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The following is a well-known version of the famous Hironaka resolution.

Let $X$ be a compact complex space. Then there exists a finite sequence of blow-ups with smooth centers $$ X_N \stackrel{\sigma_N}{\longrightarrow} \cdots \longrightarrow X_j \stackrel{\sigma_j}{\longrightarrow} X_{j-1} \longrightarrow \cdots \longrightarrow X_1 \stackrel{\text { }}{\longrightarrow} X_0=X, $$ such that $X_N$ is non-singular (i.e. a complex manifold) and the center $C_{j-1}$ of the blow-up $\sigma_j$ is contained in $\left(X_{j-1}\right)_{\text {sing. }}$. If $X$ is projective, $X_N$ is also projective.

Question: If $X$ not normal, the the singular part $X_{sin}$ of $X$ is of codimsion $1$ (or $0$). Then can finitely many blow-ups resolve $X_{sin}$ to be smooth?

My analysis: Please note that in the current context, each blow-up operation is performed with a smooth center. If a center happens to have a codimension of $1$, it automatically qualifies as a Cartier divisor. As a result, the blow-up operation remains essentially isomorphic. Consequently, we have the flexibility to consistently assume that the center of every blow-up operation possesses a codimension of at least two. So how could it be possible to resolve $X_{sin}$ to be smooth in a finite many blowups?

KReiser
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Lelong Wang
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1 Answers1

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There are no issues in the non-normal case.

Singular part of codimension zero: generally one assumes that the singular object is reduced, and then in characteristic zero, this implies the singularities are a proper closed subset and hence cannot be codimension zero.

Singular part of codimension one: it may happen that a smooth center which is codimension one is not a Cartier divisor if it is contained in the singular set (this invalidates a lot of your analysis). Consider for instance a singular point of a curve, like the node on the curve cut out by $y^2=x^3+x^2$. This cannot be a Cartier divisor, else it would be a regular point (see here, for instance).

KReiser
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