# Is it possible for a finite integral closure of a DVR to not be a PID?

Suppose that we have a point $$A$$(local ring, DVR) of an abstract curve over $$k=\bar{k}$$ given by a field $$k(X)$$. Let $$k(Y)$$ be a finite extension of $$k(X)$$ and denote by $$B$$ the integral closure of $$A$$ in $$K(Y)$$.

Is it possible for $$B$$ to not be a PID?

• Does $k(X)$ denote a field of regular functions over X? Is the question about a curve living in $K^n$ for algebraically closed field K? How do you define an integral closure of a point? – Lada Dudnikova May 15 at 20:13
• It denotes the function field of a nonsingular, proper curve X. It follows that a closed point of $X$ corresponds to a discrete valuation of $k(X)$ and by a point I mean here its valuation ring. – P. Grabowski May 15 at 20:31

I think the following more general statement is true.

Proposition. Let $$A$$ be a one-dimensional noetherian semi-local domain and let $$K$$ be its field of fractions. Consider a finite algebraic field extension $$K \subseteq L$$, and let $$B$$ be the integral closure of $$A$$ in $$L$$. Then, $$B$$ is a semi-local Dedekind domain, and is therefore a principal ideal domain.

Proof. First, by a consequence of the Krull–Akizuki theorem [Matsumura, Cor. to Thm. 11.7], we see that $$B$$ is a Dedekind domain with finitely many prime ideals $$\mathfrak{n}_1,\mathfrak{n}_2,\ldots,\mathfrak{n}_r$$ lying over each maximal ideal $$\mathfrak{m}$$ in $$A$$. We note that these $$\mathfrak{n}_i$$ are maximal ideals in $$B$$ since $$\operatorname{height}\mathfrak{n}_i \ge 1$$ (they strictly contain the prime ideal $$0 \subseteq B$$) and since $$\operatorname{height}\mathfrak{n}_i \le 1$$ (the ring $$B$$ has dimension one). Moreover, every maximal ideal of $$B$$ must lie over a maximal ideal in $$A$$, since if a prime ideal $$\mathfrak{q}$$ in $$B$$ contracts to a non-maximal ideal in $$A$$, then it must contract to the zero ideal in $$A$$, and going-up would then imply that $$\mathfrak{q}$$ is not maximal. Thus, $$B$$ is a semi-local Dedekind domain.

We now want to show that a semi-local Dedekind domain $$B$$ is a PID. Let $$\mathfrak{n}_1,\mathfrak{n}_2,\ldots,\mathfrak{n}_s$$ be the maximal ideals in $$B$$; since every ideal can be written as a product of powers of the $$\mathfrak{n}_i$$, it suffices to show that each $$\mathfrak{n}_i$$ is principal. We fix an $$i$$ in the argument below.

First, we claim that $$\mathfrak{n}_i \smallsetminus \mathfrak{n}_i^2 \ne \emptyset$$. If not, then $$\mathfrak{n}_iB_{\mathfrak{n}_i} = \mathfrak{n}_i^2B_{\mathfrak{n}_i}$$, in which case Nakayama's lemma implies $$\mathfrak{n}_iB_{\mathfrak{n}_i} = 0$$, contradicting the fact that $$B_{\mathfrak{n}_i}$$ is a DVR.

Next, we consider the surjection $$B \longrightarrow \frac{B}{\mathfrak{n}_1\cdots\mathfrak{n}_{i-1}\mathfrak{n}_i^2\mathfrak{n}_{i+1}\cdots\mathfrak{n}_s} \simeq \frac{B}{\mathfrak{n}_1} \times \cdots \frac{B}{\mathfrak{n}_{i-1}} \times \frac{B}{\mathfrak{n}_i^2} \times \frac{B}{\mathfrak{n}_{i+1}}\times\cdots\times\frac{B}{\mathfrak{n}_s},$$ where the isomorphism holds by the Chinese remainder theorem since maximal ideals are coprime. Now choose $$x \in \mathfrak{n}_i \smallsetminus \mathfrak{n}_i^2$$, which exists by the previous paragraph, and let $$b$$ be an element in the preimage of $$(1,\ldots,1,x,1,\ldots,1)$$ in the surjection above. We then see that $$b \in \mathfrak{n}_i \smallsetminus \mathfrak{n}_i^2$$ but that $$b \notin \mathfrak{n}_j$$ for all $$j \ne i$$. Writing $$(b)$$ as a product of powers of the ideals $$\mathfrak{n}_j$$ in $$B$$, we therefore see that $$(b) = \mathfrak{n}_i$$, hence $$\mathfrak{n}_i$$ is principal. $$\blacksquare$$

• Thank you very much! In the morning I have observed that the answer is no on my own. It turns out that is just a approximation lemma(A lemma from Serre's Local Fields)[One could say you have proved a version of it here.] for abstract curves. Somehow I missed that before. :D – P. Grabowski May 16 at 12:05
• @P.Grabowski Glad to help! I think that Serre's Local fields is a fine reference for these things, at least in your special case. To be precise, if $A$ is a semi-local Dedekind domain of finite type over a field, then $A$ satisfies hypothesis (F) on p. 13 by Remark 2 on p. 13, and $B$ is a semi-local Dedekind domain by Proposition I.9 and the corollary to Proposition I.10. The ring $B$ is then a PID by the corollary to the Approximation Lemma on p. 12. – Takumi Murayama May 16 at 18:24