René Pannekoek
|
Registered User
|
|
|
May 9 |
comment |
r-torsion points on elliptic curve on finite field No; let $E$ be $y^2=x^3+x$ over $\mathbf{F}_q$ with $q \equiv 3\pmod{4}$. Then we have $2 \mid q−1$ and $\# E(\mathbf{F}_q)[2] > 1$, but since $−1 \notin \mathbf{F}_q^{\ast 2}$ we have $\# E(\mathbf{F}_q)[2] = 2 < 4$. |
|
Apr 20 |
comment |
non-singular cubics are not rational The funny thing is that the arithmetic genus of a singular (e.g. nodal) cubic curve is also 1, at least if it is assumed to be geometrically integral. And such curves are actually rational. But in the non-singular case, the arithmetic and geometric genera agree, and the latter is a birational invariant (unlike the former). |
|
Apr 12 |
comment |
Counter-example to faithfully flat descent You can use \rightrightarrows. |
|
Apr 7 |
comment |
Tricks to produce examples of hypersurfaces with index greater than $1$ Thanks! I realized that you can indeed take any curve and 'pinch' some effective zero-cycle into a rational point. |
|
Apr 6 |
comment |
Tricks to produce examples of hypersurfaces with index greater than $1$ Right. But since the second part of my answer was incorrect, I have deleted it. The first part read: "A non-trivial Severi-Brauer variety of dimension $2$ over $K$ has index $3$ and is birational to a smooth cubic surface over $K$: just blow up a zero-cycle of degree $6$." |
|
Mar 28 |
awarded | ● Nice Question |
|
Mar 23 |
comment |
Rational points on a sphere in $\mathbb{R}^d$ That certainly looks interesting; I can't explain the patterns off the top of my head. Can you make an animation where you vary the height bound? I'd be interested to see it if you can. |
|
Mar 22 |
comment |
Rational points on a sphere in $\mathbb{R}^d$ I don't understand why Peter Michor's comment got two upvotes. If a quadric hypersurface $X \subset \mathbf{A}^{n+1}$ defined over $\mathbf{Q}$ has a point $P \in X(\mathbf{Q})$, then projecting away from $P$ gives a birational map $X \stackrel{\sim}{\dashrightarrow} \mathbf{A}^n$ that is defined over $\mathbf{Q}$. Restricting this birational map gives an isomorphism between open subsets of $X$ and $\mathbf{A}^n$ that is defined over $\mathbf{Q}$. In particular, the rational point son $X$ are dense in the real locus of $X$ iff the same holds for $\mathbf{A}^n$, which is trivially the case. |
|
Mar 4 |
comment |
Do isogenies with rational kernels tend to be surjective? Dear Chris. You correctly observed that, since the $(x_0y_0 + y_0 - x_0^2)d^i$ come from rational points $(x_0,y_0)$ on $E_d$, they lie in the $\widehat{\eta}$-Selmer group. This does not imply they are 5-th powers in $\mathbf{Q}_v$ for all good $v$ (if so, the $(x_0y_0 + y_0 - x_0^2)d^i$ would be 5-th powers in $\mathbf{Q}$ by Grunwald-Wang). It does say that choice of $d$ imposes restrictions on the values that $x_0,y_0$ may take. On the other hand, I don't see why these restrictions force $(x_0y_0 + y_0 - x_0^2)d^i$ to be 5-th powers more often than if $x_0,y_0$ were random. Best, René. |
|
Mar 1 |
comment |
Do isogenies with rational kernels tend to be surjective? Hmm, I am not sure I agree. Why do you think that any (let alone all) of the $(x_0y_0 + y_0 - x_0^2)d^i$ should be fifth powers almost everywhere locally? I do not see how that follows from anything. Here is a heuristic reason to believe that the pairs $(x_0,y_0)$ one gets are in fact quite random. The curves $E_d$ trace out a pencil in $\mathbf{P}^2$ if you let $d$ vary. In particular you will get all pairs $(x_0,y_0)$ as points on some $E_d$. Moreover if you believe that most elliptic curves have rank $0$ or $1$, then most $(x_0,y_0)$ will lie on an $E_d$ that has rank $1$. |
|
Mar 1 |
comment |
Do isogenies with rational kernels tend to be surjective? Moreover, I think the condition is necessary as well as sufficient. This remains true if, in the statement above, one replaces "some point of infinite order" by "a generator of $E_d(\mathbf{Q})$ modulo torsion". |
|
Mar 1 |
comment |
Do isogenies with rational kernels tend to be surjective? If you follow the approach via Galois cohomology, you get the following: "Assume that $d \in \mathbf{Q}^{\ast}$ is not a $5$-th power, and that the abelian group $E_d(\mathbf{Q})$ is of rank $1$. If for some point $(x_0,y_0) \in E_d(\mathbf{Q})$ of infinite order, and each integer $0 \leq i \leq 4$, we have that the non-zero rational number $(x_0 y_0 + y_0 - x_0^2) d^i$ is not a $5$-th power in $\mathbf{Q}^{\ast}$, then $\eta$ is surjective on $\mathbf{Q}$-points." Looks like it might be a condition that is generically satisfied, but it's hard to say. |
