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.
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. 


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 $X_{2}:=\lbrace tu= wv=0\rbrace$. 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 ${\Bbb C}^{4}$! 

