User fran&#231;ois brunault - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T20:16:02Z http://mathoverflow.net/feeds/user/6506 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/130677/link-to-a-paper-by-ramanujan/130698#130698 Answer by François Brunault for link to a paper by Ramanujan François Brunault 2013-05-15T11:15:40Z 2013-05-15T11:15:40Z <p>It's freely available on Google Books</p> <p><a href="http://books.google.fr/books?id=oSioAM4wORMC&amp;printsec=frontcover&amp;hl=fr#v=onepage&amp;q&amp;f=false" rel="nofollow">http://books.google.fr/books?id=oSioAM4wORMC&amp;printsec=frontcover&amp;hl=fr#v=onepage&amp;q&amp;f=false</a></p> <p>(on page 23).</p> http://mathoverflow.net/questions/130355/is-there-an-algebraic-curve-over-q-which-is-not-modular/130379#130379 Answer by François Brunault for Is there an algebraic curve over Q which is not modular? François Brunault 2013-05-11T23:51:51Z 2013-05-11T23:51:51Z <p>It is conjectured that there are only finitely many curves over $\mathbf{Q}$ of given genus $g \geq 2$ which are covered by a modular curve, see Conjecture 1.1 in the following paper :</p> <p>Baker, M. H. ; González-Jiménez, E. ; González, J. ; Poonen, B. . <a href="http://www-math.mit.edu/~poonen/papers/finiteness.pdf" rel="nofollow"><em>Finiteness results for modular curves of genus at least 2.</em></a> Amer. J. Math. 127 (2005), no. 6, 1325--1387.</p> <p>In fact, the authors prove a strong result towards this conjecture, namely that there are only finitely such curves which are <em>new</em> (in an appropriate sense).</p> http://mathoverflow.net/questions/129143/verifying-the-correctness-of-a-sudoku-solution/129599#129599 Answer by François Brunault for Verifying the correctness of a Sudoku solution François Brunault 2013-05-04T02:09:17Z 2013-05-04T02:09:17Z <p>The consequence relation $\models$ defined in Emil Jeřábek's answer is a matroid. In fact, it is a linear matroid.</p> <p>Let <code>$X=\{r_1,\ldots,r_9,c_1,\ldots,c_9,b_1,\ldots,b_9\}$</code> be the set of possible checks. Recall that given $S \subset X$ and $x \in X$, the notation $S \models x$ means that every Sudoku grid which is valid on $S$ is also valid on $x$.</p> <p>We may embed $X$ into the free abelian group $V$ generated by the 81 cells of the Sudoku grid, by mapping a check $x \in X$ to the formal sum of the cells contained in $x$. The span of $X$ has rank $21$, and the kernel of the natural map $\mathbf{Z}X \to V$ is generated by the six relations of the form $r_1+r_2+r_3-b_1-b_2-b_3$.</p> <p><strong>Proposition</strong>. We have $S \models x$ if and only if $x \in \operatorname{Vect}(S)$.</p> <p>Proof. By Proposition 2 from Emil's answer, the consequence relations $\models$ and $\vdash$ coincide, so we may work with $\vdash$. Let us prove that $S \vdash x$ implies $x \in \operatorname{Vect}(S)$. By transitivity, we may assume <code>$S=D \backslash \{x\}$</code> for some $D \in \mathcal{D}$. It is straightforward to check that <code>$x \in \operatorname{Vect}(D \backslash \{x\})$</code> in each case (i)-(iv).</p> <p>Conversely, let us assume $x=\sum_{s \in S} \lambda_s s$ for some $\lambda_s \in \mathbf{Z}$. Since the elements of $X$ have degree 9, we have $\sum_{s \in S} \lambda_s = 1$. Any Sudoku grid provides a linear map $\phi : V \to E$, where $E$ is the free abelian group with basis <code>$\{1,\ldots,9\}$</code> (map each cell to the digit it contains). If the grid is valid on $S$ then $\phi(s)=[1]+\cdots+[9]$ for every $s \in S$, and thus $\phi(x)=[1]+\cdots+[9]$, which means that the grid is valid on $x$. QED</p> <p>Note that a set of checks $S$ is complete if and only if $\operatorname{Vect}(S)=\operatorname{Vect}(X)$. In particular, the minimal complete sets are those which form a basis of $\operatorname{Vect}(X)$, and it is now clear that every such set has cardinality $21$.</p> <p>We also obtain a description of the independent sets : these are exactly the sets which are linearly independent when considered in $V$. Any independent set may be extended to a minimal complete set (we may have worked with $\mathbf{Q}$-coefficients instead of $\mathbf{Z}$-coefficients above).</p> http://mathoverflow.net/questions/129509/elliptic-curve-with-a-degree-2-isogeny-to-itself/129513#129513 Answer by François Brunault for elliptic curve with a degree 2 isogeny to itself? François Brunault 2013-05-03T10:34:33Z 2013-05-03T10:34:33Z <p>Yes, but the elliptic curve needs to have complex multiplication, since the multiplication-by-$n$ map has degree $n^2$. For an explicit example, you can take $E=\mathbf{C}/(\mathbf{Z}+i\mathbf{Z})$ with the isogeny being multiplication by $1+i$.</p> http://mathoverflow.net/questions/128739/modular-forms-on-gamma-04-with-nebentypus/128815#128815 Answer by François Brunault for Modular Forms on $\Gamma_0(4)$ with Nebentypus François Brunault 2013-04-26T10:50:46Z 2013-04-26T13:18:51Z <p><strong>EDIT</strong> : In what follows the weight $k$ is assumed to be an integer.</p> <p>The matrix <code>$\begin{pmatrix} 1 &amp; 1/2 \\ 0 &amp; 1 \end{pmatrix}$</code> normalizes the group $\Gamma_0(4)$, so in your case $g$ is still a modular form on $\Gamma_0(4)$ with trivial Nebentypus. </p> <p>More generally if you start with $f$ on $\Gamma_0(N)$ and twist it by $1/m$ then you get a form on $\Gamma_0(N') \cap \Gamma_1(m)$ with $N'=\operatorname{lcm}(N,m^2)$.</p> <p>In general, when you consider $g(z)=f(z+1/m)$, you are twisting $f$ by the <em>additive</em> character $\alpha(n)=\exp(2\pi i n/m)$. You can always write $\alpha$ as a linear combination of (not necessarily primitive) Dirichlet characters $\chi$ of level dividing $m$. Thus you can write $g$ as a linear combination of twists $f \otimes \chi$ for such $\chi$ (up to bad Euler factors). If you want to do this completely explicitly then the formulas are quite complicated in general, see Merel, <a href="http://www.math.jussieu.fr/~merel/M70.pdf" rel="nofollow">Symboles de Manin et valeurs de fonctions L</a>, Section 2.5. At some point, it may be useful to switch to the adelic language and work with the automorphic representation associated to the newform $f$.