|
Feb 28 |
comment |
Do isogenies with rational kernels tend to be surjective? Perhaps you could do the same experiment for the Legendre family $E_\lambda:y^2=x(x−1)(x−\lambda)$ with the role of $P$ played by the $2$-torsion point $(0,0)$. This gives $2$-isogenies $\eta$ for each $E_\lambda$. Barring mistakes on my part, the surjectivity of $\eta$ on $\mathbf{Q}$-points can be rephrased (a small condition on $\lambda$ aside) in terms of the image of an element in $E_\lambda(\mathbf{Q})$ of infinite order under the coboundary map associated to $\widehat{\eta}$ (dual of $\eta$). It seems much easier to make this explicit for $2$-isogenies than for $5$-isogenies. |
|
Feb 27 |
comment |
Do isogenies with rational kernels tend to be surjective? Similar phenomena are discussed in the highest-ranking answer to this question: mathoverflow.net/questions/113968/…. |
|
Feb 14 |
comment |
Properties of quotient variety As Dmitri points out, it is not so much the singularities of Y as the branch locus of π that you have to worry about. Outside of that, π will be étale and therefore the inverse image of the part of C that lies outside of the branch locus will be non-singular. In the case where π:X→Y is the map from an abelian variety X to its quotient by -1, the non-étale locus is given by π(X[2]), which is a union of $2^{2 \operatorname{dim}(X)}$ points. Finding out what the inverse image of C looks like above any of the points in π(X[2]) should be a perfectly local problem. |
|
Feb 13 |
revised |
E an elliptic curve over Z[1/N], how many p such that E(Z/p^2) = (Z/p)^2? deleted 31 characters in body; edited title |
|
Feb 11 |
revised |
Do torsors give a long exact sequence of cohomology? added 1040 characters in body |
|
Feb 7 |
comment |
How many curves in a family possess a rational point? I guess the answer is trivially yes, because the hypotheses are never fulfilled. If $B(\mathbb{Z})$ contains an element $b$, then $X_b$ must be a proper smooth curve over $\operatorname{Spec}(\mathbb{Z})$ of genus $>1$, and such curves don't exist. |
|
Feb 4 |
comment |
Do torsors give a long exact sequence of cohomology? Thank you, Will, this is indeed part of the motivation for my question. I should have included it in the post. |
|
Feb 4 |
comment |
Do torsors give a long exact sequence of cohomology? Thank you, this is very interesting! Actually however, I was thinking along the lines of Will's comment: if X,Y are k-group schemes and f is a homomorphism, then we are able to extend the exact sequence: if G is abelian we have a long exact sequence in the right sense of the word, whereas in general with the help of non-abelian cohomology we can get at least 2 more terms. I was hoping that one could do this in more generality. |
|
Feb 4 |
revised |
Do torsors give a long exact sequence of cohomology? deleted 74 characters in body |
|
Feb 4 |
revised |
Do torsors give a long exact sequence of cohomology? deleted 28 characters in body; deleted 63 characters in body |
|
Feb 4 |
revised |
Do torsors give a long exact sequence of cohomology? deleted 5 characters in body |
|
Feb 4 |
revised |
Do torsors give a long exact sequence of cohomology? G has to be finite-type, so removed parentheses |
|
Feb 4 |
revised |
Do torsors give a long exact sequence of cohomology? some elaboration |
|
Feb 4 |
asked | Do torsors give a long exact sequence of cohomology? |
|
Feb 3 |
comment |
Are rational varieties simply connected? I believe that the Godeaux surface is a quotient of your quintic surface, not the quintic itself. |
|
Jan 27 |
revised |
Elliptic fibration of K3 surface added 43 characters in body |
|
Jan 27 |
answered | Elliptic fibration of K3 surface |
|
Dec 14 |
comment |
How do you compute the primes of bad reduction? Dear François: Thank you very much. I must say though I wonder: in terms of complexity, one might be worse off with the Gröbner basis calculation than if you would upper-bound $S$ using my method and then for each $p \in S$ would check if $X_{\mathbf{F}_p}$ is smooth. I can't prove this, of course, after all you do have to factor an integer $N$ over whose size you don't really seem to have much control |
|
Dec 13 |
revised |
How do you compute the primes of bad reduction? added 84 characters in body |
|
Dec 13 |
