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?

  • $\begingroup$ 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? $\endgroup$ – Lada Dudnikova May 15 at 20:13
  • $\begingroup$ 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. $\endgroup$ – 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. $\blacksquare$

  • $\begingroup$ 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 $\endgroup$ – P. Grabowski May 16 at 12:05
  • 1
    $\begingroup$ @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. $\endgroup$ – Takumi Murayama May 16 at 18:24

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