User leonardo - MathOverflow

Answer by Leonardo for sums of rational squares

2012-03-01T10:51:43Z

Dear Michael,

concerning your first question, I think that Franz's proof is really in the spirit of Fermat's techniques. Concerning the second question, here is a short, elementary proof, inspired by a theorem of Thue (cf. exercice 1.2 in Franz's book Reciprocity laws). I tried to write it using only notions known to Fermat.

We want to prove that if there exist positive integers $m,n,a$ and $b$ such that $$m^2n=a^2+b^2,$$ then $n$ is itself the sum of two integer squares. It is easily seen that it is sufficient to prove this assertion under the hypothesis that $a$ and $b$ (and therefore $a$ and $n$) are coprime and that $n$ is not a square. Let $t$ be the unique positive integer such that $t^2< n<(t+1)^2$. Since there are $(t+1)^2>n$ integers of the form $au+bv$ with $0\leq u,v\leq t$, it follows that $n$ divides the difference $a(u-u')+b(v-v')$ of two of them. Setting $x=u-u'$ and $y=v-v'$, we have the inequalities $|x|,|y|\leq t$. The integer $n$ then divides $a^2x^2-b^2y^2$; since it also divides $a^2y^2+b^2y^2$, it divides their sum, which is equal to $a^2(x^2+y^2)$. Now, the integers $a$ and $n$ being coprime, it follows that $n$ divides $x^2+y^2$. The inequalities $0< x^2+y^2<2n$ finally imply that $n=x^2+y^2$.

Answer by Leonardo for an easy example of valuation ring which is not noetherian？

2011-11-28T22:17:24Z

Consider the subring $A$ of $\Bbb Q_p(X)$ consisting of rational functions defined at $X=0$ and such that $f(0)\in\Bbb Z_p$. In other words, let $B$ denote the localization of the ring $\Bbb Q_p[X]$ at the maximal ideal $(X)$ and set $A=\Bbb Z_p+XB$. It is a two-dimensional valuation ring which is therefore not noetherian (cf. Damian Rössler's comment above).

Answer by Leonardo for Etale covers of the affine line

2011-05-03T02:33:05Z

As stated by David Speyer and Clark Barwick, the awnser to the second question is the following:

Any smooth projective curve $C$ defined over a field $k$ of positive characteristic $p$ can be realized as a finite cover of the projective line only ramified above one point.

Here is a short constructive proof only based on Riemann-Roch theorem. It can be considered as an illustration of Kedlaya's proof, only dealing with curves.

First of all, there exists a generically étale finite cover $C\to\mathbf P^1$, induced by a rational function $f\in k(C)$ (in fact, any element of $k(C)-k(C)^p$ will do the job).
Denote by $R\in$ Div$(C)$ the (reduced) ramification divisor of the above cover (i.e. the ramified points are couted without multiplicity). From Riemann-Roch theorem, for large $n$, there exist a rational function $g\in k(C)$ having a pole of order $n$ at each point of the support of $R$. We may take $n$ strictly greater than $\frac{\deg(f)}p$.
Then, the rational function $h=f+g^p$ induces a cover $C\to\mathbf P^1$ only ramified above infinity.

Answer by Leonardo for Generalizations of Belyi's theorem

2011-05-02T07:25:33Z

There is a nice generalization of Belyi's theorem in positive characteristic, proved by M. Saïdi in his paper Revêtements modérés et groupe fondamental de graphe de groupes. (Compositio Math. 107 (1997), no. 3), Théorème 5.6:

Let $C$ be a smooth projective curve defined over a field $K$ of characteristic $p>2$. The following conditions are equivalent:

The curve $C$ can be defined over $\bar{\mathbf F}_p$,
There exists a finite cover $C\to\mathbf P^1$ tamely ramified above $\infty,0$ and $1$ (and étale elsewhere).

