User leonardo - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T01:34:42Z http://mathoverflow.net/feeds/user/14809 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/88539/sums-of-rational-squares/89948#89948 Answer by Leonardo for sums of rational squares Leonardo 2012-03-01T10:51:43Z 2012-03-01T21:30:45Z <p>Dear Michael,</p> <p>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 <em>Reciprocity laws</em>). I tried to write it using only notions known to Fermat.</p> <p>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 $t^2&lt; n&lt;(t+1)^2$. 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 $0&lt; x^2+y^2&lt;2n$ finally imply that $n=x^2+y^2$.</p> http://mathoverflow.net/questions/82052/an-easy-example-of-valuation-ring-which-is-not-noetherian/82119#82119 Answer by Leonardo for an easy example of valuation ring which is not noetherian？ Leonardo 2011-11-28T22:17:24Z 2011-11-28T22:17:24Z <p>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).</p> http://mathoverflow.net/questions/868/etale-covers-of-the-affine-line/63764#63764 Answer by Leonardo for Etale covers of the affine line Leonardo 2011-05-03T02:33:05Z 2011-05-03T07:13:37Z <p>As stated by David Speyer and Clark Barwick, the awnser to the second question is the following:</p> <p><em>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.</em></p> <p>Here is a short constructive proof only based on Riemann-Roch theorem. It can be considered as an illustration of Kedlaya's <a href="http://arxiv.org/abs/math/0303382" rel="nofollow">proof</a>, only dealing with curves.</p> <ul> <li>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).</li> <li>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 $\frac{\deg(f)}p$.</li> <li>Then, the rational function $h=f+g^p$ induces a cover $C\to\mathbf P^1$ only ramified above infinity.</li> </ul> http://mathoverflow.net/questions/46258/generalizations-of-belyis-theorem/63686#63686 Answer by Leonardo for Generalizations of Belyi's theorem Leonardo 2011-05-02T07:25:33Z 2011-05-02T09:44:51Z <p>There is a nice generalization of Belyi's theorem in positive characteristic, proved by M. Saïdi in his paper <em>Revêtements modérés et groupe fondamental de graphe de groupes.</em> (Compositio Math. 107 (1997), no. 3), Théorème 5.6:</p> <p>Let $C$ be a smooth projective curve defined over a field $K$ of characteristic $p>2$. The following conditions are equivalent:</p> <ul> <li>The curve $C$ can be defined over $\bar{\mathbf F}_p$,</li> <li>There exists a finite cover $C\to\mathbf P^1$ <em>tamely</em> ramified above $\infty,0$ and $1$ (and étale elsewhere).</li> </ul> <p>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 <em>Curves with only triple ramification</em>, Ann. Inst. Fourier (Grenoble) 53 (2003), no. 7).</p> http://mathoverflow.net/questions/1909/what-are-dessins-denfants/63687#63687 Answer by Leonardo for What are dessins d'enfants? Leonardo 2011-05-02T07:52:32Z 2011-05-02T07:52:32Z <p>Historically, one of the first papers on th subject is <em>Drawing curves over number fields</em>, 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 <em>Triangulations</em>, 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 <em>The Grothendieck theory of dessins d'enfants</em> (London Mathematical Society Lecture Note Series, 200), which is now the main reference on the subject.</p> http://mathoverflow.net/questions/44082/elliptic-curves-over-finite-fields/63646#63646 Answer by Leonardo for Elliptic curves over finite fields Leonardo 2011-05-01T20:43:25Z 2011-05-01T23:51:29Z <p>I completely agree with the earlier answers. Just two remarks...</p> <ul> <li>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$(q-1,m)$.</li> <li>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 $$\varphi(t)=\left((t^2-1)^{2n+1},t\right).$$</li> </ul> http://mathoverflow.net/questions/57438/branch-locus-of-the-galois-closure-of-a-belyi-morphism/63565#63565 Answer by Leonardo for Branch locus of the Galois closure of a Belyi morphism Leonardo 2011-04-30T22:19:32Z 2011-04-30T23:22:37Z <p>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 $B'$.</p> http://mathoverflow.net/questions/82052/an-easy-example-of-valuation-ring-which-is-not-noetherian/82119#82119 Comment by Leonardo Leonardo 2011-11-28T22:20:09Z 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 $p$. http://mathoverflow.net/questions/49315/what-group-is-a-b-a2b2/49320#49320 Comment by Leonardo Leonardo 2011-11-20T17:29:24Z 2011-11-20T17:29:24Z Even simpler: x=a, y=b^{-1}a. http://mathoverflow.net/questions/46258/generalizations-of-belyis-theorem/63686#63686 Comment by Leonardo Leonardo 2011-06-27T00:55:33Z 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 Sa&#239;di's proof (Theorem 4.6). You can find the preprint at <a href="http://www.math.harvard.edu/~wushi/" rel="nofollow">math.harvard.edu/~wushi</a> http://mathoverflow.net/questions/57438/branch-locus-of-the-galois-closure-of-a-belyi-morphism/63565#63565 Comment by Leonardo Leonardo 2011-05-03T12:01:25Z 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'.