10
$\begingroup$

Is there a resolution of singularities for flat families?

More precisely, if $X \rightarrow \mathbb{A} ^n$ is a flat map, does there exist a map $Y \rightarrow X$ such that, for every $p \in \mathbb{A} ^n$, the fiber map $Y_p \rightarrow X_p$ is a resolution of singularities? Can one require, moreover, that the map $Y \rightarrow \mathbb{A} ^n$ is smooth?

$\endgroup$

2 Answers 2

11
$\begingroup$

I assume you want $Y \to X$ to be proper. The answer is a definite no, in general. For example, take a polynomial $f: \mathbb A^2 \to \mathbb A^1$; such a $Y$ would have to be finite over $\mathbb A^2$, and birational, so $Y = \mathbb A^2$. There are lots of counterexamples in higher dimension too: for example, it follows from the purity theorem that you usually can't have a simultaneous resolution when $X$ is smooth. Thus, for example, in the very simple example $f\colon \mathbb A^3 \to \mathbb A^1$, $f(x, y, z) = x^2 + yz$, in which the only singular fiber is over the origin, and it has the simplest kind of surface singularity, of type $A_1$, you don't have a simultaneous resolution.

There are some non-trivial results, but they require the base to be 1-dimensional, and they require a base change on the base to get the resolution. For example, in the example above if one makes a base change $t \mapsto t^2$ on the base and normalizes, one has a simultaneous resolution. This is particular case of a theorem of Brieskorn: see for example M. Artin, Algebraic construction of Brieskorn's resolutions, Journal of Algebra 29 (1974). This is only possible in very particular cases, though.

$\endgroup$
7
$\begingroup$

This is false in general. Take $f:\mathbb A^2\to \mathbb A^1$, $(x,y)\mapsto xy$. Any attempt to resolve the singular fiber will bring in a new component in the fiber, so it remains singular.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.