</p> http://mathoverflow.net/questions/128318/embeddings-of-finite-groups-into-gln-q-p Embeddings of finite groups into GL(n,Q_p) François Brunault 2013-04-22T07:10:40Z 2013-04-24T14:52:44Z <p>This question is inspired by some interesting comments on <a href="http://mathoverflow.net/questions/128194/upper-bound-on-order-of-finite-subgroups-of-gl-nz-p" rel="nofollow">this recent question</a>.</p> <p>Fix an integer $n \geq 1$ and a finite subgroup $G$ of $\mathrm{GL}_n(\mathbf{C})$. It is known that there are infinitely many primes $p$ such that $G$ embeds into $\mathrm{GL}_n(\mathbf{Q}_p)$. This follows from the Cassels Embedding Theorem : every finitely generated field of characteristic 0 embeds into $\mathbf{Q}_p$ for infinitely many primes $p$ (see Cassels's book <em>Local Fields</em> or Cassels's article <em>An embedding theorem for fields</em>. Bull. Austral. Math. Soc. 14 (1976), 193-198; see also Chapter 4 of <a href="http://math.uga.edu/~pete/8410FULL.pdf" rel="nofollow">Pete L. Clark's notes</a>). </p> <p>Representation theoretic arguments actually show that $G$ occurs in $\mathrm{GL}_n(\overline{\mathbf{Q}})$ and thus in $\mathrm{GL}_n(K)$ for some number field $K$. From this and Chebotarev's density theorem, we deduce that the set $S(G)$ of primes $p$ such that $G \hookrightarrow \mathrm{GL}_n(\mathbf{Q}_p)$ has positive density.</p> <p>Does the set $S(G)$ have a natural density?</p> <p>Is the set $S(G)$ a Galoisian set of prime numbers in the sense of <a href="http://mathoverflow.net/questions/11747/galoisian-sets-of-prime-numbers" rel="nofollow">this question</a>?</p> http://mathoverflow.net/questions/128458/a-mathbbq-rational-canonical-model-for-xn/128474#128474 Answer by François Brunault for A $\mathbb{Q}$-rational canonical model for $X(N)$? François Brunault 2013-04-23T12:55:09Z 2013-04-23T12:55:09Z <p>This is done by Shimura in <em>Introduction to the Arithmetic Theory of Automorphic Functions</em>, Chapter 6.</p> <p>By definition, the field $\mathbf{Q}(X(N))$ of rational functions with respect to this model is the field of modular functions for the congruence subgroup $\Gamma(N)$ whose Fourier expansions belong to $\mathbf{Q}(\zeta_N)((q^{1/N}))$.</p> http://mathoverflow.net/questions/53048/cube-cube-cube-cube/127710#127710 Answer by François Brunault for cube + cube + cube = cube François Brunault 2013-04-16T15:11:55Z 2013-04-16T15:11:55Z <p>The solution mentioned in the question consists in fact of $9$ pieces, as shown in the following picture I took yesterday. I don't know whether a solution with $8$ pieces exists.</p> <p><img src="http://perso.ens-lyon.fr/francois.brunault/P1000848b.jpg" alt="Solution"></p> http://mathoverflow.net/questions/126969/lower-bounds-for-petersson-inner-products-of-cuspforms-with-integral-fourier-coef/127123#127123 Answer by François Brunault for Lower bounds for Petersson inner products of cuspforms with integral Fourier coefficients François Brunault 2013-04-10T17:57:23Z 2013-04-10T17:57:23Z <p>It's not hard to see that the answer to (a) is yes. There is a basis of $S_2^{\textrm{new}}(\Gamma_0(N))$ consisting of newforms. These newforms come into Galois orbits <code>$\{f^\sigma\}_{\sigma}$</code>. Here $\sigma$ runs through the embeddings of $K_f$ into $\mathbf{C}$, where $K_f$ is the field generated by the Fourier coefficients of $f$. A basic fact is that the $\mathbf{C}$-span of a given Galois orbit is generated by cusp forms with integral coefficients. This follows from considering the forms $\sum_{\sigma} \sigma(a) f^\sigma$ where $a$ runs through a $\mathbf{Z}$-basis of the ring of integers of $K_f$.</p> <p>It follows that if the newform $f$ has integral coefficients, then its orthogonal complement $f^\perp$ is generated by cusp forms with integral coefficients. Now consider the linear map $\lambda_f : S_2(\Gamma_0(N),\mathbf{Z}) \to \mathbf{R}$ given by $\lambda_f(g)=(f,g)$. By the previous remark, the kernel of $\lambda_f$ has rank one less than the rank of $S_2(\Gamma_0(N),\mathbf{Z})$, which implies that the image of $\lambda_f$ is of the form $c_f \mathbf{Z}$ for some $c_f >0$. Thus we can take for $c$ the minimum of all the $c_f$. In fact $c_f = (f,f)/m_f$ for some integer $m_f$ measuring the congruences of $f$ with other cusp forms.</p> http://mathoverflow.net/questions/126699/status-of-beilinson-conjectures/126766#126766 Answer by François Brunault for Status of Beilinson conjectures? François Brunault 2013-04-07T08:57:19Z 2013-04-07T10:10:34Z <p>In addition to Andreas's excellent answer, we should also mention the Tamagawa number conjecture of Bloch and Kato, which predicts the undetermined rational factor arising in Beilison's conjectural description of the $L$-value. The Bloch-Kato conjecture was later reformulated and generalized by Fontaine and Perrin-Riou to the case of motives with coefficients in an arbitrary number field. Here are some references :</p> <p>Bloch, Kato, <a href="http://www.ams.org/mathscinet/search/publdoc.html?pg1=MR&amp;s1=1086888" rel="nofollow">L-functions and Tamagawa numbers of motives.</a></p> <p>Fontaine, Perrin-Riou, <a href="http://www.ams.org/mathscinet/search/publdoc.html?r=1&amp;pg1=CNO&amp;s1=1265546&amp;loc=fromrevtext" rel="nofollow">Autour des conjectures de Bloch et Kato: cohomologie galoisienne et valeurs de fonctions L.</a></p> <p>Colmez, <a href="http://www.math.jussieu.fr/~colmez/851bourbaki.pdf" rel="nofollow">Fonctions L p-adiques.</a></p> <p>Kings, <a href="http://archive.numdam.org/ARCHIVE/JTNB/JTNB_2003__15_1/JTNB_2003__15_1_179_0/JTNB_2003__15_1_179_0.pdf" rel="nofollow">The Bloch-Kato conjecture on special values of L-functions. A survey of known results.