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Hi.

Let $X$ be a pure $n$-dimensional complex subspace of manifold $Z$. It is true that $X$ has no embedded components? (perhaps that is obvious with Weierstrass preparation theorem or Noether normalisation theorem...)

Thank you.

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Plenty of algebraic examples give analytic examples upon analytification; e.g., $z^2 = zw = 0$ in the $(z,w)$-plane $Z$ is topologically the $w$-axis (so pure dimension 1) and has the origin as an embedded point (in the sense that the maximal ideal of the local ring at the origin is the annihilator of $z$ in that local ring; this even holds in the completion of the local ring). –  Boyarsky Jul 4 '10 at 11:42
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I think it often depends also on what the writer means when he says "pure dimensional". Boyarsky's example is certainly right though. On the other hand, in Eisenbud's book "Commutative algebra with a view towards algebraic geometry", Eisenbud defines "pure codimension 1" in terms of associated primes, not just minimal primes. In otherwords, Boyarsky's example is not pure \emph{codimension} 1 from Eisenbud's terminology. –  Karl Schwede Jul 4 '10 at 13:26
    
The question involves "pure dimension", which I have never seen used in the literature with a meaning that is sensitive to associated primes (and likewise for "pure codimension"). Does the algebraic geometry (as opposed to commutative algebra) literature use such alternative terminology? –  Boyarsky Jul 4 '10 at 17:05
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Usually when I see pure dimensional, it is in reference to reduced schemes/reducible varieties and so the issue of embedded primes would not show up. This is how it's done in Harris for example. I don't have strong feelings on this though, so you could be right about pure dimensional for schemes (but if it's non-standard...?) I've done a quick search of several canonical algebraic references and didn't find anything more illuminating unfortunately (I looked in the red book, Hartshorne, Bruns and Herzog, both Matsumuras, Nagata, Eisenbud and Harris, Eisenbud, Harris, Atiyah-Macdonald, etc). –  Karl Schwede Jul 4 '10 at 18:45
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1 Answer

The claim is false.

1) In analytic geometry, a complex space $X$ is said to be pure dimensional if all irreducible component of the reduction of $X$ have the same dimension (as Krull dimension of local ring).

In fact, we can produce many examples of pure dimensional space with embedded component obtained by base change. We can consider the simple and classical example of union of two plane which project on $C^{2}$:

Let $X:=\lbrace(u,v,t,w)\in {\Bbb C}^{4}: uv=uw=ut=tw=0\rbrace$ be the union of two planes which can rewriting (by change of coordinates) as the union of $X_{1}:=\lbrace t=w=0\rbrace$ and $X_{2}:=\lbrace t-u= w-v=0\rbrace$. Consider the projection $f:X\rightarrow {\Bbb C}^{2}$, $(u,v,t,w)\rightarrow (u,v)$.

Then $X$ is $2$-pure dimensional, $f$ is finite, open (universally open in Alg.geom) and surjective (but no flat because $X$ is not Cohen Macaulay!). denote by $Y$ the fiber product $X\times_{{\Bbb C}^{2}} X$. Then the canonical morphism (deduced by base change) $g:Y\rightarrow {\Bbb C}^{2}$ is finite, open and surjective too with $Y$ pure dimensional but not reduced and with an embedded component: the origin in ${\Bbb C}^{4}$!

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