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Let $X=V(I)\subset \mathbb{C}^{n}$ be the vanishing set of an ideal of complex polynomials, let $\pi \colon \mathbb{C}^{n} \to \mathbb{C}^{n-1}$ be the projection onto the first $n-1$ coordinates and denote by $\tilde{X}$ the image of $X$ under $\pi$.

Question: Is it true that the set of regular points of $\tilde{X}$ is dense in $\tilde{X}$? By a regular point I mean a point at which $\tilde{X}$ looks like a complex-analytic manifold.

According to Wikipedia this is true in the reals case in the following sense: The projection of a vanishing set of real polynomials yields a semi-algebraic set and the smooth points (points at which the set is a subvariety) are dense in a semi-algebraic set.

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The set $\tilde{X}$ contains a dense open subset $U$ of an algebraic set $Z$: that is, $U \subset \tilde{X} \subset Z$. (This is because $\tilde{X}$ is a constructible set, by Chevalley's theorem.)

So density of the smooth points in $\tilde{X}$ is equivalent to the same statement for $Z$.

Now it is a basic theorem, which I am guessing you already know (if not, see Chapter 2 of Shafarevich), that the smooth points are dense in any variety over the complex numbers.

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  • $\begingroup$ In general $\tilde{X}$ is only constructible in $\mathbb C^{n-1}$, right ? $\endgroup$ – Cantlog Feb 21 '14 at 17:21
  • $\begingroup$ Dear @Cantlog, you're right. I thought I had an argument that in the special case of a projection, what I wrote is actually true, but I think that's not right. I'll edit. $\endgroup$ – user64687 Feb 22 '14 at 9:41
  • $\begingroup$ Just to be sure that I got everything right: The fact that I have a projection is not important. If $X \subset \mathbb{C}^{n} $ is an algebraic variety and $f \colon X \to \mathbb{C}^{k}$ is a polynomial map, then the image is constructible, which yields that a subset $U$ of $f(X)$ is open and dense in the Zariski-closure of $f(X)$. $\endgroup$ – Sebastian Feb 24 '14 at 11:14
  • $\begingroup$ Dear @Sebastian: yes, that's true: the fact that you consider a projection doesn't play a role in my argument, or in the conclusion. (Of course, there may be a different argument for the case of a projection that doesn't invoke Chevalley's theorem.) $\endgroup$ – user64687 Feb 24 '14 at 11:48
  • $\begingroup$ @Asal Following your argument I only get that the smooth points are dense are dense in the Zariski topology. Do you have an argument whether the smooth points are dense in the Euclidean topology as well? $\endgroup$ – Sebastian Feb 24 '14 at 17:46
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It is also true that the smooth points are dense in the Euclidean topology: Let $X = X_{1} \cup \dots \cup X_{l}$ be the decomposition into irred. components and $f \colon \mathbb{C}^{n} \to \mathbb{C}^{k}$ a polynomial map. Then $f(X_{i})= : \tilde{X}_{i}$ is a constructible and irreducible set containing a set $U_{i}$ which is open and dense in its Zariski closure $Z_{i}$ of $\tilde{X}_{i}$, which is irreducible since irreducibility is preserved by taking the closure. Open subset are constructible sets, and the Euclidean and the Zarisi closure of constructible subsets in irreducible varieties coincides (see Mumford, p.60). So both $U_{i}$ and $Z_{i}^{reg}$ are Euclidean-dense, open subsets of $Z_{i}$, hence the same holds for the intersection, which is an open and dense subset of $X_{i}$.

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Yes. Recall (Tarski-Seidenberg Theorem) that projection of semi-algebraic set is semi-algebraic set. Then $\tilde{X}$ is semi-algebraic set, since a complex algebraic set is a real algebraic set. Now semi-algebraic sets has stratification smooth, say $\tilde{X}=\bigcup\limits_{d=1}^k X_d$, ($X_k\not= \emptyset$) with $X_d$ being a finite union of cell of dimension $d$ or empty (http://perso.univ-rennes1.fr/michel.coste/polyens/RASroot.pdf, Theorem 1.3), in particular, $Reg(\tilde{X})=X_k$. But $\mathcal{H}_k(A)=0$ if $A$ is submanifold with $\dim A<k$, then $$ \mathcal{H}_k(X\setminus X_k)\leq \sum \limits_{d=0}^{k-1}\mathcal{H}_k(X_d)=0. $$ Therefore $X_k$ is dense in $\tilde{X}$.

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