</a></p> <p>Flach, <a href="http://math.caltech.edu/papers/baltimore-final.pdf" rel="nofollow">The equivariant Tamagawa number conjecture : A survey.</a></p> <p>Gealy, <a href="http://thesis.library.caltech.edu/5020/1/thesis-final.pdf" rel="nofollow">On the Tamagawa Number Conjecture for Motives Attached to Modular Forms.</a></p> <p>Bellaïche, <a href="http://people.brandeis.edu/~jbellaic/BKHawaii5.pdf" rel="nofollow">An introduction to the conjecture of Bloch and Kato.</a></p> http://mathoverflow.net/questions/125684/algebraic-independence-of-e-2-e-4-and-e-6/125808#125808 Answer by François Brunault for Algebraic independence of $E_2$, $E_4$ and $E_6$ François Brunault 2013-03-28T09:32:38Z 2013-03-28T09:32:38Z <p>Here is a reference for the algebraic independance of $E_2,E_4,E_6$ over $\mathbf{C}$ :</p> <p><a href="http://smf4.emath.fr/Publications/SeminairesCongres/2005/12/pdf/smf_sem-cong_12_1-117.pdf" rel="nofollow">MR2186573 (2007a:11065) Martin, François ; Royer, Emmanuel . Formes modulaires et périodes. (French) [Modular forms and periods] Formes modulaires et transcendance, 1--117, Sémin. Congr., 12, Soc. Math. France, Paris, 2005.</a></p> <p>See Lemme 117 p. 80. The proof is along the lines suggested by Emerton.</p> http://mathoverflow.net/questions/125694/computing-the-order-of-the-image-of-0-under-the-modular-parametrization-map-for-a/125702#125702 Answer by François Brunault for computing the order of the image of 0 under the modular parametrization map for an elliptic curve François Brunault 2013-03-27T08:33:04Z 2013-03-27T08:33:04Z <p>Here is an expanded version of g6hq's answer. Your question is indeed sensitive to the Manin constant of the modular parametrization $X_0(N) \to E$. This in turns depends on whether the elliptic curve $E$ is the so-called "strong Weil curve" in its isogeny class, which by definition means that the kernel of $\phi_* : J_0(N) \to E$ is connected.</p> <p>For example when $E=11a1=X_0(11)$, then $\phi$ is an isomorphism so everything is fine. But when $E=11a3=X_1(11)$, the modular parametrization $\phi : X_0(11) \to X_1(11)$ of least degree has degree 5 and in fact satisfies $\phi^* \omega_E = c_E \omega_f$ where $f=f_E \in S_2(\Gamma_0(11))$ is the newform associated to $E$, with a non-trivial Manin constant $c_E=5$. Thus the torsion point $I(0)$ has, in fact, order $c/c_E=5$.</p> <p>In general, given any elliptic curve $E/\mathbf{Q}$, there is an optimal parametrization $\phi : X_1(N) \to E$. Such a parametrization conjecturally satisfies $\phi^* \omega_E = \omega_f$ (Stevens' conjecture). But the class of the divisor $[0]-[\infty]$ in the Jacobian of $X_1(N)$ is defined only over $\mathbf{Q}(\mu_N)^+$, not over $\mathbf{Q}$.</p> http://mathoverflow.net/questions/125165/critical-values-of-motives/125181#125181 Answer by François Brunault for critical values of motives François Brunault 2013-03-21T15:50:35Z 2013-03-21T15:50:35Z <p>The factor $L_\infty(M,s)$ is holomorphic and non-vanishing for $\operatorname{Re}(s)$ large enough, so it is definitely necessary to also ask that $L_\infty(\hat{M},1-s)$ has no pole at the given integer. As an example, for the Riemann zeta function $\zeta(s)$, only the even integers $n \geq 2$ are critical.</p> <p>I haven't done the computation of critical integers for $L$-functions of $K3$ surfaces, but there should be no difficulty in this computation as the answer depends only on the behaviour of the motive at infinity, which in this case just means (the cohomology of) the complex variety.</p> http://mathoverflow.net/questions/124328/does-split-l-function-imply-split-jacobian/124403#124403 Answer by François Brunault for Does split L-function imply split jacobian François Brunault 2013-03-13T11:47:46Z 2013-03-13T11:47:46Z <p>Faltings proved that two abelian varieties $A$ and $B$ defined over a number field are isogenous if and only if they have the same $L$-function. See Korollar 2 p. 361 in</p> <p><a href="http://dx.doi.org/10.1007/BF01388432" rel="nofollow">Faltings, G. <em>Endlichkeitssätze für abelsche Varietäten über Zahlkörpern.</em> Invent. Math. 73 (1983), no. 3, 349--366.</a></p> <p>So if you assume $L(A,s)=L(B,s) L(C,s)$ for some abelian varieties $A,B,C$ defined over a number field $K$, you can deduce that $A$ is $K$-isogenous to $B \times C$.</p> http://mathoverflow.net/questions/122037/an-expression-for-the-function-f-e-that-appears-in-the-weil-pairing/122144#122144 Answer by François Brunault for An expression for the function $f_e$ that appears in the Weil Pairing François Brunault 2013-02-18T08:05:04Z 2013-02-18T08:05:04Z <p>To expand on wccanard's comment : see my answer to <a href="http://mathoverflow.net/questions/121291/calculate-function-from-its-divizor/" rel="nofollow">this question</a> for a general method to compute a rational function with given (principal) divisor.</p> <p>This will give you $f_e$ as a rational function of $x$ and $y$. You then just need to expand this in terms of the standard formal coordinate $z=-x/y$.</p> http://mathoverflow.net/questions/121678/the-value-pm-1-for-the-square-root-of-wilsons-theorem-p-1-2-mod-p/121703#121703 Answer by François Brunault for The value $\pm 1$ for the square root of Wilson's theorem, ((p-1)/2)! mod p François Brunault 2013-02-13T11:20:52Z 2013-02-13T11:20:52Z <p>Regarding your question (1), here are two articles I know dealing with the problem of computing factorials :</p> <p>Crandall, Dilcher, Pomerance. <em>A search for Wieferich and Wilson primes.</em> Math. Comp. 66 (1997), no. 217, 433--449. (MR1372002)</p> <p>Costa, Gerbicz, Harvey. <em><a href="http://arxiv.org/abs/1209.3436" rel="nofollow">A search for Wilson primes</a></em>. </p> <p>(Recall that Wilson primes are those primes $p$ such that $(p-1)! \equiv -1 \mod{p^2}$.)