# Induced resolution of singularities

I am not a specialist in singularity theory but currently I have to touch resolution of singularities and I'd like to know whether I have understood Hironaka's theorem correctly.

Let $$k$$ be a field of characteristic zero and $$X$$ an $$k$$-variety and $$Z$$ a closed subscheme of $$X$$. Then Hironaka's theorem claims that there exists a log-resolution $$h: Y \longrightarrow X$$ of the pair $$(X,Z)$$, by this I mean:

• $$Y$$ is a smooth $$k$$-variety.
• $$h$$ is projective, proper birational morphism.
• $$h$$ induces an isomorphism outside $$X_{\mathrm{sing}} \cup Z$$.
• If $$E$$ is the exceptional locus then $$h^{-1}(Z) + E$$ has strict normal crossings.

and $$h$$ itself can be taken to be a composition of successive blowing-ups along smooth centers.

I am interested in the following situation: let's assume we still have $$(X,k)$$ as above and we are given a morphism $$f: X \longrightarrow \mathbb{A}^1_k$$. Let $$U$$ be any smooth open subscheme of $$X$$ and $$F = X \setminus U$$, hence $$X_{\mathrm{sing}} \subset F$$ so we could take a log-resolution of the pair $$(X,F \cup f^{-1}(0))$$ (one may assume $$f^{-1}(0)$$ is nowhere dense if necessary).

Let $$E_i$$ ($$i \in I$$) be the smooth irreducible components of $$h^{-1}(F \cup f^{-1}(0))$$ with corresponding multiplicities $$N_i \neq$$ ($$i \in I$$) , i.e. $$h^{-1}(F \cup f^{-1}(0)) = \sum_{i \in I} N_i E_i.$$ What I want to know is can we control the multiplicities in the way that there exists $$C \subset I$$ such that $$h^{-1}(f^{-1}(0)) = \sum_{i \in C} N_i E_i \ \ \ \text{and} \ \ \ h^{-1}(F) = \sum_{i \in I \setminus C} N_i E_i.$$ If it is the case, then does restricting to $$U$$ (which gives a new resolution since blowing-ups commute with flat base-changes) lose some multiplicities? $$\require{AMScd}$$ $$\begin{CD} Y_{\mid U} @>{}>> X\\ @VVV @VVhV\\ (U,f^{-1}(0) \cap U) @>{}>> (X,F \cup f^{-1}(0)) \end{CD}$$

This has nothing to do with your morphism to $$\mathbb A^1$$. You are asking if there is another closed subset $$F'$$ (which is secretly equal to $$h^{-1}(0)$$), then can you separate the exceptional divisors according to the pre-images of the two sets $$F$$ and $$F'$$. This depends on how $$F$$ and $$F'$$ intersect with respect to the resolution. If none of the centers of the blow ups in the given resolution is contained in $$F\cap F'$$, then this holds. Then $$C=\{i\in I\vert h(E_i)\subseteq F'\}$$ (and then, of course, $$I\setminus C=\{i\in I\vert h(E_i)\subseteq F\}$$). In this case, when you restrict to $$U$$, you do not "lose" any multiplicities. That could only happen if you lose one of the $$E_i$$, but that would only happen if $$h(E_i)\cap U=\emptyset$$, but that would mean that $$h(E_i)\subseteq F\cap F'$$, which we assumed does not happen.
The problem is that if $$h(E_i)\subseteq F\cap F'$$, then that particular $$E_i$$ appears in both $$h^{-1}(F)$$ and $$h^{-1}(F')$$. This happens for sure if for instance $$F\subseteq F'$$.
• So according to what you wrote, we may lose $N_i$ if we take base-change along an arbitrary open immersion? Commented Jan 5, 2023 at 21:09
• Yes, but I would rather say that you lose the $E_i$, not the $N_i$. The latter is a consequence of the former. Commented Jan 6, 2023 at 4:50