The proof, very short and elegant, relies on a result of Fulton on the existence of covers with only double ramification. Unfortunately, the argument does not apply in characterisctic $2$, for which the question is still open (some recent progress in this situation can be found in S. Schröer's paper Curves with only triple ramification, Ann. Inst. Fourier (Grenoble) 53 (2003), no. 7).

Answer by Leonardo for What are dessins d'enfants?

2011-05-02T07:52:32Z

Historically, one of the first papers on th subject is Drawing curves over number fields, by G.B. Shabat and V.A. Voevodsky (The Grothendieck Festschrift, Vol. III, 199–227, Progr. Math., 88), which I strongly recommend. Another nice and historical paper is Triangulations, by M. Bauer and C. Itzykson (many references: R.C.P. 25, Vol. 44 (1992), Discrete Math. 156 (1996), no. 1-3 or L. Schneps's book below). Both papers are also concerned with the (combinatorial) cellular decomposition of moduli spaces of curves. They appeared before L. Schneps book The Grothendieck theory of dessins d'enfants (London Mathematical Society Lecture Note Series, 200), which is now the main reference on the subject.

Answer by Leonardo for Elliptic curves over finite fields

2011-05-01T20:43:25Z

I completely agree with the earlier answers. Just two remarks...

For question 3, If $K$ is finite of cardinality $q$, then $E(K)$ is isomorphic to $\mathbf Z/n\mathbf Z\times\mathbf Z/m\mathbf Z$, where $n$ divides gcd$(q-1,m)$.
Concerning your last question, here is a simple example where you can explicitly 'see' all the $K$-rational points without direct computations; I hope it may interest you. Consider the elliptic curve $$E:Y^2=X^3+1$$ defined over the finite field $K=\mathbf F_p$, where $p=3n+2$ is a prime number. Then, there is a bijection $\varphi:K\to E(K)-\lbrace O\rbrace$ (where $O$ is the point at infinity) given by $$\varphi(t)=\left((t^2-1)^{2n+1},t\right).$$

Answer by Leonardo for Branch locus of the Galois closure of a Belyi morphism

2011-04-30T22:19:32Z

As correctly awnsered by Dimitri, given a Belyi map $f:X\to Y$, there exists a finite cover $g:Z\to X$ such that the composition $h=f\circ g:Z\to Y$ is Belyi and Galois (this is true in the more general setting of finite covers of curves, namely, the Galois closure of a cover $X\to Y$ does not change the branch locus in $Y$). Working in characteristic $0$, the description of the branch locus $B'$ of $g$ essentially follows from Abhyankar's lemma: first of all, $B'$ is contained in the set $f^{-1}(\lbrace\infty,0,1\rbrace)\supset R$. Let $e_\infty$ be the lcm of the ramification indices of the points of $R$ lying in the fiber $f^{-1}(\infty)$. Then a point point $P\in f^{-1}(\infty)$ belongs to $B'$ if and only if its ramifiaction index is strictly less than $e_\infty$. The same applies also for the points in $f^{-1}(0)$ and $f^{-1}(1)$ and leads to a complete description of $B'$.

Comment by Leonardo

2011-11-28T22:20:09Z

You can also replace $\Bbb Q_p$ by $\Bbb Q$ and $\Bbb Z_p$ by $\Bbb Z_{(p)}$, the localization of $\Bbb Z$ at the prime $p$.

Comment by Leonardo

2011-11-20T17:29:24Z

Even simpler: x=a, y=b^{-1}a.

Comment by Leonardo

2011-06-27T00:55:33Z

Wushi Goldring wrote an expository paper focussed on Belyi's theorem and its generalizations. It will appear on Serge Lang memorial volume and it contains Saïdi's proof (Theorem 4.6). You can find the preprint at math.harvard.edu/~wushi

Comment by Leonardo

2011-05-03T12:01:25Z

You're welcome! And yes, you're right, no contradiction with the nice example of Francesco: the Galois closure of his degree 3 cover is ramified over 6 points (and not 3, but the criterion works in full generality), with ramification index 2 (the gcd of the ramification indices over each branch point). Therefore, g is unramified over R and ramifies exaclty over R'.