</p> <p>It turns out that there exists an algorithm to compute the product of $N$ terms in arithmetic progression in no more than $O(N^{\alpha+\epsilon})$ multiplications, with $\alpha = \frac{\sqrt{5}-1}{2} = 0.618...$ (see p. 441 in the first article). They also mention an algorithm to compute $(p-1)! \mod{p^2}$ in time $O(p^{1/2+\epsilon})$.</p> <p>The second article seems to use a somewhat different method and achieves an <em>average</em> polynomial time algorithm to compute this quantity, making use of the fact that there are redundant products when considering many values of $p$.</p> <p>Both articles always work mod $p^2$. It is certainly faster to work only mod $p$, but I'm not sure whether this has an impact on the theoretical running time of the algorithms.</p> http://mathoverflow.net/questions/121291/calculate-function-from-its-divizor/121449#121449 Answer by François Brunault for calculate function from its divizor François Brunault 2013-02-11T07:53:38Z 2013-02-11T07:53:38Z <p>Every principal divisor is a finite ${\bf Z}$-linear combination of divisors of the form $D=[P]+[Q]-[P+Q]-[O]$ with $P,Q \in C$. For such $D$, let $L:\lambda=0$ be the line through $P$ and $Q$.</p> <p>If $P+Q=O$ then $L$ is vertical and $D={\rm div}(\lambda)={\rm div}(x-x_P)$.</p> <p>If $P+Q \neq 0$ then ${\rm div}(\lambda)=[P]+[Q]+[-P-Q]-3[O]$ so that $D={\rm div}\bigl(\frac{\lambda}{x-x_{P+Q}}\bigr)$.</p> <p>This method is ok if the coefficients of your divisor are reasonably small. Otherwise, you might consider using some kind of fast exponentiation. The idea is that a divisor $2N[P]$ is equivalent to $N[2P]+N[O]$ modulo the principal divisors. So you just need to compute a function $f$ such that ${\rm div} f=2[P]-[2P]-[O]$ and then compute $f^N$ by fast exponentiation. At each step the coefficient in your divisor has been divided by $2$, thus giving a logarithmic (instead of linear) number of steps.</p> <p>I'm currently writing a PARI/GP script which takes as input a divisor on elliptic curve, checks if it's principal and if so outputs a rational function with this divisor. If you or other people are interested I could put the link here.</p> <p>Finally a precision on the decomposition $f=f_1(x)+yf_2(x)$ I mentioned in my comment. Since the function $x$ has degree $2$ on $C$, the degree of $f_1$ as a usual rational function is in fact $\leq \deg(f)$. A more careful analysis also shows $\deg(f_2) \leq \deg(f)$ since there is compensation when you divide by $y$.</p> http://mathoverflow.net/questions/118376/are-there-noncongruence-subgroups-of-finite-index-of-the-modular-group-generate/118510#118510 Answer by François Brunault for Are there noncongruence subgroups (of finite index) of the modular group generated only by 2 or 3 elements? François Brunault 2013-01-10T08:30:24Z 2013-01-10T08:30:24Z <p>A. J. Scholl has written a series of articles on modular forms on noncongruence subgroups. In the article </p> <p><a href="http://ams.u-strasbg.fr/mathscinet/search/publdoc.html?arg3=&amp;co4=AND&amp;co5=AND&amp;co6=AND&amp;co7=AND&amp;dr=all&amp;pg4=AUCN&amp;pg5=TI&amp;pg6=PC&amp;pg7=ALLF&amp;pg8=ET&amp;review_format=html&amp;s4=scholl&amp;s5=noncongruence&amp;s6=&amp;s7=&amp;s8=All&amp;vfpref=html&amp;yearRangeFirst=&amp;yearRangeSecond=&amp;yrop=eq&amp;r=2&amp;mx-pid=1446557" rel="nofollow">On the Hecke algebra of a noncongruence subgroup</a>, Bull. London Math. Soc. 29 (1997), no. 4, 395–399</p> <p>he gives two examples of index 7 subgroups of $\mathrm{SL}_2(\mathbf{Z})$ generated by 3 elements :</p> <p><code>$$\Gamma_{4,3} = \Bigl\langle \begin{pmatrix} 1 &amp; 4 \\ 0 &amp; 1 \end{pmatrix},\begin{pmatrix} 2 &amp; 1 \\ 1 &amp; 1 \end{pmatrix},\begin{pmatrix} 1 &amp; -1 \\ 2 &amp; -1 \end{pmatrix} \Bigr\rangle$$</code> </p> <p><code>$$\Gamma_{5,2} = \Bigl\langle \begin{pmatrix} 1 &amp; 5 \\ 0 &amp; 1 \end{pmatrix},\begin{pmatrix} 0 &amp; -1 \\ 1 &amp; 0 \end{pmatrix},\begin{pmatrix} 2 &amp; 3 \\ 1 &amp; 2 \end{pmatrix} \Bigr\rangle$$</code> </p> <p>The subscripts refer to the widths of the cusps (both groups have exactly two cusps).</p> http://mathoverflow.net/questions/116797/constructive-proof-of-projective-implies-proper Constructive proof of "Projective implies proper" François Brunault 2012-12-19T16:41:45Z 2012-12-20T08:13:41Z <p>For every ring $A$, the structural morphism of schemes $\pi_A : {\bf P}^n_{A} \to {\rm Spec}{A}$ is a closed map. The usual proof of this fact is not constructive : given equations of a closed subset $Z$ of ${\bf P}^n_{A}$, it doesn't produce equations for $\pi_A(Z)$.</p> <p>In the case $A$ is a polynomial ring over an algebraically closed field $k$, this result is none other than the fundamental theorem of elimination theory : the image of a Zariski-closed subset of ${\bf P}^n(k) \times k^m$ under the second projection is a Zariski-closed subset of $k^m$. The first proofs of this theorem (Cayley, Kronecker, Sylvester) used resultants and thus were constructive.</p> <p>In fact, the proof using elimination theory is universal in the following sense. Given integers $n,r \geq 1$, $d_1,\ldots,d_r \geq 1$, consider the universal homogenous polynomials $P_1,\ldots,P_r$ of degree $d_1,\ldots,d_r$ in the indeterminates $T_0,\ldots,T_n$, having coefficients in the polynomial ring $\widetilde{A} = \mathbf{Z}[Y_{i,\alpha} : 1 \leq i \leq r]$, where the indeterminates $Y_{i,\alpha}$ are the coefficients of $P_i$. Then there exists an explicit "resultant system" $R_1,\ldots,R_s \in \widetilde{A}$ such that $\pi_{\widetilde{A}}(V_+(P_1,\ldots,P_r))=V(R_1,\ldots,R_s)$. This means that specializations of $P_1,\ldots,P_r$ in some algebraically closed field $k$ have a common root in ${\bf P}^n(k)$ if and only if the corresponding specializations of $R_1,\ldots,R_s$ all vanish. Of course $s$ has to depend on $n,r,d_i$, but everything is explicit (at least from a theoretical point of view).