16
$\begingroup$

Let $X$ be a smooth projective irreducible curve defined over an algebraically closed field $\mathbb{K}$ (of arbitrary characteristic), and let $p\in X$ be a closed point. Denote by $\mathcal{O}_p(X)$ the local ring of rational functions which are regular at $p$. Then, is it true that the completion of $\mathcal{O}_p(X)$ (with respect to its maximal ideal) is isomorphic to the ring $\mathbb{K}[[x]]$ of formal power series in one variable ? I think that it should follow from Cohen theorem, but I cannot find a reference for this. Most of the results on this are in commutative ring theory books (for example Matsumura) but not really in the language of algebraic geometry. Can someone please give me a reference? I need this result in a research article. Thanks in advance.

$\endgroup$
  • $\begingroup$ You should specify that your point is a "closed point". $\endgroup$ – Jason Starr Dec 29 '14 at 19:49
  • $\begingroup$ Yes, that was what I had in mind, I specified it. Thanks. $\endgroup$ – Jérémy Blanc Dec 29 '14 at 19:53
  • $\begingroup$ Your ring is a Noetherian regular local ring of dimension $1$ with residue field $\mathbb{K}$, so its completion is a complete Noetherian regular local ring with the same properties (all of this is in Atiyah-MacDonald). Therefore, by Cohen theorem, it must be a ring of formal power series over the residue field. Since the dimension is $1$ there is precisely one indeterminate, i.e. $$\hat{\mathcal{O}}_{p, X} = \mathbb{K}[[x]].$$ Am I missing something? $\endgroup$ – Francesco Polizzi Dec 29 '14 at 20:26
  • 1
    $\begingroup$ Just to be more precise: since $p \in X$ is a closed point, the residue field of the local ring is a finite extension of $\mathbb{K}$, hence it is isomorphic to $\mathbb{K}$ because $\mathbb{K}$ is algebraically closed. $\endgroup$ – Francesco Polizzi Dec 29 '14 at 20:34
  • 1
    $\begingroup$ You don't need Cohen's theorem. Define a map $K[t]\to O_{p,X}$ sending $t$ to a coordinate function of $X$ at $p$, and it is straightforward to show that it induces an isomorphism on the completions, using the fact that the residue field of $O_{p,X}$ is $K$ as said Francesco. $\endgroup$ – Cantlog Dec 29 '14 at 21:15
14
$\begingroup$

Let me expand and generalize my comments above. We can prove the following

Proposition. Let $X$ be a projective scheme of dimension $n$ which is defined over an algebraically closed field $k$. If $p \in X$ is a closed, regular point and $\mathcal{O}_{X, \, p}$ is the local ring of $X$ at $p$, then there is an isomorphism $$\widehat{\mathcal{O}}_{X, \, p}= k[[x_1, \ldots, x_n]].$$

Proof. The ring $\mathcal{O}_{X, \, p}$ is a Noetherian regular local ring of dimension $n$, whose residue field is $k$ since $p \in X$ is a closed point and $k$ is algebraically closed. Therefore its $\mathfrak{m}$-adic completion $\widehat{\mathcal{O}}_{X, \,p}$ is a Noetherian complete regular ring with the same residue field and the same dimension, see

Atiyah-MacDonald: Introduction to Commutative Algebra, Proposition 10.15, Theorem 10.26, Corollary 11.19, Proposition 11.24.

By Cohen structure theorem it follows that $\widehat{\mathcal{O}}_{X, \,p}$ is a formal power series over its residue field. Since its dimension is $n$, this implies that it is isomorphic to $k[[x_1, \ldots, x_n]]$ and we are done.

| cite | improve this answer | |
$\endgroup$
  • 6
    $\begingroup$ It is interesting to note that if $k$ imperfect of characteristic $p$ and $x \in X = \mathbf{A}^1_k$ is the closed point corresponding to $t^p-a$ for $a \in k - k^p$ then $O_{X,x}^{\wedge} \simeq k(a^{1/p})[\![u]\!]$ as abstract local rings by Cohen's theorem but this cannot be arranged to be an isomorphism of $k$-algebras (good exercise!). $\endgroup$ – user74230 Dec 30 '14 at 2:25
12
$\begingroup$

My answer is most likely going to be rephrasing Francesco's answer above. Here's how I think about your question. IMHO, the Cohen Structure Theorem is too big a thermonuclear weapon to invoke, because from memory, the hardest part of the proof is the existence of a coefficient field. In your case, we get a coefficient field almost for free! What we need to prove is this:

Proposition: Let $A$ be an $n$-dimensional regular ring that is an integral domain. Furthermore, suppose that $A$ is a finite-type $k$-algebra with $k$ algebraically closed. Then for any maximal ideal $\mathfrak{m}$, we have $\widehat{A_{\mathfrak{m}}} \cong k[[t_1,\ldots,t_n]]$.

Proof: Let $x_1,\ldots,x_n$ be a regular system of parameters of $\mathfrak{m}A_{\mathfrak{m}}$. By the universal property of the power series ring, we have a ring homomorphism $\varphi : k[[t_1,\ldots,t_n]] \to \widehat{A_{\mathfrak{m}}}$, sending $t_i$ to the image of $x_i$ in $\widehat{A_{\mathfrak{m}}}$. Since the residue field of $\widehat{A_{\mathfrak{m}}}$ is isomorphic to $k$ by Zariski's Lemma, it follows by Nakayama's Lemma and Stacks 031D that $\varphi$ is surjective.

The ring $\widehat{A_{\mathfrak{m}}}$ is a regular local ring, a fortiori an integral domain. Hence $\varphi$ is a surjective ring homomorphism between two integral domains of the same dimension and thus is an isomorphism. Done!

Added: The proposition above is not true if we delete the regularity hypothesis. Consider the union of two lines $X = \operatorname{Spec} k[x,y]/(xy)$. Let $p$ denote the origin of the plane $\Bbb{A}^2$. Since $x$ and $y$ are non-zero in $\mathcal{O}_{X,p}$, they are non-zero in the completion since $\mathcal{O}_{X,p} \to \widehat{\mathcal{O}_{X,p}}$ is injective by Krull's Intersection Theorem. It follows that $\widehat{\mathcal{O}_{X,p}}$ is not isomorphic to the power series ring in one variable.

| cite | improve this answer | |
$\endgroup$
  • 4
    $\begingroup$ On the other hand, the hypothesis that $k$ be algebraically closed is not necessary. For example, assuming that the residual extension $k\to A/\mathfrak m$ is finite and separable, one can use Hensel's lemma to lift the residue field $A/\mathfrak m$ into a subfield of $\widehat{A_{\mathfrak m}}$. $\endgroup$ – ACL Dec 30 '14 at 11:26
  • $\begingroup$ @ACL I didn't know this. Thanks for your comment. $\endgroup$ – Ben Lim Dec 30 '14 at 14:50
8
$\begingroup$

Actually, if $x$ is a nonsingular point of an irreducible $n$-dimensional algebraic variety $X$ over an algebraically closed field $K$ then a choice of local parameters $u_1, \dots , u_n$ at $x$ gives rise to an embedding of the local ring $O_{X,x}$ at $x$ to the ring $K[[T_1, \dots , T_n]]$ of formal power series such that each $u_i$ goes to $T_i$. This embedding induces an isomorphism between the completion of $O_{X,x}$ and the ring of formal power series. See Ch. 2, Sect. 2 of ''Basic Algebraic Geometry 1" by Shafarevich.

| cite | improve this answer | |
$\endgroup$
3
$\begingroup$

One possible reference is Mumford, The red book of varieties and schemes, chap. III, § 6.

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

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.