# “Noetherian” and “finitely generated” for polynomial algebras

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?

• 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. – François Brunault Apr 30 '19 at 15:05
• Meta discussion here: meta.mathoverflow.net/questions/4200/flood-of-new-users – Steven Landsburg May 2 '19 at 15:01

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.