</p> <p>Now let $A$ be any ring and let $I=(f_1,\ldots,f_r)$ be an homogenous ideal <em>of finite type</em> of $A[T_0,\ldots,T_n]$. Then the resultant system above specialized at $f_1,\ldots,f_r$ provides explicit equations for $\pi_A(V_+(I))$ (this can be seen by studying the geometric fibers of $\pi_A$). In particular if $A$ is noetherian, then the map $\pi_A$ is closed, and we have a constructive proof for that. But in general, a closed subset of ${\bf P}^n_A$ need not be defined by finitely many equations. This raises the following questions :</p> <ol> <li><p>Is there a way to prove that the map $\pi_A$ is closed for every ring $A$, by some clever reduction to the noetherian case?</p></li> <li><p>If $Z$ is a closed subset of ${\bf P}^n_A$, given to us by infinitely many explicit equations $(f_i)_{i \in I}$, is there a way to produce explicit equations for $\pi_A(Z)$? In other words, is there a constructive proof of the fact that $\pi_A$ is closed?</p></li> <li><p>Regarding question 2, an obvious thing to do is to look at all finite subfamilies $(f_i)_{i \in J}$, where $J$ is a finite subset of $I$, and to consider the associated resultant systems. Are all these equations sufficient to define $\pi_A(Z)$?</p></li> </ol> <p><strong>EDIT.</strong> Will Sawin has proved that the answers to all these questions is yes. Following Daniel Litt's comment, we can also consider $\pi_A(Z)$ as a closed subscheme of $\operatorname{Spec} A$, namely the closed subscheme defined by the kernel of the morphism $A \to \mathcal{O}_Z(Z)$. Do the resultant systems generate this ideal of $A$?</p> http://mathoverflow.net/questions/116273/how-do-you-compute-the-primes-of-bad-reduction/116303#116303 Answer by François Brunault for How do you compute the primes of bad reduction? François Brunault 2012-12-13T17:57:23Z 2012-12-13T17:57:23Z <p>Assume for simplicity that $Y$ has pure relative dimension $d$. By considering the standard affine cover of $\mathbf{P}^n_{\mathbf{Z}}$, one easily reduces to the case where $Y=V(F_1,\ldots,F_k)$ is a closed subscheme of $\mathbf{A}^n_{\mathbf{Z}}$. Then the special fiber <code>$Y_p = V(F_{1,p},\ldots,F_{k,p}) \subset \mathbf{A}^n_{\mathbf{F}_p}$</code> is smooth if and only if for every <code>$x \in Y_p(\overline{\mathbf{F}}_p)$</code>, the Jacobian matrix $(\frac{\partial F_{i,p}}{\partial X_j}(x))$ has rank $n-d$. Note that $Y_p$ is $d$-dimensional by the flatness assumption, so all geometric tangent spaces have dimension $\geq d$, which implies that the rank of the Jacobian matrix is everywhere $\leq n-d$. Thus $Y_p$ is smooth if and only if the ideal generated by $F_{1,p},\ldots,F_{k,p}$ together with all $(n-d)$-minors of the Jacobian matrix is equal to $\mathbf{F}_p[X_1,\ldots,X_n]$.</p> <p>Now consider the ideal $I$ of $R=\mathbf{Z}[X_1,\ldots,X_n]$ generated by $F_1,\ldots,F_k$ together with all $(n-d)$-minors of the Jacobian matrix. Since the generic fiber of $Y$ is smooth, we have $I \cap \mathbf{Z}=N\mathbf{Z}$ for some integer $N \geq 1$, and the prime factors of $N$ are precisely the primes of bad reduction of $Y$.</p> <p>Proof. If $p$ doesn't divide $N$ then $(I,p)$ contains $N$ and $p$, so $(I,p)=R$ and $Y_p$ is smooth. If $p$ divides $N$, write $N=p^k M$ with $p$ not dividing $M$. If we had $(I,p)=R$, then we would also have $(I,p^k)=R$. Then $M \in (MI,p^k M)=(MI,N) \subset I$, a contradiction. Thus $Y_p$ is not smooth.</p> <p>You can compute the integer $N$ using Gröbner bases for polynomials over $\mathbf{Z}$, see e.g. <a href="http://magma.maths.usyd.edu.au/magma/handbook/text/1112#12189" rel="nofollow">http://magma.maths.usyd.edu.au/magma/handbook/text/1112#12189</a> for the Magma implementation. I'm not expert in Gröbner bases, so I would appreciate if someone could confirm whether it is always possible to find $I \cap \mathbf{Z}$ when a set of generators of $I$ is given, and whether the Magma implementation works in all cases.</p> http://mathoverflow.net/questions/115403/dimension-of-polynomial-algebras/115481#115481 Answer by François Brunault for Dimension of polynomial algebras François Brunault 2012-12-05T08:28:06Z 2012-12-05T08:28:06Z <p>The following reference should be of interest :</p> <p>Brewer, Montgomery, Rutter, Heinzer, <em>Krull dimension of polynomial rings</em>.</p> <p>For example, they prove, see Corollary 2 p. 30, that any semi-hereditary ring (all finitely generated ideals are projective) satisfies the dimension formula above. This generalizes Seidenberg's result since a Prüfer ring is a semi-hereditary integral domain.</p> <p>It might be interesting to study the class of rings $R$ satisfying the following condition : for every prime ideal $P$ of $R$ and every $n \geq 1$, we have $\textrm{height}(P[X_1,\ldots,X_n]) = \textrm{height}(P)$. This condition implies $\dim R[X_1,\ldots,X_n] = \dim R +n$ for every $n$ (this can be deduced from Thm 1 of this paper), but I don't know about the converse.</p> <p>The authors also discuss the class of strong $S$-rings introduced by Kaplansky (see the paper for the definition). This class contains the Noetherian rings and the Prüfer rings, and is stable by localizations and quotients. Kaplansky proved that a strong $S$-ring $R$ satisfies $\mathrm{height}(P[X]) = \textrm{height}(P)$ for every prime ideal $P$ of $R$, and thus $\textrm{dim}(R[X])=\textrm{dim}(R)+1$. But a strong $S$-ring doesn't necessarily satisfy the dimension formula for every $n$. In the other direction, the authors give an example of a ring which satisfies the height formula $\textrm{height}(P[X_1,\ldots,X_n]) = \textrm{height}(P)$ for every prime ideal $P$, but which is not a strong $S$-ring. </p> http://mathoverflow.net/questions/114828/is-this-height-transcendence-degree-inequality-true-without-ac/114978#114978 Answer by François Brunault for Is this height-transcendence-degree inequality true without AC ? François Brunault 2012-11-30T13:38:58Z 2012-12-04T15:37:29Z <p>Here is a proof of (**) by induction on the height of $P$.