Can every variety appear as singular locus?

Question

let $V$ be a projective variety. Does there always exist another variety $X$ such that $\operatorname{Sing}X=V$? Here $\operatorname{Sing}X$ means the singular locus of $X$ with reduced structure.

Answer 1

The answer is yes. Let $Y$ be any projective algebraic variety ($V$ in the OP). Then for some sufficiently large integer $n$, $\mathbb P^n$ contains two disjoint copies of $Y$ as a closed sub-scheme, i.e., $Y\amalg Y\cong Y'\subset \mathbb P^n =: X'$. I've chosen the notation to match Thm. 5.4 of [Ferrand, Daniel. Conducteur, descente et pincement. Bulletin de la Société Mathématique de France, Volume 131 (2003) no. 4, pp. 553-585. doi : 10.24033/bsmf.2455], according to which the locally ringed space $X:=X'\amalg_{Y'}Y$, obtained by glueing $\mathbb P^n$ to itself along the two copies of $Y$, is a scheme (and a fortiori a proper variety). Moreover, the morphism $f\colon X'\to X$ is finite, $Y\subset X$ is closed and $f$ is an isomorphism away from $Y$, i.e., $f\colon X'-Y' \xrightarrow{\cong} X-Y$.

Since $f$ is finite, generically an isomorphism and $X' = \mathbb P^n$ is smooth, hence normal, $f\colon X'\to X$ is the normalisation of $X$ and the singular locus of $X$ is exactly the locus above which $f$ is not an isomorphism. By construction, this is exactly $Y$, for every point of $Y$ has two pre-images under $f$ and $f$ is an isomorphism away from $Y$.

Note that there is nothing special here about $\mathbb P^n$; the construction works with any smooth projective variety $X'$ containing two disjoint copies of $Y$. But if $X'$ and $Y$ are not at least quasi-projective, then the assumptions of Ferrand's pinching theorem above may not be satisfied. Also, note that $X$ is proper, but not necessarily projective.

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[Edit: Below you find a previos attempt resulting in a reducible variety, which I did not want to delete in case it's helpful to anybody.]

This does not quite answer the question, for I presume that the varieties in question are assumed to be irreducible, but I have an answer in case $X$ is allowed to be reducible.

Suppose $V\subset \mathbb P^n$ is the vanishing set of the homogeneous polynomials $f_1,\dots,f_s\in k[x_0,\dots,x_n]$, say of degree $d_i:=\deg(f_i)$, and consider the product of the corresponding Veronese embeddings $$\nu=\nu_{d_1}\times\dots\times\nu_{d_s}\colon \mathbb P^n\to \mathbb P^{N_1}\times \dots \times \mathbb P^{N_s}.$$

In each factor $\mathbb P^{N_i}$, there is a hyperplane $H_i$ such that $\nu_{d_i}^{-1}(H_i)=V(f_i)\subset \mathbb P^n$. Hence $V=\bigcap_i V(f_i)=\bigcap_i \nu_{d_i}^{-1}(H_i)=\nu^{-1}(\prod_i H_i)\cong \nu(\mathbb P^n)\cap \prod_i H_i$, which happens to be the singular locus of the union of the two smooth varieties $\nu(\mathbb P^n)\cup \prod_i H_i$.