comment |
How do you compute the primes of bad reduction? That makes a lot of sense, thanks! Your second remark agrees with what I thought myself, I just thought it made the notation simpler if I restricted to $k=1$. |
|
Dec 13 |
asked | How do you compute the primes of bad reduction? |
|
Dec 2 |
comment |
Solved cubic Thue equation @Richard: You should convince people who read your article that solving that equation is a routine matter - given the bounds established by Baker, say. Indeed, all that Mathematica uses is the existence of these bounds on $x,y$, which are part of the (by now) standard theory. The fact that you learned about this only recently doesn't force you to belabor the point; on the contrary, if you spend too many words on this your readers might get the false impression that there's something unusual going on here (which there ain't). |
|
Dec 2 |
comment |
How does “modern” number theory contribute to further understanding of $\mathbb{N}$? @Johnny: even such a fundamental statement as "every elliptic curve over Q has finitely generated Mordell-Weil rank", which doesn't refer to algebraic number theory in the least, needs algebraic number theory to prove it. But you see the same principle in much more basic examples as well: try finding all integer solutions to $x^2+4=y^3$ without using the Gaussian integers. (I'm not saying it can't be done, but it's completely routine once you're using some basic ANT.) |
|
Dec 1 |
comment |
Solved cubic Thue equation May I be so bold as to ask how you know these are the only ones? |
|
Nov 30 |
awarded | ● Enthusiast |
|
Nov 26 |
revised |
The boundedness of the rank of twists of a fixed curve. added 39 characters in body |
|
Nov 26 |
revised |
The boundedness of the rank of twists of a fixed curve. added 177 characters in body |
|
Nov 26 |
revised |
The boundedness of the rank of twists of a fixed curve. added 409 characters in body |
|
Nov 26 |
revised |
The boundedness of the rank of twists of a fixed curve. added 82 characters in body; added 11 characters in body |
|
Nov 26 |
answered | The boundedness of the rank of twists of a fixed curve. |
|
Nov 26 |
comment |
Does finite+reduced fibers+connected fibers imply isomorphism? Fiber above the cusp is not reduced. |
|
Nov 25 |
comment |
Surjectivity of reduction maps of elliptic curves over Q So far, all your counterexamples seem to come from isogenies of degree $3$. Any reason to expect isogenies $\phi$ of degree $2$ not to satisfy $\mathbf{Q}(\phi^{-1}E(\mathbf{Q})) = \mathbf{Q}$? |
|
Nov 24 |
answered | reference for (co)homology theories |
|
Nov 24 |
revised |
Surjectivity of reduction maps of elliptic curves over Q added 6 characters in body |
|
Nov 23 |
comment |
Elliptic Curves and Torsion Points Dear Joe: Out of curiosity, can you prove your last statement without appealing to Mazur? I like the result, but at the same time it seems too simple a statement for me to require something so deep as Mazur's theorem... |
|
Nov 21 |
comment |
Are ranks of Jacobians over number fields unbounded? @Noam Elkies: A belated remark. You not only want the projections to be non-constant, but you'll want them to be "independent". For instance, simply choosing $C$ to be the diagonal $E \subset E^r$ clearly won't do the job. So I'm guessing the condition must be that the images of the $r$ pull-back maps $H^0(E,\Omega_E) \rightarrow H^0(C,\Omega_C)$ must span an $r$-dimensional subspace. Alternatively, someone suggested to me that one could look at curves $C$ in $\prod_{i=1}^r E_i$, where the $E_i$ are non-isogenous elliptic curves of positive rank, with non-trivial projections to each factor. |
|
Nov 21 |
revised |
Surjectivity of reduction maps of elliptic curves over Q added 340 characters in body; edited title |
|
Nov 21 |
comment |
Surjectivity of reduction maps of elliptic curves over Q Hi Felipe, I don't mind getting all these answers, they've been very helpful :) I've checked out the reference to Gupta-Murty, which does indeed prove what you say. Rank $\ge 6$ is large though, so I'll check for improvements later. (Although if Maarten Derickx's calculation proves correct, I guess some assumption on the size of $E(\mathbf{Q})$ on top of rank $>0$ is necessary.) I don't understand your solution to 3., do you mean to start by picking an isogeny that maps to the reduced curve? If so, I fail to see how you can get anything from that unless the isogeny lifts to $\mathbf{Q}$. |