</p> <p>If $P=0$, the inequality (**) is obvious. Let $P$ be a prime ideal of $R$ of height $d \geq 1$, and consider a chain of prime ideals $0=P_0 \subset P_1 \subset \ldots \subset P_d = P$ of length $d$ in $R$. The domain $R/P_1$ has finite Krull dimension, and the prime ideal $P/P_1$ has height $d-1$ so by the induction hypothesis \begin{equation*} {\rm height}(P/P_1) + {\rm trdeg}_k (R/P) \leq {\rm trdeg}_k(R/P_1). \end{equation*} It remains to prove that ${\rm trdeg}_k(R/P_1) \leq{\rm trdeg}_k(R)-1$. Let $x_1,\ldots,x_e$ be elements of $R$ whose images in $R/P_1$ are algebraically independent over $k$. Let $y$ be any element of $P_1 \backslash \{0\}$. Consider the map $\phi : k[X_1,\ldots,X_e,Y] \to R$ sending $X_i$ to $x_i$ and $Y$ to $y$. If $Q = \sum_j Q_j(X_1,\ldots,X_e) Y^j$ lies in the kernel of $\phi$, then reducing modulo $P_1$ gives $Q_0(x_1,\ldots,x_e)=0$, so that $Q_0=0$. Since we are working in a domain and $y \neq 0$, an easy induction gives $Q=0$. Thus $x_1,\ldots,x_e,y$ are algebraically independent over $k$, which concludes the proof.</p> http://mathoverflow.net/questions/114715/is-a-domain-all-of-whose-localizations-are-noetherian-itself-noetherian/114750#114750 Answer by François Brunault for Is a domain all of whose localizations are noetherian itself noetherian ? François Brunault 2012-11-28T10:02:48Z 2012-11-28T11:41:46Z <p>The ring of integers $\mathcal{O}_{\mathbf{C}_p}$ of $\mathbf{C}_p$ is not noetherian, but its only nontrivial localization is $\mathbf{C}_p$, which is noetherian.</p> <p><strong>EDIT</strong> This doesn't answer the question : the ring $\mathcal{O}$ is local, so its localization at the maximal ideal is $\mathcal{O}$ itself, which isn't noetherian.</p> <p>The nonzero ideals of $\mathcal{O}$ are of the form $I_{\geq \alpha} = \{x \in \mathcal{O} : v(x) \geq \alpha\}$ with <code>$\alpha \in \mathbf{Q}_{&gt;0}$</code>, and $I_{> \alpha} = \{x \in \mathcal{O} : v(x) > \alpha\}$ with <code>$\alpha \in {\bf R}_{\geq 0}$</code>. Here $v$ is the $p$-adic valuation on <code>$\mathbf{C}_p$</code>. The ring $\mathcal{O}$ is one-dimensional : its only prime ideals are $(0)$ and the maximal ideal $I_{>0}$.</p> http://mathoverflow.net/questions/114619/existence-of-an-r-basis-with-at-least-one-unit-in-it/114740#114740 Answer by François Brunault for Existence of an $R$-basis with at least one unit in it? François Brunault 2012-11-28T08:00:51Z 2012-11-28T10:45:14Z <p>First some general remarks. You're asking whether $S/R$ is free as a $R$-module. If $S$ is a $R$-algebra which is free of finite rank, then the map $R \to S$ splits as a map of $R$-modules. This fact was already noticed by Florian Eisele in <a href="http://mathoverflow.net/questions/89040/does-s-being-a-free-rank-n-r-algebra-imply-that-s-r-is-free-rank-n-1/89048#89048" rel="nofollow">his answer</a> to the other MO question.</p> <p>Now for an explicit counter-example to your question. Consider the ring $R={\bf Z}[x,y,z]/(x^2+y^2+z^2-1)$. It is an integral domain. It is known that there exists a $R$-module $M$ which is not free such that $R \oplus M \cong R^3$. For a nice construction, see e.g. <a href="http://www.math.uconn.edu/~kconrad/blurbs/linmultialg/stablyfree.pdf" rel="nofollow">Keith Conrad's notes</a>. Explicitly we can take $M=\{(f,g,h) \in R^3 : xf+yg+zh=0\}$. Note that we can embed $M$ in $R^2$ by $(f,g,h) \mapsto (f,g)$, and the cokernel $R^2/M$ is a torsion module, so there exists $F \in R \backslash \{0\}$ such that $F \cdot R^2 \subset M$.</p> <p>Now, we would like to construct a $R$-algebra structure on $R \oplus M$. We can do this by considering the $R$-algbera $S_0 = R \otimes_{\mathbf{Z}} \mathcal{O}$ where $\mathcal{O}$ is an order of a cubic field $K$. It is an integral domain, since the polynomial $x^2+y^2+z^2-1$ is irreducible over any field of characteristic not $2$. Let $(1,\alpha,\beta)$ be a $\mathbf{Z}$-basis of $\mathcal{O}$. Embed $R \oplus M$ in $S_0$ by $(f,(g,h)) \mapsto f+g\alpha+h\beta$. This won't be a subring of $S_0$ in general, but $S=R \oplus FM$ is a subring of $S_0$ since $(FM) \cdot (FM) \subset F^2 S_0 \subset R \oplus FM$. So we have constructed an integral domain $S$ over $R$ such that $S/R \cong M$ is not free over $R$.</p> <p>I don't know whether it's possible to find a counterexample where $R \to S$ splits as a map of rings, in other words where $S=R \oplus I$ where $I$ is an ideal of $S$.</p> http://mathoverflow.net/questions/113856/what-are-some-consequences-of-the-mumford-tate-conjecture/114036#114036 Answer by François Brunault for What are some consequences of the Mumford-Tate conjecture? François Brunault 2012-11-21T08:16:35Z 2012-11-21T08:16:35Z <p>Here is a result in a somewhat different direction. Assuming MTC, Hindry and Ratazzi <a href="http://arxiv.org/abs/0911.5505" rel="nofollow">recently proved</a> a bound for the torsion subgroup of abelian varieties of ${\rm GSp}$-type over arbitrary finite extensions.</p> http://mathoverflow.net/questions/112595/models-of-the-modular-curve-y-1n/112615#112615 Answer by François Brunault for Models of the modular curve $Y_1(N)$ François Brunault 2012-11-16T21:07:08Z 2012-11-16T21:17:40Z <p>In the model you describe, the cusp $\infty$ of $X_1(N)$ is not defined over ${\bf Q}$ (but the cusp $0$ is). A way to see this is that the marked elliptic curve $({\bf C}/({\bf Z}+\tau{\bf Z}),1/N)$ is isomorphic to the marked Tate curve $E_q=({\bf C}^\times/q^{\bf Z},e^{2\pi i/N})$ with $q=e^{2\pi i\tau}$. When you let $\tau \to \infty$, you get $q \to 0$ so that $E_q \to ({\bf G}_m,e^{2i\pi/N})$, which is not defined over ${\bf Q}$. This fact is explained in Diamond-Im, <em>Modular forms and modular curves</em>, see 9.3.5 and 9.3.6.