '11 at 4:06

$\begingroup$ Daniel Lift: it is part of the definition of a local algebra that the residue field is the field $k$. Since $k$ is finite over itself, my proof works. Thanks. $\endgroup$ – Alexander Isaev Apr 6 '11 at 4:33