# multiplicities of flat pullbacks

Let $f:X\to Y$ be a flat morphism between two Noetherian, integral, regular schemes. Let $D\in \operatorname{Div}(Y)$ any (Weil or Cartier) divisor, in this setting it has sense to consider the pullback $\,f^\ast D\in\operatorname{Div }(X)$.

Define the ramification index of $f$ at $x$ as: $$e_x:=\operatorname{length}_{\mathcal O_{X,x}}\frac{\mathcal O_{X,x}}{\mathfrak m_{f(x)}\mathcal O_{X,x}}$$

I'd like to show that the following formula is true:

$$f^\ast D=\sum_{x\in\star} e_x\operatorname{mult}_{f(x)}(D)\overline{\{x\}}$$ where $\star$ is the set of points $x\in X$ of codimension $1$ such that also $f(x)$ has codimension $1$. This formula is very important indeed it gives an explicit way to calculate the pullback of a divisor.

Can you give an hint for the proof or give any references? For instance the Stack project reference given here is wrong (probably the document has been updated).

• The correct reference in the cited Stack-project document is 51.13, so maybe it was a typo. – Jürgen Böhm Mar 13 '17 at 22:49
• Ok, it seems reasonable. But how can I use it? – notsure Mar 13 '17 at 22:59
• I am just writing up an answer. – Jürgen Böhm Mar 13 '17 at 23:00

$\require{AMScd}$ $\newcommand{\spec}[1]{\mathrm{Spec}(#1)}$ $\newcommand{\tensor}{\otimes}$ $\newcommand{\ideal}[1]{\mathfrak{#1}}$

Let $f^*D$ be the left side of the equation and $f^\sharp D$ the right side of the equation. Then we note first that both sides are linear in $D$.

Consider the square $$\begin{CD} X @<j'<< X' \\ @VVfV @VVf'V \\ Y @<j<< Y' \end{CD}$$ where $j:Y' \to Y$ and $j':X' \to X$ are open immersions, and $f' = f|_{X'}$.

We assume, that $f'^*D = f'^\sharp D$ and we note

$$j'^\sharp f^\sharp D = f'^\sharp j^\sharp D$$

and

\begin{align} j^* D = j^\sharp D \\ j'^* E = j'^\sharp E \end{align} for all divisors $D$ on $Y$ and $E$ on $X$.

Then $$$$j'^* f^* D = f'^* j^*D = f'^\sharp j^*D = f'^\sharp j^\sharp D = j'^\sharp f^\sharp D = j'^* f^\sharp D$$$$ As $X'$ and $Y'$ can vary over a cover of $f:X \to Y$ we have $f^*D = f^\sharp D$ provided this holds for every $f':X' \to Y'$ where $X'=\spec{B}$ and $Y'=\spec{A}$ are open affine subschemes of $X$ and $Y$.

So we can assume that $X=\spec{B}$, $Y=\spec{A}$ and (because of linearity and locality) that $D=(a)$ where $a$ is a nonzerodivisor of $A$. Then $f^*D = (a)$ where $a$ is considered as element of $B$. It is a non-zero-divisor because $-\tensor_A B$ is exact on $0 \to A \xrightarrow{\cdot a} A$.

Now choose a minimal prime $\ideal{q} \supseteq aB$, so that $\ideal{p} = \ideal{q} \cap A \supseteq (a)$ is a minimal prime over $aA$ (this follows from the going down property of the flat $B/A$).

The equation to be proved then reduces to the form

$$\mathrm{len} B_\ideal{q}/a B_\ideal{q} = \mathrm{len} (B_\ideal{q}/\ideal{p} B_\ideal{q}) \mathrm{len} (A_\ideal{p}/a A_\ideal{p})$$

But this follows from the cited Stacks 51.13 with $A = A_\ideal{p}$, $B = B_\ideal{q}$ and $M = A/aA$.