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Let $X$ be a scheme locally of finite type over a field $k$, and let $p \in X$ be a $k$-point with ${\rm dim}\; {\Omega_{X}}_{|p}={\rm dim}\; \mathfrak{m}_{p}/\mathfrak{m}^{2}_{p}=m$ (the 'embedding dimension' of the local ring at $p$).

Then a paper that I'm reading asserts that there is a Zariski open neighbourhood $ p \in U \subseteq X$ and a closed immersion $ U \hookrightarrow M$, where $M$ is a regular (maybe even smooth) scheme over $k$ of dimension $m$. (Clearly $m$ is the minimal dimension for which such an embedding is possible.)

I would like a reference or explanation for this fact, which I have not been able to find in standard commutative algebra books (though it is probably there somewhere). Presumably this is the reason that ${\rm dim}\;\mathfrak{m}_{p}/\mathfrak{m}^{2}_{p}$ is called the embedding dimension of a local ring.

I would also like a reference or explanation showing that one cannot always take $M=\mathbb{A}^{m}$.

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My impression was that this is not true in the greatest generality you could hope for. Given a local ring $R$, it is not always a quotient of an $(\text{emb dim} R)$-dimensional regular local ring, unless $R$ is complete (the latter positive statement is in Matsumura somehwere if I recall correctly). In particular, formal neighborhoods are always ok. –  Karl Schwede May 18 '12 at 14:38
    
I'm happy to assume, as in the situation of the question, that R is a localization of a finitely generated $k$-algebra. Presumably it is true here. I'd rather not pass to a completion, since I'm really looking for a Zariski local statement, although even etale local would be useful. –  A. Pascal May 18 '12 at 15:13
    
Closed point. As in the question, I want to show "there is a Zariski open neighbourhood $p \in U \subseteq X$ and a closed immersion $U \hookrightarrow M$, where M is a regular scheme over k of dimension m." –  A. Pascal May 18 '12 at 19:10

3 Answers 3

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One algebraic version of this statement is that if $A$ is a local ring with embedding dimension $m$ and $A$ is the quotient of a regular local ring, then $A$ is the quotient of a regular local ring of dimension $m$. Write $A = R/I$ where $R$ is a regular local ring and suppose that $R$ has dimension $n$. Then we have a surjection of the Zariski cotangent spaces of the two rings $m_R/m_R^2 \rightarrow m_A/m_A^2$. We can find a basis for the kernel of this map consisting of $n-m$ vectors. Lift these to elements $f_1, \ldots, f_{n-m}$ in the kernel of $A \rightarrow R$ and let $S = R / \langle f_1, \ldots, f_{n-m}\rangle$ so that $A = S/J$ for some ideal $J$. Then I just have to show that $S$ is a regular local ring of dimension $e$. By construction, its Zariski cotangent space has dimension exactly $m$, and by the Krull principal ideal theorem, it has dimension at least $m$, so it must be a regular local ring of dimension $m$.

In the setting that you asked, every local ring of a scheme of finite type over a field is the quotient of a regular local ring since it is the quotient of a localization of affine space. If you apply the argument above then you get equations $f_1, \ldots, f_{n-m}$ which define a regular scheme at $p$ and thus in a neighborhood of $p$. This is your scheme $M$.

As for your question about whether $M$ can always be chosen to be $\mathbb A^m$, the answer is certainly not. Just take $p$ to be a smooth point on any non-rational variety. Then pretty much by definition, there is no neighborhood of $p$ which is isomorphic to an open subset of affine space. It would take some more work to think of an example where $p$ is singular, but it seems like a sufficiently singular point of a variety in a non-rational variety would work. (Although, I believe any zero-dimensional scheme of embedding dimension $m$ and finite type over a field has a closed immersion into affine space of dimension $m$.)

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Thanks. All of the answers were helpful, but this was the simplest and most general. About $\mathbb{A}^{n}$: of course you are right. –  A. Pascal May 21 '12 at 14:33

Let me give you a proof in the complex category.

Assume that $U \subset X$ is an open neighborhood of $p \in X$ which is a closed complex subvariety of a domain $D \subset \mathbb{C}^n$ with coordinates $z_1, \ldots, z_n$. Let $f_1, \ldots, f_k$ be functions on $D$ such that $$\mathcal{O}_{X,p}=\mathcal{O}_{D, p}/(\overline{f}_1, \ldots, \overline{f}_k),$$ where $\overline{f}_i$ denotes the germ of $f_i$ at the point $p \in D$.

