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I saw a claim that states for a field $k$ and an irreducible of finite type $k$ scheme $X$, $\textrm{dim}X=\textrm{dim} \mathcal{O}_{X,x}$ for any closed point $x$.

The proof starts with reducing the case $X$ is integral affine scheme, but I cannot understand how can we reduce the case. Is there any good idea?

User0829
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1 Answers1

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The dimension does not depend on the reducible structure of $\mathcal O_X.$ I.e., integral = reduced + irreducible, and we can compute dimension from the reduced induced structure.

To compute the dimension, use the following fact: since the scheme is irreducible, any open subscheme is dense (why?). In particular, this holds for an open affine.

Andrew
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  • You mean that $\textrm{dim}X=\textrm{dim}X_{red}$ where $X_{red}$ is induced reduced scheme of $X$?. Actually, I wonder why the dimension of open affine subset $U$ that contains a closed point should equal to dimension of $X$. I think density of $U$ is not enough for same dimension. – User0829 Jun 18 '13 at 16:12
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    @User0829 It is because dimension is defined as a topological property and as topological spaces $X$ and $X_{red}$ have the same topology. – Matt Jun 18 '13 at 16:21
  • @Matt Thanks. But I think that is not enough for reducing to affine. Dimension of $X$ is the supremum of dimension of $U$ (maximum in above case) hence we may choose open affine $U$ which has same dimension with $X$. Then any closed points cotained in that $U$? – User0829 Jun 18 '13 at 16:33
  • Dear @User0829, the definition of dimension I'm referring to here is on p.5 of Hartshorne, namely, it's the supremum over all lengths of chains of irreducible closed subspaces. Using this definition, we compare chains in $U$ versus in $X.$ – Andrew Jun 18 '13 at 16:43
  • @Andrew That is the definition exactly what I know, and I tried to compare chains. What I'm worrying is that there is am example that $\textrm{dim}U < \textrm{dim}X$ for dense open $U$. Can we say that such problem does not occur in this case? (I think I am confusing/missing something) – User0829 Jun 18 '13 at 16:54
  • Dear @User0829, I think I see. Are you thinking that maybe we take a maximal chain for $X$ that happens to lie on the boundary of $U$? This would be a problem. But for the reduction, we only need some open affine. So we can choose any open affine containing the closed point in the chain, and this will intersect all elements of the chain nontrivially. – Andrew Jun 18 '13 at 17:17
  • ps -- there will be a maximal chain, since our scheme is finite type over $k.$ – Andrew Jun 18 '13 at 17:21
  • @Andrew For using your argument, I think we have to find the maximal length chain of irreducible closed sets that starts with given closed point, don't we? But how can we find such a chain? With this, when we consider the induced reduced structure,the stalk of reduced structure is not isomorphic to the stalk of original structure. I think this could arise a problem(even though they are not ismorphic, they may have same dimension(as a ring), but I didn't check it) – User0829 Jun 18 '13 at 17:32
  • Of course, we can find maximal length chain. But how can we guarantee it starts with given closed point? – User0829 Jun 18 '13 at 17:36
  • Suppose given a maximal length chain. There exists an open affine $U$ containing the closed point, and $\dim X = \dim U.$ Thus we prove the claim first for $x\in U,$ i.e., suppose $X$ is affine. Once we prove this case, suppose $V$ is any other affine open in $X$. By what we just proved for affines, we know that $\dim V$ is the dimension of the stalk at a closed point in $U\cap V$ (which is open, dense), so equal to $\dim U$, but also we could take the stalk at any other point in $V$ including those outside of $U.$ Since $X$ is covered by open affines which intersect nontrivially, we're done. – Andrew Jun 18 '13 at 18:19
  • @Andrew I'm sorry for repating question. The thing that I counldn't understand in your argument is the existence of such open affine $U$. Actually, I asked it again and again in above comments but I think that I didn't convey it well, so you may not catch it. – User0829 Jun 18 '13 at 23:16
  • Dear @User0829, no need to be sorry! As I mentioned, we know there exists a maximal chain. The minimal element is a closed point. Any point in a scheme is contained in an open affine subscheme, say $U.$ Now we use the density of $U$ to argue that the maximal chain determines a chain of the same length in $U,$ and moreover that $\dim U = \dim X.$ Now we prove the theorem first for the open affine $U.$ But if there is some closed point outside of $U,$ what can we say? Well, we know there exists some open affine $V$ containing it. Also, $U\cap V$ is open and dense. Since the theorem holds for $V$ – Andrew Jun 19 '13 at 02:32
  • we know: first, that $\dim V = \dim U$ by comparing $\dim V$ to the dimension of a stalk at $x\in U\cap V,$ and thus, second, that the dimension at any other stalk in $V$ is also $\dim U$, even at $x\in V\setminus U.$ Does that make sense? So we begin with a maximal chain, and take a neighbourhood of the closed point. We might miss some points. But then irreducibility saves the day, once we prove the theorem for affines. In other words, if we want $\dim\mathcal O_{X,x}$, the equalities go: $\dim X = \dim U = \dim\mathcal O_{U,u}$ for all $u\in U$ which equals $\dim\mathcal O_{V,u}$ for all – Andrew Jun 19 '13 at 02:42
  • $u\in U\cap V$ which equals $\dim V = \dim\mathcal O_{V,x} = \dim\mathcal O_{X,x}.$ – Andrew Jun 19 '13 at 02:43
  • let us continue this discussion in chat – User0829 Jun 19 '13 at 02:44