Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $\phi:Y\to X$ be a morphism of $\Bbbk$-varieties. In Görtz and Wedhorn's Algebraic Geometry 1, the ramification index is defined by

$$e_Q(Y) := \mathrm{length}_{\mathcal{O}_{Y,Q}}\left(\mathcal{O}_{Y,Q}\right)$$

and then $e_{Q/P} := e_Q(Y_P)$ where $Y_P$ is the scheme-theoretic fiber of $P\in X$ under $\phi$, i.e. $Y_P=Y\times_X\mathrm{Spec}(\Bbbk(P))$. For points of codimension one and assuming that $X$ and $Y$ are smooth, I know a different definition: Namely, let $f$ be a uniformizing parameter at $P$, i.e. $\mathfrak{m}_P=(f)$ is the unique maximal ideal of $\mathcal{O}_{X,P}$. Let $v_Q:\Bbbk(Y)\to\mathbb{Z}$ be the valuation corresponding to $\mathcal{O}_{Y,Q}$. Then,

$$e_{Q/P} = v_Q(\phi^\sharp_Q(f))$$

where $\phi^\sharp_Q: \mathcal{O}_{X,P} \to \mathcal{O}_{Y,Q}$ is the induced map. I would like to know why these two definitions coincide (in codimension one). I frankly don't even have an idea where to start.

share|cite|improve this question
up vote 3 down vote accepted

In fact, I thought for a while and it became rather clear to me. Let's reduce to the affine case $Y=\mathrm{Spec}(B)$ and $X=\mathrm{Spec}(A)$. We can assume $f\in A$ and $P$ is a prime ideal of $A$. Then, we consider

$$\mathcal{O}_{Y_P,Q}=(B\otimes_A\Bbbk(P))_Q = (B\otimes_A(A_P/\mathfrak{m}_P))_Q = B_Q/(\mathfrak{m}_P\cdot B_Q) = B_Q/(f) $$

Now, if $g$ is a uniformizer at $Q$, then $f=u\cdot g^e$ where $u$ is a unit and $e=e_{Q/P}$ in the sense of the second definition. But by the above, $\mathcal{O}_{Y_P,Q}=B_Q/(g^e)$, which has length $e$ over itself.

I am not accepting this for a while, please comment if you notice any mistake.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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