TITLE: trying to understand the support of the sheaf of relative differentials
QUESTION [5 upvotes]: So I'm trying to understand a proof of Belyi's theorem from http://eprints.soton.ac.uk/29785/1/b45h1koe.pdf
specifically lemma 3.4.
The setup is as follows: Let $X/\mathbb{C}$ be a curve, and let $t : X\rightarrow\mathbb{P}^1_\mathbb{C}$ be a meromorphic function on $X$ thought of as a covering map of degree $n$. Further, let $\text{Crit}(t)$ denote the critical points of the cover $t$ - ie, the points in $\mathbb{P}^1_\mathbb{C}$ which have fewer than $n$ pre-images under $t$.
Then, he claims that $\text{Crit}(t) = t(\text{supp}(\Omega^1_{X/\mathbb{P}^1_\mathbb{C}}))$.
Now, it's my understanding that the critical points of $t$ should be the images of the ramification points of $t$ under $t$, so I've been trying to understand why it should be the case that the sheaf of relative differentials of $X/\mathbb{P}^1_\mathbb{C}$ should be nonzero only on the ramification points (or at least only above the critical points).
To this end, I'm trying to understand the definition given in Hartshorne (III.8), namely:
$\Omega^1_{X/\mathbb{P}^1_\mathbb{C}} = \Delta^*(\mathcal{I}/\mathcal{I}^2)$, where $\Delta : X\rightarrow X\times_{\mathbb{P}^1_\mathbb{C}} X$ is the diagonal map, and $\mathcal{I}$ is the sheaf of ideals of the image $\Delta(X)$ in some open subset $W\subset X\times_{\mathbb{P}^1_\mathbb{C}} X$.
I kind of understand sheaves of ideals (they're essentially functions on the ambient space that vanish on the closed subscheme), but I'm still not very comfortable with the notion of $\Delta^*(\mathcal{I}/\mathcal{I}^2)$ (in this case defined to be $\Delta^{-1}\mathcal{I}/\mathcal{I}^2\otimes_{\Delta^{-1}\mathcal{O}_{X\times X}}\mathcal{O}_X$, where the fibred product is taken over $\mathbb{P}^1$).
Any comments on how I should think of $\Delta^*(...)$ and why the sheaf of relative differentials only have nonzero stalks at ramification points would be awesome!
thanks.

REPLY [4 votes]: Ravi Vakil has a good explaination for the definition $\Delta^*(I/I^2)$ in his notes. See his AG notes here or here (chapter 23). In particular, I guess thinking about this locally makes it a little clearer what's going on, in terms of derivations etc. Also, when $X$ is smooth, it is instructive to see that this really gives the cotangent bundle on $X$. 
As for your question about ramification points: Let $f:X\to Y$ be a finite morphism of curves (I will assume that these are smooth in the following). It is useful to have in mind the exact sequence $$0\to f^*\Omega_{Y}\to \Omega_X \to \Omega_{X|Y}\to 0.$$(This is exact at the right in the smooth case, but not in general). Note that $\Omega_{X|Y}$ is a torsion sheaf since the two other sheaves are locally free of the same rank (they are line bundles on $X$). At a point $q\in Y$ and $p\in X$ in the preimage of $q$, let $dx$ denote a generator for $\Omega_{Y,q}$ as a $O_{Y,q}$-module. Now, $(\Omega_{X|Y})_P=0$ if and only if $f^*dx$ is a generator of $\Omega_{X,p}$, which happens if and only if $f$ pulls back a local parameter to a local parameter, that is $p$ is unramified. Moreover, the exact sequence above shows that the ramification index is exactly the length of the sheaf $\Omega_{X|Y}$. Finally, note that this sequence gives the Riemann hurwitz formula, relating the canonical divisors of $X$ and $Y$ and the ramification divisor of $f$.

REPLY [2 votes]: Let $\pi:X\to Y$ be a finite morphism of curves.
Then, for any point $x$ in $X$ lying over $y$ in $Y$, the coefficient $v_x(\pi)$ of $\Omega_{\pi}$ is the valuation of the different of the extension of dvr's $\mathcal{O}_{y}\subset \mathcal{O}_x$. If you are working in characteristic zero, then $$v_y(\pi) = e_x-1,$$ where $e_x$ is the ramification index. So you see that $\Omega_{\pi}$ is supported on the ramification points.
Also, you have a short exact sequence (it's on page 2 of Chapter IV.2 in Hartshorne) which relates $\Omega_\pi$ with $\Omega_X$ and $\Omega_Y$. The above actually shows the important Riemann-Hurwitz formula: $$K_X = \pi^{\ast} K_Y + R.$$ Here $R$ is the ramification divisor. This equals $\Omega_{\pi}$ in this case.