Simple normal crossings divisor and locally monomial functions

Question

Let $X$ be a smooth algebraic variety with $\varphi$ a regular function on $X$. Let $X_0=\varphi^{-1}(0)$ and suppose that $X_0\subset X$ is a reduced, singular subvariety. Let $\pi:\widetilde{X}\to X$ be a strong resolution of the pair $(X,X_0)$. Denote $\widetilde{X}_0=\pi^{-1}(X_0)$. Denote $\widetilde{\varphi}=\varphi\circ\pi$.

Since $\widetilde{X}$ is a strong resolution of the pair $(X,X_0)$ we have that $\widetilde{X}_0$ is a simple normal crossings divisor- this means that in a Zariski neighborhood of any point $p\in \widetilde{X}_0$, one may choose coordinates so that the divisor $\widetilde{X}_0$ looks like an intersection of coordinate hyperplanes. This all makes sense to me.

Here's where my confusion comes in. I have seen the following claim: there exists a coordinate system near $p$ such that $\widetilde{\varphi}$ is monomial, ie there exist an invertible $\alpha$, coordinates $x_1,\cdots,x_n$, and nonnegative integers $a_1,\cdots,a_n$ such that on a neighborhood of $p$, we have $$\widetilde{\varphi}=\alpha x_1^{a_1}\cdots x_n^{a_n}$$

It seems intuitively to me like the proof of this statement should be more or less a one-liner. Unfortunately, the source does not contain it, I can't find that argument elsewhere, and I can't think of any good arguments to prove this. The best I can do is to say the following:

Since $\widetilde{\varphi}$ vanishes on $\widetilde{X}_0$, it should be in $I(\widetilde{X}_0)$, which is locally principal, so I can write it as $\sum f_ig$ where $g$ is locally a generator of $I(\widetilde{X}_0)$. I can then collect terms... but this argument seems wrong (what about something like $x^2y+xy^2$, for instance?).

Why is the claim about $\widetilde{\varphi}$ being locally monomial true?

Answer 1

I figured this out after a while of sitting on it.

The solution I started to present need a little addition to it, but is largely correct. In the notation of the post,

Since $\widetilde{\varphi}$ vanishes on $\widetilde{X}_0$, it should be in $I(\widetilde{X}_0)$, which is locally principal, so I can write it as $\sum f_ig$ where $g$ is locally a generator of $I(\widetilde{X}_0)$. I can then collect terms to write the function as $h(x_1^{a_1}\cdots x_m^{a_m})$ where $a_i$ are maximal non-negative integers. Then $h$ cannot vanish on any subvarieties intersecting $\widetilde{X}_0$ by the stipulation that $V(\widetilde{\varphi})=\widetilde{X}_0$. So $h$ is locally a unit, and the proof is finished.

Answer 2

The resolution $\pi$ is obtained by a sequence of blowups in smooth centers transversal to the actual exceptional divisor $E$ at each stage (this divisor is already assumed to be normal crossings). The blowup thus preserves the nc property of the exceptional divisor.

The variety $X_0$ you wish to resolve undergoes the following transformation. Let $J$ be its ideal in $X$. At each step, it receives a principal monomial factor $M$ supported by the exceptional divisor, say, $J$ admits a factorization $J = MI$, for some ideal $I$. One now proves that the order (at each point) of the ideal $I$ eventually goes down to $0$, so at the end we will have $J = M$ (in the local rings). This just means that the variety $X_0$ has become a normal crossings divisor.

Consider the following example: $X_0 = V(x^3 - y^2)$ in $k^2$. The first blowup makes $X_0$ smooth (a parabola), but it will be tangent to the exceptional curve $E$. The next blowup makes $X_0$ transversal to the two components of $E$, but as three smooth curves meet, this is not yet nc. A final third blowup achieves this.