Let $k$ be a field, and $R$ a $k$-algebra. Suppose that $R_{\mathfrak{p}}$ is a finitely generated $k$-algebra for all prime ideals $\mathfrak{p}$ of $R$. Does this imply that $R$ is also a finitely generated $k$-algebra? (I think this is false, but couldn't find a counterexample.)
4
-
$\begingroup$ Since any finitely generated $k$-algebra is Jacobson, it can only be local if it is zero-dimensional. Are you sure this is the question you want to ask? $\endgroup$– user2035Commented Oct 27, 2011 at 14:47
-
$\begingroup$ Actually, doesn't $R=\prod\_{n=1}^\infty k$ work? $\endgroup$– George LowtherCommented Oct 27, 2011 at 14:54
-
$\begingroup$ @George: it depends on $k$. For $k=\mathbf F_2$, it works, for $k=\mathbf Q$ it does not. $\endgroup$– user2035Commented Oct 27, 2011 at 15:02
-
1$\begingroup$ @a-fortiori: I guess thats because you need $k$ to be finite to ensure that $R_\mathfrak{p}\cong k$ at primes $\mathfrak{p}$ corresponding to nonprincipal ultrafilters of $\mathbb{N}$. You could look at $$R=\left\{x\in\prod\_{n=1}^\infty k\colon x_n{\rm\ is\ eventually\ constant}\right\}$$instead. $\endgroup$– George LowtherCommented Oct 27, 2011 at 15:27
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
Take any compact Hausdorff space $X$ and let $R$ consist of the locally constant functions $X\to k$. Then, $R$ is a $k$-algebra. Its prime ideals are all of the form $\mathfrak{p}=\left\{f\in R\colon f(x)=0\right\}$ for $x\in R$, so $R_{\mathfrak{p}}\cong k$ is trivially finitely generated as a $k$-algebra. If $X$ has infinitely many connected components then $R$ is not finitely generated as a $k$-algebra. So, taking $X$ to be the one-point compactification $\mathbb{\bar N}=\mathbb{N}\cup\{\infty\}$ of $\mathbb{N}$ gives the required counterexample. Equivalently, $$ R=\left\{x\in\prod\_{n=1}^\infty k\colon x_n{\rm\ is\ eventually\ constant}\right\}. $$
answered Oct 27, 2011 at 16:48