# François Brunault

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 Name François Brunault Member for 2 years Seen 10 hours ago Website Location Lyon Age 33
I am a number theorist working at the École normale supérieure de Lyon in France. My research interests are elliptic curves, L-functions and zeta functions, especially the study of their special values (conjectures by Beilinson, Bloch, Kato, Zagier...).
 10h comment Permutations of $(Z/pZ)^*$Victor is right, one should replace the "relabeling" condition by the one given in my previous comment. Then for p=5, the statement is true. 11h comment Permutations of $(Z/pZ)^*$There are other modifications which preserve condition (A), for example replace the map $a$ with $a'$ defined by $a'(i) = a(i) \circ b$ where $b$ is a fixed permutation of (Z/pZ)^*. 12h comment Permutations of $(Z/pZ)^*$I let Magma compute the maps satisfying (A) in the case p=5. It turns out that some of them do not have the identity permutation in the image, which seems to indicate that the property you ask is not true in general. As an example, take the four permutations of (Z/5Z)^* whose tables of values are [4,2,3,1], [1,3,2,4], [2,1,4,3] and [3,4,1,2] respectively. Note also that any map satisfying (A) defines a Latin square indexed by (Z/pZ)^* (but the converse is not true). May17 comment Is there any proof that you feel you do not “understand”?@Deane : See also Terence Tao's blog post terrytao.wordpress.com/2007/09/14/… May15 answered link to a paper by Ramanujan May15 comment Possible ratios of Pythagorean fractionsFor 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 perso.ens-lyon.fr/francois.brunault/recherche/… The equations are not precisely the same, but you can easily adapt and find the right change change of variables. May15 comment Possible ratios of Pythagorean fractionsThe conclusion is that 4/9 is not the ratio of any two Pythagorean fractions. May15 comment Possible ratios of Pythagorean fractionsSorry, this is 120a2 and not 120a1. May15 comment Possible ratios of Pythagorean fractionsThis 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 $y=0$. May14 comment What is the Zariski closure of a locally closed set, when “locally” means the Euclidean topology?@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. May14 comment Proof of the weak Goldbach ConjectureFor the first question, I would simply read the introduction of Helfgott's paper. May12 comment Simple automorphism groups of field extensions of infinite transcendence degreeI don't have access to Lascar's article neither, so I cannot check this. May12 comment Simple automorphism groups of field extensions of infinite transcendence degreeA MathSciNet search gives the article Automorphism groups of fields, and their representations by Rovinskii, according to which Lascar actually proves the result for any extension of algebraically closed fields of uncountable transcendence degree. May12 awarded ● Nice Answer May11 answered Is there an algebraic curve over Q which is not modular? May10 comment How I can prove that Λ(C,s) have infinitely many simple zeros at non-positive integers?Use the functional equation and the existence of Euler product for s>1. May10 comment Tenacious structure I meant $2^{(3^d-1)/2}$ in the last comment May10 comment Tenacious structure 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 $d \geq 3$. May10 comment Tenacious structure Isn't it possible to just compare the number of such disjoint unions vs. the number of such sets? May9 accepted What is this subgroup of $\mathfrak S_{12}$ ? May8 comment Concrete examples of noncongruence, arithmetic subgroups of SL(2,R)Another useful reference are K. Conrad's notes : math.uconn.edu/~kconrad/blurbs/grouptheory/… May6 comment Elliptic curves over QQ with identical 13-isogenyBut these elliptic curves seem to have irreducible mod 13 representations. May6 comment Elliptic curves over QQ with identical 13-isogenyLooking for weight 2 newforms whose Fourier coefficient are almost always congruent modulo 13 gives the pairs (52a,988b) and (208c,3952c) (notations from Cremona's tables), but I haven't checked whether this yields elliptic curves with isomorphic Galois modules. May6 comment Elliptic curves over QQ with identical 13-isogenyAs indicates the title of the article mentioned by Chandan, this questions was raised by Mazur, see the article Questions about Numbers math.harvard.edu/~mazur/papers/scanQuest.pdf page 44. May6 comment Galois group of constructible numbers@Cristos A. Ruiz : I don't think so, since the latter group is not (pro)solvable : it has a linear group in the composition series. On the other hand, the Galois group you mention is only a subgroup of $\operatorname{GSp}_{2n}$ in general, so it might be possible that it is pro-2 in some cases, especially when $n=1$ i.e. $A$ is an elliptic curve. I don't know explicit examples though. May6 comment Galois group of constructible numbersIt follows from a well known result of Galois theory that $\operatorname{Gal}(\mathcal{C}/\mathbf{Q})$ is the maximal pro-2-quotient of the absolute Galois group of $\mathbf{Q}$. May5 awarded ● Nice Answer May4 comment Verifying the correctness of a Sudoku solution @Sam : Good question, I don't know the answer. In fact, I should say that all the difficult work here is contained in Emil's answer. In particular we need $\models$ = $\vdash$ (his Proposition 2) in order to prove the matroid property, and this is done by a case-by-case analysis, so it is not clear what happens for higher values of $n$. May4 comment Is there a Sudoku matroid?Some neighbours are partying very late, which explains the entanglement despite the time difference :) May4 comment Is there a Sudoku matroid?I just answered to this in the original question :) mathoverflow.net/questions/129143/… May4 answered Verifying the correctness of a Sudoku solution May3 comment Verifying the correctness of a Sudoku solution All the consequence relations coming from $\mathcal{D}$ are linear, in the sense that $x \in \operatorname{Vect}(D \backslash \{x\})$ for every $x \in D \in \mathcal{D}$ (here we view $r_i$, $c_j$, $b_k$ as formal linear combinations of Sudoku cells). I think we can deduce from this that the Steinitz exchange axiom holds, and thus $\models$ is indeed a matroid. May3 answered elliptic curve with a degree 2 isogeny to itself? May2 comment Verifying the correctness of a Sudoku solution Nice! It is not hard to see that any minimal complete set is a maximal independent set. Do you know whether the converse holds? Apr30 comment Verifying the correctness of a Sudoku solution Symmetric Sudokus can be used to show that $s \geq 4$. For example, start from the grid given in Figure 5 of this article : math.cornell.edu/~connelly/bcc_sudokupaper.pdf and swap the digits 4 and 6 in each of the following rows : 2, 5, 8. Then all rows, all columns as wells as the three squares along the diagonal are correct, but the six other squares are incorrect. You can adapt this for other sets of squares of size 3. Apr30 comment Verifying the correctness of a Sudoku solution @Tony : If we only check 6 rows and 6 columns, then there are 9 cells which can be altered indifferently, so we need to check all 9 squares. Using this kind of reasoning, it seems you can improve your lower bound to 18. Apr29 comment Verifying the correctness of a Sudoku solution If we consider the $c_i$, $r_j$ and $s_k$ as elements in the free abelian group with basis $\{1,\ldots,9\}$, then relations of the form $r_1+r_2+r_3=s_1+s_2+s_3$ show that e.g. correctness of $r_1,r_2,r_3,s_1,s_2$ implies correctness of $s_3$. Apr26 accepted Modular Forms on $\Gamma_0(4)$ with Nebentypus Apr26 revised Modular Forms on $\Gamma_0(4)$ with Nebentypusadded 75 characters in body Apr26 revised Modular Forms on $\Gamma_0(4)$ with Nebentypusadded 8 characters in body Apr26 answered Modular Forms on $\Gamma_0(4)$ with Nebentypus Apr23 answered A $\mathbb{Q}$-rational canonical model for $X(N)$? Apr23 revised Embeddings of finite groups into GL(n,Q_p)Added reference suggested by Pete L. Clark Apr22 asked Embeddings of finite groups into GL(n,Q_p) Apr19 revised Solutions to $\binom{n}{5} = 2 \binom{m}{5}$Added number theory tag Apr16 awarded ● Nice Answer Apr16 awarded ● Necromancer Apr16 answered cube + cube + cube = cube Apr11 accepted Lower bounds for Petersson inner products of cuspforms with integral Fourier coefficients Apr10 answered Lower bounds for Petersson inner products of cuspforms with integral Fourier coefficients