Quail Entry: Hurwitz Thm

Around 12/21/09 I looked into how to prove Hurwitz thm (but I didn’t write down some of the details until 1/1/10).  I more or less followed Hartshorne, but the proof comes as the corollary of a number of a number of other results whose full strength is not needed for just Hurwitz thm.  Below is roughly how I unfolded the proof of Hurwitz’s thm…

Hurwitz: X,Y curves (complete over alg. cl. field (also nonsingular!)) and f \colon X \to Y is finite.  Then

2g_x - 2 = \deg f (2g_y - 2) + \deg R (1)

Where R is the ramification divisor defined as R = \sum_{p \in X} length(\Omega_{X/Y})_p \cdot p.

The first step is to interpret Hurwitz as a statement about sheaves of differentials, this also leads to a better understanding of ramification divisor.  \deg \Omega_{X/k} = 2g_x -2 (this can be proved via Riemann Roch), in any case Hurwitz now can be written as

\deg \Omega_{X/k} = \deg f \deg \Omega_{Y/k} + \deg R (2)

Any by definition R is closely related to \Omega_{X/Y}.  These three sheaves of differentials fit into a s.e.s

0 \to f^*\Omega_{Y/k} \to \Omega_{X/k} \to \Omega_{X/Y} \to 0 (3)

pf: general theory of differentials associated to ring maps A \to B \to C says exactness holds everywhere except possibly the injection

*Note the exact seq. of differentials associated to A \to B \to C is in some ways analogous to the l.e.s of a triple in homology.

to show the first map is injection the key is to show the maps is nonzero at the generic point; indeed is the map is not injective, the something nonzero maps to zero, but at the generic point everything nonzero is invertible to this would say a unit mapped to zero, hence the map would be zero.  Now the result follows because

\Omega_{X/Y} is 0 at the generic point, means the first map is surjective so certainly not zero.  For \zeta the generic point, \Omega_{X/Y, \zeta} = \Omega_{K(X)/K(Y)}.  An algebra thm says that the dimension of this space is \ge tr. deg K(X)/K(Y) = 0, also K(X)/K(Y) is separable, and in the separable case the inequality is an equality, so the space in equation is zero.

Now describe the sheaves of differentials and maps in terms of local parameters. Let t be a loc. parm at p \in X and s a loc. par. for q = f(p).  Now

\Omega_{X/k, p} = O_{X,p} \cdot dt.  The fact that its free of rank 1 follows from an alg. result of the form \Omega_{B/k} is a free B module of rank \dim B for B a loc. ring whose residue field is isomorphic to k; also require k to be perfect and B to be a localization of a finitely generated k algebra.  That dt is a generator follows from Nakayama’s lemma and that dt becomes generator after modding out by max ideal.

Get something similar for \Omega_{Y/k,q}.  Localize (3) at p.  Have O_{Y,q} \to O_{X,p} and its local map, means s \mapsto ut^{e_p} for some unit u; the natural number e_p is called the ramification index.  If the characteristic doesn’t divide e_p then the ramification is tame; otherwise its wild.

When all the ramifications are tame, \deg R = \sum_p (e_p -1): first map in localized version of (3) is ds \mapsto d(ut^{e_p}) = du \cdot t^{e_p} + ue_p t^{e_p - 1} dt.  If ramification tame, then image of injection is t^{e_p -1}dt hence quotient \Omega_{X/Y,p} has length e_p -1 given by dt \subset (dt, tdt) \subset ... \subset (dt, ..., t^{e_p - 2}dt).  But if the ramification is wild, then ds = t^{e_p}du and du = t^kdt for k\ge 0; e.g. our unit could be u = 1 + t^{k+1} then du = (k+1)t^k dt.  So in general the length is \ge e.

The upshot is considering R as a closed subscheme, get O_R \cong \Omega_{X/Y}


Now use (3) and state via divisors. The canonical divisors are defined such that O(K_X) = \Omega_{X/k} So tensoring (3) with the inverse of this sheaf we get

0 \to f^*O(K_Y)\otimes O(K_X) \to O_X \to O_R \to 0

Would like to say f^*O(K_Y)\otimes O(K_X) \cong O(f^*K_Y - K_X) \cong O(-R) and hence conclude K_X \sim f^*K_Y + R and be done by taking degrees, but for this need iso

Compatibility (ex. II.6.8): f,X,Y above and D \in CL(Y) then f^*O(D) \cong O(f^*D).

Recall on divisors f^* is first defined on a point via f^*q = \sum_{p \mapsto q} e_p \cdot p and then extended by linearity.  Note the curves being nonsingular implies they are locally factorial so can use Weil = Cartier = Pic. The problem essentially asks if the following two maps agree

D \mapsto f^*D \mapsto O(f^*D)

D \mapsto O(D) \mapsto f^*O(D) (4)

Can reduce to the case D = pt

If it holds for a point, then for general D = \sum_i n_i \cdot p_i have f^*O(D) = f^*\bigl(\otimes_i O(p_i)^n_i \bigr).

It holds very generally (for sheaves of modules) that f^*(F \otimes G) \cong f^*F \otimes f^*G.  Enough to show iso on the level of presheaves because then the left side is the sheafifacation of the presheaf on the right which is the right side.  And on the level of presheaves it boils down to the following isomorphism

B \otimes_A(M\otimes_AN) \cong (B \otimes_A M) \otimes_B (B\otimes_AN)

and this can be proved by showing the left side has the universal property of the right side.

the upshot f^*O(D) = \otimes_i [f*O(p_i)]^{n_i} = \otimes_i O(f^*p_i)^{n_i} = O(\sum_i n_i f^*p_i) = O(f^*D)

(point to justify!; made some remarks in ogus excerpts) f \colon X \to Y is flat, this implies f^* is exact (note f^{-1} is always exact).  So for q \in Y both of the following two sequences are exact

0 \to O(-q) \to O_Y \to O_q \to 0

0 \to f^*O(-q) \to O_X \to f^*O_q \to 0

Remains to show f^*O_Q \cong O_{f^*Q} note this makes sense because f^*D is effective; the subschemes x, 2x are differentiated in that the associated ideal sheafs are m_x, m_x^2

Clear to see f^*O_Q only supported on f^{-1}Q.  Say P \mapsto Q, .  Now take stalks of the s.e.s (4).  Seems like a general fact that (f^{-1}F)_p = F_p this arises essentially because

\varinjlim_{V \ni p}\varinjlim_{U \supset f(V)} = \varinjlim_{U \ni f(p)}

parametrize the same sets.  It follows that (f^*O(-q))_p = O_{X,p} \otimes O(-q)_q, now O(-q)_q = m_q = (s) more or less because m_q is the ideal sheaf that cuts out the subscheme q; using the local parameters above, i.e. that s \mapsto ut^{e_p} it follows that

(f^*O(-q))_p = O_{X,p} \otimes_{O_{Y,q}} m_q \cong (t^{e_p}) = m_p^{e_p}

So localizing (4) at each p \in f^{-1}q produces

0 \to O(-e_p \cdot p) \to O_{X,p} \to f^*O_q \to 0

hence f^*O(-q) = O(-f^*q) which also shows f^*O_q = O_{f^*q}

This gives all the pieces of Hurwitz thm.


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