</p> <p>There is an alternative model $Y_\mu(N)$ classifying elliptic curves $E$ together with a closed immersion $\mu_N \hookrightarrow E$ (see loc. cit. 8.2.2). In this model the cusp $\infty$ is defined over ${\bf Q}$, so it gives an affirmative answer to your second question.</p> <p>You can switch from one model to another with the Atkin-Lehner involution $W_N$, which becomes an isomorphism defined over ${\bf Q}$ — it is only defined over ${\bf Q}(\mu_N)$ when considered as an involution of either $X_1(N)$ or $X_{\mu}(N)$. But I don't see a nice way to characterize those functions which are rational for the canonical model in terms of the $q$-expansion at $\infty$.</p> http://mathoverflow.net/questions/110727/on-a-theorem-of-galois On a theorem of Galois François Brunault 2012-10-26T06:19:03Z 2012-11-09T08:58:12Z <p>I am currently teaching Galois theory and this week, I mentioned the following theorem of Galois :</p> <p><em>Let $P(x) \in \mathbf{Q}[x]$ be an irreducible polynomial of prime degree. Then $P$ is solvable by radicals if and only if the splitting field of $P$ is generated by any two roots of $P$.</em></p> <p>I was asked by a student whether this theorem can be generalized to polynomials whose degree is composite, maybe allowing the splitting field to be generated by more than two roots. I know that the proof of Galois's theorem relies on determining the solvable subgroups of $\mathfrak{S}_p$, but I don't know enough group theory to tell what can be proved in the case where the degree of the polynomial is composite, say $pq$ where $p$ and $q$ are (possibly equal) primes.</p> <p>Does such a generalization of Galois's theorem exist? Or is there a conceptual reason why such a generalization cannot hold? In the latter case, do there already exist generalizations of Galois's theorem, possibly in different directions?</p> http://mathoverflow.net/questions/114139/can-we-ascertain-that-there-exist-an-epimorphism-g-rightarrow-h/111118#111118 Answer by François Brunault for Can we ascertain that there exist an epimorphism $G\rightarrow H?$ François Brunault 2012-11-01T07:10:41Z 2012-11-04T10:35:41Z <p>As explained in the comments, the result is true if $H$ is abelian.</p> <p>Here is an argument which shows that the result is true in the somewhat orthogonal case where $H$ has trivial center <strong>[EDIT] and is indecomposable [/EDIT]</strong>.</p> <p>Write the epimorphism $G \times G \to H \times H$ as <code>$\begin{pmatrix} a &amp; b \\ c &amp; d \end{pmatrix}$</code> with $a,b,c,d : G \to H$. Let $A,B,C,D$ be the respective images of $a,b,c,d$ in $H$.</p> <p>The groups $A$ and $B$ commute elementwise in the sense that $xy=yx$ for every $x \in A$ and $y \in B$. Moreover, they generate $H$ by assumption. So we have an exact sequence</p> <p>\begin{equation*} 1 \to A \cap B \to A \times B \to H \to 1 \end{equation*}</p> <p>and similarly for $C,D$. Note that $A \cap B$ commutes with $A$ and $B$, so it must lie in the center of $H$, thus it should be trivial. Therefore $H=A \times B = C \times D$. It follows that $A=\{e\}$ or $B=\{e\}$, thus $a$ or $b$ is surjective.</p> <p>The same argument also works in some cases where the center of $H$ is not trivial, for example when $H$ is a group of order $p^3$ with $p$ prime.</p> http://mathoverflow.net/questions/108543/over-which-fields-does-the-mordell-weil-theorem-hold Over which fields does the Mordell-Weil theorem hold? François Brunault 2012-10-01T13:01:11Z 2012-11-02T11:05:34Z <p>According to a well-known theorem of Mordell, the group of rational points $E(\mathbf{Q})$ of an elliptic curve $E/\mathbf{Q}$ is finitely generated. Weil generalized this theorem to abelian varieties over number fields.</p> <p>Less well-known is the following generalization, due I believe to Néron : if $K$ is a field of finite type (that is, finitely generated over its prime field) and $A$ is an abelian variety over $K$, then $A(K)$ is finitely generated. There is an even more general statement, the Lang-Néron theorem, for relative field extensions which are finitely generated (see <a href="http://math.stanford.edu/~conrad/papers/Kktrace.pdf" rel="nofollow">Brian Conrad's article</a> for the precise statement and a proof of this theorem).</p> <p>Q1. Are there other fields $K$ for which the group of $K$-rational points of an abelian variety over $K$ is always finitely generated?</p> <p>In the other direction, there exist fields $K$ for which $A(K)$ is clearly never finitely generated whenever $\operatorname{dim}(A) \geq 1$. For example $K=\mathbf{C}$, in which case it follows from the description af abelian varieties as complex tori. If $K$ is a finite extension of $\mathbf{Q}_p$, then $A(K)$ contains a finite-index subgroup isomorphic to $\mathcal{O}_K^{\operatorname{dim} A}$, so $A(K)$ is again never finitely generated. Other examples I can think of are complete discretely valued fields and algebraically closed fields. Note that we often have the stronger result that $A(K) \otimes \mathbf{Q}$ is infinite-dimensional (except when $K=\overline{\mathbf{F}}_p$, in which case $A(K)$ is a torsion group).</p> <p>Q2. Are there other fields $K$ for which the group of $K$-rational points of a non-trivial abelian variety over $K$ is never finitely generated?</p> http://mathoverflow.net/questions/110846/conjugating-the-lyness-5-cycle-into-a-rotation-of-the-plane/110847#110847 Answer by François Brunault for Conjugating the Lyness 5-cycle into a rotation of the plane François Brunault 2012-10-27T17:51:09Z 2012-10-27T17:51:09Z <p>Yes, this map is conjugate to an automorphism of $\mathbf{P}^2$. See</p> <p>A. Beauville, J. Blanc, <a href="http://arxiv.org/abs/math/0402037" rel="nofollow"><em>On Cremona transformations of prime order</em></a>, C.R. Acad. Sci. Paris 339 (2004), no4, 257-259.