10
$\begingroup$

Let $k$ be a field. Does there exist a positive integer $n$ such that there is $k$-subalgebra of $k[x_1, \dots, x_n]$ which is Noetherian but not finitely generated?

$\endgroup$
  • 4
    $\begingroup$ If $A \subset k[x_1,\ldots,x_n]$ is a graded subalgebra then a standard fact is that Noetherian implies finitely generated: a system of generators is obtained by taking generators of the ideal $I = A \cap k[x_1,\ldots,x_n]_{\geq 1}$ (polynomials with no constant term). I don't know about the general case. $\endgroup$ – François Brunault Apr 30 '19 at 15:05
  • 3
    $\begingroup$ Meta discussion here: meta.mathoverflow.net/questions/4200/flood-of-new-users $\endgroup$ – Steven Landsburg May 2 '19 at 15:01
15
$\begingroup$

The following example of Eakin [Eak72] says that $n = 2$ already suffices. I have tried to fill in some details to (hopefully) make the example independent from the rest of Eakin's paper.

Example [Eak72, Example on p. 79]. Let $k$ be a field, and consider the formal power series ring $k[[t]]$ in one variable over $k$. By an argument of Schmidt (see [Sch33, Hilfssatz 5] or [MLS39, Second Proof of Lemma 1]), there exist two algebraically independent elements $x,y \in k[[t]]$ that have the same positive valuation with respect to the valuation on $k[[t]]$. For a specific example, Schmidt's result implies that one can choose $$x = t \qquad\text{and}\qquad y = \sum_{n=1}^\infty t^{n!}.$$ Now let $V = k[[t]] \cap k(x,y)$ and $R = k[x,1/y] \cap V$. Setting $X = x$ and $Y = 1/y$, we have $$k \subseteq k[X,XY] \subseteq R \subseteq k[X,Y].$$ We then show the following:

Claim. $R$ is noetherian but not finitely generated as a $k$-algebra.

To show that $R$ is noetherian, we note that $R$ is a Krull ring by [Mat89, Theorem 12.4(ii)] since both $k[x,1/y]$ and $V$ are Krull. A theorem of Heinzer [Hei69, Theorem 9] (see also [Eak72, footnote 1 on p. 78]) then implies that $R$ is noetherian.

To show that $R$ is not finitely generated as a $k$-algebra, we first note that the prime ideal $\mathfrak{p} = \mathfrak{m}_V \cap R$ has height one in $R$ and that $V = R_{\mathfrak{p}}$, since $V$ is a DVR that must appear when $R$ is written as the intersection of DVR's in $k(X,Y)$ [Mat89, Theorem 12.3]. The residue field of $V$ is $k$, since it must simultaneously contain $k$ and also be contained in the residue field of $k[[t]]$, which is $k$. Thus, $\mathfrak{p}$ is a prime ideal of height one in $R$ whose residue field is not transcendental over $k$, which cannot be the case if $R$ is finitely generated over $k$ [ZS75, Corollary on p. 92].

References

[Eak72] Paul Eakin. "A note on finite dimensional subrings of polynomial rings." Proc. Amer. Math. Soc. 31 (1972), 75–80. DOI: 10.2307/2038515. MR: 289498.

[Hei69] William Heinzer. "On Krull overrings of a Noetherian domain." Proc. Amer. Math. Soc. 22 (1969), 217–222. DOI: 10.2307/2036956. MR: 254022.

[Mat89] Hideyuki Matsumura. Commutative ring theory. Second ed. Translated from the Japanese by M. Reid. Cambridge Stud. Adv. Math. 8. Cambridge Univ. Press, Cambridge, 1989. DOI: 10.1017/CBO9781139171762. MR: 1011461.

[MLS39] Saunders Mac Lane and O. F. G. Schilling. "Zero-dimensional branches of rank one on algebraic varieties." Ann. of Math. (2) 40, (1939), 507–520. DOI: 10.2307/1968935. MR: 158.

[Sch33] Friedrich Karl Schmidt. "Mehrfach perfekte Körper." Math. Ann. 108 (1933), no. 1, 1–25. DOI: 10.1007/BF01452819. MR: 1512831.

[ZS75] Oscar Zariski and Pierre Samuel. Commutative algebra. Vol. II. Reprint of the 1960 edition. Grad. Texts in Math. 29. Springer-Verlag, New York-Heidelberg, 1975. DOI: 10.1007/978-3-662-29244-0. MR: 389876.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy