# Jérémy Blanc

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 Name Jérémy Blanc Member for 1 year Seen 4 hours ago Website Location Basel Age 31
 6h answered Why don’t more mathematicians improve Wikipedia articles? May14 comment Usage of complex moments in complex planeI am not expert but I would think about something like analysis or complex geometry. May14 comment What is the Zariski closure of a locally closed set, when “locally” means the Euclidean topology?@ tkluck if $X$ is smooth then it can be viewed as a differential manifold of dimension $m$, which means that it is locally diffeo to $\mathbb{R}^m$. Take a small open ball $U$ and intersects it with $X$, then it is a smooth manifold of dimension $m$ again. Hence, no regular function on $X$ vanishes on $U\cap X$. May14 comment What is the Zariski closure of a locally closed set, when “locally” means the Euclidean topology?I have never seen the notion of absolutely irreducible, but on the web it seems to mean irreducible over $\mathbb{C}$, so $x^2+y^2+y^3+x^3$ is again a counterexample, and the same with any variety with isolated singularities. May14 comment Usage of complex moments in complex planeWhy is this question tagged algebraic geometry? May14 comment What is the Zariski closure of a locally closed set, when “locally” means the Euclidean topology?Probably the equation of $x^2+y^2+y^3+x^3=0$ should be what you are looking for. This is a curve with an isolated singular point, and the intersection of any small ball around the origin only consist of the point. If the variety is smooth and contains real points, then I think that the answer is yes and it just follows from dimension arguments: the intersection $X\cap U$ has the same dimension as $X$ and is therefore Zariski dense. May5 awarded ● Enlightened May5 accepted non-singular cubics are not rational May2 comment Automorphisms of Plane Curves are linearFor degree $4$, this is because the embedding is canonical. For higher degree, I do not see why it should be true. Apr24 comment Continuous automorphisms of Q*Why is this question closed? It is written "off topic". Is it so easy for all people? Apr24 comment Continuous automorphisms of Q*1) As Yoav Kallus said, your $Phi$ is certainly not an automorphism, as it is not bijective (except for $n=1$ where it is the identity). 2) Could you say what you mean by an automorphism? Is it an homeomorphism from $\mathbb{Q}^*$ to itself? 3) Why do you talk about $GL_n$? It does not seem to be related to your question. Apr24 comment Automorphisms of Generic Abelian VarietiesI have a stupid question: what do you mean by automorphisms?? Because for me, an elliptic curve over an algebraically closed always have an infinite group of automorphisms (take translations). So is your question about the quotient of the automorphism group by the group of translations? Or you mean the automorphisms which preserve the group structure of the variety? Sorry if I am pointing out something that was clear for everybody anyway. Apr21 comment non-singular cubics are not rationalThanks. Sorry I did not understood it. Now it is fixed. Apr21 revised non-singular cubics are not rationaladded 2 characters in body Apr21 awarded ● Nice Answer Apr20 comment non-singular cubics are not rational@Ashwath Arbindranath, general means that there is an open set in the moduli space of cubic threefolds where it is not rational. And you are right that the precise description of this set is given in Clemens and Griffiths. @ Chandan Singh: thanks for the other way of seeing the non-rationality. Apr19 answered non-singular cubics are not rational Apr14 awarded ● Fanatic Apr5 comment Infinite dimensional algebraic geometryIn fact, my reference is mostly for affine ind-varieties, not projective ones. I do not know what is the good reference for Proj of a limit of algebras, and on how the grading of the limit comes from the grading of the $A_n$ (in a unique way??) Apr4 comment Infinite dimensional algebraic geometryThe good reference for ind-varieties is Chapter IV of S. Kumar, "Kac-Moody groups, their flag varieties and representation theory." Progress in Mathematics, 204. Birkhäuser Boston, Inc., Boston, MA, 2002. Apr2 accepted Complete intersection space curves Apr1 answered Complete intersection space curves Apr1 comment inverse image of a pencil of hyperplanes @ leffe: yes, this is exactly this. Apr1 comment inverse image of a pencil of hyperplanes As a comment, another example in dimension $2$ (which is almost the same as the one of Francesco): let $X\to \mathbb{P}^2$ be the double covering ramified over a smooth quartic. Then, $X$ is a smooth del Pezzo surface of degree $2$ and the pull-back of an hyperplane section is the anticanonical divisor of $X$, which is ample but not very ample (the morphism associated is exactly the double covering). Mar27 comment Fixed points of group actionOn $\mathbb{P}^2_{\mathbb{C}}$, the action of a cyclic group is linearizable. Looking at the possible "eigenvalues" (up to multiple) you find either three points fixed or one line and one isolated fixed point. In both cases, the action of the anti-holomorphic involution has to fix one point, so you find a fixed real point. This does not work in dimension odd, as pointed by Peter Mueller. By the way, you should accept his answer to your question by clicking on the mark at the left. Mar26 comment Fixed points of group actionThere is no relation between the order of the group and the fact that you have fixed points. The important thing is that the extension $\mathbb{C}:\mathbb{R}$ is of degree $2$. Hence, the two fixed points on $\mathbb{P}^1$ may be defined over $\mathbb{C}$ but not over $\mathbb{R}$. In contrast, acting on $\mathbb{P}^2$ with a cyclic group you always have a fixed point. Mar22 comment Rational points on a sphere in $\mathbb{R}^d$Yes. And the formula of the stereographic projection is not more complicated in high dimension, so this also answers to question 2. Mar20 comment Producing $(-2)$ curves on a smooth surfaceYou dont need an infinite field in fact, you just need a smooth point defined on your field. You can take a smooth point on the curve, then blow-up the point infinitely near to it corresponding to the tangent direction and so on. Mar17 comment Set isomorphism of smooth varieties implies isomorphism as varietiesSince it is injective, it is birational with its image (compute the degree of field extension: it is $1$). Because it is surjective, it is birational. Mar16 comment Set isomorphism of smooth varieties implies isomorphism as varietiesIt is in EGA IV, 3ème partie, Théorème 8.12.6, page 45, it is called " ,,Main Theorem'' de Zariski." It says that your map decomposes into an open immersion, followed by a finite morphism. In your case, you should be able to see that the both are isomorphisms. Mar10 comment On the equation defining a surfaceNow, I gave an (plenty of) example which works. Mar10 revised On the equation defining a surfaceadded 1358 characters in body Mar10 answered About del Pezzo surfaces Mar10 accepted Crepant Birational Map on the Blow-up Mar9 comment On the equation defining a surfaceThanks. Sorry for the stupid mistake. Mar9 revised On the equation defining a surfacedeleted 342 characters in body; added 10 characters in body Mar9 comment “Degree” of an algebraic varietyI have never seen this notion but it seems funny. It is not at all the usual degree, but why not? Can you do something with this notion? Like for the usual degree, compare degree of intersection of two algebraic sets with the degree of the two sets? Or relate it to something else? Mar9 answered On the equation defining a surface Mar8 comment Rational points on surfaces of general typeThanks Antoine for the comment and the reference, I downloaded the article and will read it. Mar7 revised Rational points on surfaces of general typedeleted 3 characters in body Mar7 comment Rational points on surfaces of general typeThanks for the answer, I like it. However, I would like to have other results. For example if you take hypersurfaces of $\mathbb{P}^3$ of degree $d>4$, with singularities which are not too bad (so that it is of general type), can we say something? And apart from the proofs in some particular cases (which is the second question), is there a reason a priori why general type should imply non-Zariski dense rational points? (first question) Mar7 asked Rational points on surfaces of general type Mar7 comment Any other definition for algebraic number than the root of algebraic equation?What do you mean? Do you want a definition equivalent to the one you give or you want to ask if there are some texts where algebraic number means something else? Mar6 comment Crepant Birational Map on the Blow-upIn fact $\mapstochar$ does not work on MO. Mar6 revised Crepant Birational Map on the Blow-upadded 51 characters in body Mar6 comment Crepant Birational Map on the Blow-up@Sándor Kovács. Normally I use $\dasharrow$ and it works in LaTeX but not on this website, thanks for the hint. I fixed it. You can also use \ mapstochar\ dashrightarrow in fact. @Andrew: By Segre embedding you probably mean from $\mathbb{P}^1\times \mathbb{P}^1$ to the quadric in $\mathbb{P}^3$. But in this case it is the blow-up of two pts of $\mathbb{P}^2$, followed by the contraction of the line passing through the two points. Mar6 answered Crepant Birational Map on the Blow-up Mar5 comment Blowing up a projective surfaceIf you do not fix the embedding of Y into a projective space, the degree can be almost anything (with fixed d). Feb28 awarded ● Enlightened Feb28 awarded ● Nice Answer