</p> <p>See also T. de Fernex, <em>On planar Cremona maps of prime order</em>, Nagoya Math. Journal, Vol. 174 (2004), 1–28, which contains a classification of planar Cremona maps of prime order up to conjugation. According to Remark 1.3.4 therein, the fact above was already known to Iskovskikh.</p> http://mathoverflow.net/questions/130883/is-there-any-proof-that-you-feel-you-do-not-understand/130898#130898 Comment by François Brunault François Brunault 2013-05-17T07:43:08Z 2013-05-17T07:43:08Z @Deane : See also Terence Tao's blog post <a href="http://terrytao.wordpress.com/2007/09/14/pythagoras-theorem/" rel="nofollow">terrytao.wordpress.com/2007/09/14/&hellip;</a> http://mathoverflow.net/questions/130624/possible-ratios-of-pythagorean-fractions/130651#130651 Comment by François Brunault François Brunault 2013-05-15T09:42:21Z 2013-05-15T09:42:21Z For those interested in putting an intersection of two quadrics into cubic and then Weierstrass form, you can have a look at the first 2 pages of the following note I wrote <a href="http://perso.ens-lyon.fr/francois.brunault/recherche/biquadratique.pdf" rel="nofollow">perso.ens-lyon.fr/francois.brunault/recherche/&hellip;</a> The equations are not precisely the same, but you can easily adapt and find the right change change of variables. http://mathoverflow.net/questions/130624/possible-ratios-of-pythagorean-fractions/130651#130651 Comment by François Brunault François Brunault 2013-05-15T09:32:43Z 2013-05-15T09:32:43Z The conclusion is that 4/9 is not the ratio of any two Pythagorean fractions. http://mathoverflow.net/questions/130624/possible-ratios-of-pythagorean-fractions/130651#130651 Comment by François Brunault François Brunault 2013-05-15T09:30:45Z 2013-05-15T09:30:45Z Sorry, this is 120a2 and not 120a1. http://mathoverflow.net/questions/130624/possible-ratios-of-pythagorean-fractions/130651#130651 Comment by François Brunault François Brunault 2013-05-15T09:30:20Z 2013-05-15T09:30:20Z This elliptic curve is 120a1 in Cremona's tables, it has rank 0 and its Mordell-Weil group is isomorphic to $\mathbf{Z}/4\mathbf{Z} \times \mathbf{Z}/2\mathbf{Z}$. So it seems that the only rational points on this intersection of 2 quadrics are those with $x=0$ or $y=0$. http://mathoverflow.net/questions/130486/what-is-the-zariski-closure-of-a-locally-closed-set-when-locally-means-the-euc Comment by François Brunault François Brunault 2013-05-14T11:52:28Z 2013-05-14T11:52:28Z @tkluck: You should be careful when speaking of the Zariski topology on $\mathbf{R}^n$. Usually we only consider Zariski topology on (the underlying set of) schemes like $\mathbf{A}^n_{\mathbf{R}$, which is bigger than $\mathbf{R}^n$ set-theoretically. Of course, you can consider the topology induced on the set of real points, but some care is needed. In the examples above, the set $X(\mathbf{R})$ is dense in $X$ so is still irreducible by general topology. So the isolated singularities provide counterexamples to your question as stated. http://mathoverflow.net/questions/130545/proof-of-the-weak-goldbach-conjecture Comment by François Brunault François Brunault 2013-05-14T07:39:46Z 2013-05-14T07:39:46Z For the first question, I would simply read the introduction of Helfgott's paper. http://mathoverflow.net/questions/130401/simple-automorphism-groups-of-field-extensions-of-infinite-transcendence-degree Comment by François Brunault François Brunault 2013-05-12T18:18:14Z 2013-05-12T18:18:14Z I don't have access to Lascar's article neither, so I cannot check this. http://mathoverflow.net/questions/130401/simple-automorphism-groups-of-field-extensions-of-infinite-transcendence-degree Comment by François Brunault François Brunault 2013-05-12T18:11:55Z 2013-05-12T18:11:55Z A MathSciNet search gives the article <i>Automorphism groups of fields, and their representations</i> by Rovinskii, according to which Lascar actually proves the result for any extension of algebraically closed fields of uncountable transcendence degree. http://mathoverflow.net/questions/130261/how-i-can-prove-that-c-s-have-infinitely-many-simple-zeros-at-non-positive-int Comment by François Brunault François Brunault 2013-05-10T16:10:57Z 2013-05-10T16:10:57Z Use the functional equation and the existence of Euler product for s&gt;1. http://mathoverflow.net/questions/130227/tenacious-structure Comment by François Brunault François Brunault 2013-05-10T08:39:17Z 2013-05-10T08:39:17Z I meant $2^{(3^d-1)/2}$ in the last comment http://mathoverflow.net/questions/130227/tenacious-structure Comment by François Brunault François Brunault 2013-05-10T08:37:38Z 2013-05-10T08:37:38Z If I'm not mistaken, the number of disjoint unions $P_0 \cup \cdots \cup P_{d-1}$ satisfying your condition is $(3^d-1) (3^{d-1}-1) \cdots (3-1)$, and the number of possible $X$'s is $2{\^}((3^d-1)/2)$ which is larger for $d \geq 3$. http://mathoverflow.net/questions/130227/tenacious-structure Comment by François Brunault François Brunault 2013-05-10T08:22:52Z 2013-05-10T08:22:52Z Isn't it possible to just compare the number of such disjoint unions vs. the number of such sets? http://mathoverflow.net/questions/130062/concrete-examples-of-noncongruence-arithmetic-subgroups-of-sl2-r/130063#130063 Comment by François Brunault François Brunault 2013-05-08T20:05:10Z 2013-05-08T20:05:10Z Another useful reference are K. Conrad's notes : <a href="http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/SL(2,Z).pdf" rel="nofollow">math.uconn.edu/~kconrad/blurbs/grouptheory/&hellip;</a> http://mathoverflow.net/questions/129818/elliptic-curves-over-qq-with-identical-13-isogeny Comment by François Brunault François Brunault 2013-05-06T12:46:47Z 2013-05-06T12:46:47Z But these elliptic curves seem to have irreducible mod 13 representations.