Then one proves that $$m=\textrm{embdim}_p X= n- \textrm{rank}J_p(f_1, \ldots, f_k),$$ where $J_p(f_1, \ldots, f_k)=\big( \frac{\partial f_i}{\partial z_j}(p) \big)$ is the Jacobian matrix of the $f_i$ at $p$.

Then by the Implicit Function Theorem it follows that a neighborhood of $p \in X$ can be realized as a closed complex subvariety of $\mathbb{C}^{n- \textrm{rank}J_p}=\mathbb{C}^m$.

For a reference with more details, see for instance [Okuma, Plurigenus of Surface Singularities, Chapter 1, p. 1-2].

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I'm thinking about how to adapt this to the algebraic setting. Any hints? –  A. Pascal May 18 '12 at 15:24
    
In particular, I don't know what to use in place of the implicit function theorem. –  A. Pascal May 18 '12 at 15:32
    
If you want to adapt this proof, you need an algebraic analog of Implicit Function Theorem, like the Hensel's Lemma. This is elaborated, for instance, in Kuhlman's paper "Valuation theoretic and model theoretic aspects of local uniformization". It is known that complete local rings, such as the ring of p-adic integers and rings of formal power series over a field, are Henselian. For non-complete rings the situation is more delicate and I actually do not know in which cases you can adapt the proof above. –  Francesco Polizzi May 18 '12 at 15:49
    
You can find Kuhlman's paper in the arXiv: arxiv.org/abs/1003.5689. Moreover, this post on MSE can be useful for you: math.stackexchange.com/questions/48419/… –  Francesco Polizzi May 18 '12 at 15:52

Presumably the algebraic analogue of something that uses the Implicit Function Theorem involves saying the word "etale" a lot. Thus:

Let $x_1,...,x_m$ be a lift of a basis of $m_p/m_P^2$ to the ring regular functions of some open affine neighborhood $U$ of $P$. Then the map $U \to \textrm{Spec} k[x_1,..,x_m]$ has image some closed subscheme $\textrm{Spec} k[x_1,..,x_m]/I$. This map is etale at $P$ since it induces an isomorphism on formal completions of local rings, so it is etale in some neighborhood of $P$, so it is locally the vanishing set of $n$ equations in $n$ variables over $k[x_1,...,x_m]/I$ whose Jacobian is a unit. Lift those equations arbitrarily from $k[x_1,...,x_m]/I$ to $k[x_1,...,x_m]$. Because everything in $I$ is degree two or higher at zero/$P$, the Jacobian is still invertible at zero/$P$, thus invertible on an open neighborhood.

The vanishing set of the lift of those equations is a cover of $\textrm{Spec}k[x_1,..,x_m]$, etale in a neighborhood of the origin, thus smooth and dimension $m$ in a neighborhood of the origin. The closed subscheme cut out by the ideal $I$ is the same as the vanishing set of the original equations over $k[x_1,...,x_m]/I$, which was locally isomorphic to $U$. Thus $U$ is locally a closed subscheme of a variety smooth of dimension $m$.

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Thanks. So the statement I want is true etale locally. But do you think it is true Zariski locally (with a more complicated proof)? –  A. Pascal May 18 '12 at 16:42
    
I think everything I said works Zariski locally. Maybe I missed something though. –  Will Sawin May 18 '12 at 22:33
    
Upon more careful reading, there are some things I don't understand. First, do you mean "has image some closed subscheme" or rather "factors through a closed subscheme"? Also, I presume you mean formal completion instead of profinite completion. Given that, we have the the map is etale at $P$. But then I don't know what "it is given by a set" means. "It" would seem to refer to the map, but I don't follow here. The map is already given, no? Finally, where does the cover come from and why does the map from $U$ lift to the cover? –  A. Pascal May 19 '12 at 5:23
    
1. Taking the smallest closed subscheme it factors through is fine, because that satisfies all the relations between $x_1,...,x_m$ that hold in the local ring. 2. Yes, I mean the formal completion. 3. The map is the vanishing set of a set of #n# equations in $n$-space. This is a characterization of etale maps, from Milne. 4. The cover comes from lifting the equations whose vanishing set is locally $U$ to $k[x_1,...,x_m]$ from $k[x_1,..,x_m]/I$, this produces a cover of $\textrm{Spec}k{x_1,...,x_m]$. Restricting the cover to $V(I)$ gives $U$, so $U$ factors through the cover. –  Will Sawin May 19 '12 at 15:02
    
I hope that is more clear. I will edit the answer to make it more clear when I have time. –  Will Sawin May 19 '12 at 15:02

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