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<-------- Over there, you'll find my website.


Dec
19
reviewed Close Linear algebra over principal rings 1
Dec
19
reviewed Close Isoceles Triangles on a Grid Proof
Dec
17
comment Open problems with monetary rewards
As for why 2) is no longer valid, I suspect it's because MacWilliams died in 1990. en.wikipedia.org/wiki/Jessie_MacWilliams
Dec
17
reviewed Close Graduate program applications that require questionnaires and other non-letter material
Dec
15
reviewed Close Approximating an integral
Dec
15
comment What is a discrete shape
Whatever it is, it has nothing to do with algebraic geometry, so you are better off changing the tag.
Dec
14
reviewed Leave Closed personal relationships
Dec
14
reviewed Close On a paper by Yoneda
Dec
9
comment Algebraic equivalence vs linear equivalence
For smooth projective complex varieties, this is the same as the Picard variery (or equivalently the Albanese) is zero, almost by definition. As the dimension of the Picard variety is the first Betti number, here is your criterion: $b_1=0$.
Dec
9
comment Have there been any new developments in the Firoozbakht conjecture?
@KarlSchwede As explained in Carlo's answer, this is slighty stronger than a 80-year old conjecture of Cramer, which is NOT a consequence of the Riemann hypothesis. I don't think you should expect new developments every few years and, if there are, you'll hear about them.
Dec
9
reviewed Close Have there been any new developments in the Firoozbakht conjecture?
Dec
6
reviewed Leave Closed The resolution of which conjecture/problem would advance Mathematics the most?
Dec
4
comment Good lecture notes/books on Jacobian of hyperelliptic curve
If you want to view things from a more applied standpoint, then Galbraith "Mathematics of public key cryptography" is a good reference.
Dec
4
comment Why there are two point at infinity on certain elliptic curve
This is not an MO question. The function field of the quartic has two embeddings in $k((1/x))$ corresponding to $y = \pm x^2 + \cdots$, so two places at infinity.
Dec
4
comment minimizing an integral over integer-coefficient polynomials $\displaystyle \inf_{f \in \mathbb{Z}[x]} \int_a^b f(x)^2 \, dx $
Integer valued polynomials is not the same as polynomials with integer coefficients. Which one do you want?
Dec
3
awarded  Yearling
Dec
3
comment Adjoining torsion points from abelian varieties
@LaurentMoret-Bailly I guess you are right, unless the curve is defined over a subfield (not necessarily $\mathbb{Q}$). Still, one needs to be careful about the point that is picked. In the restriction of scalars, $(\lambda,0)$ corresponds to $\prod (\lambda^{\sigma},0) \in \prod E^{\sigma}$ which is rational. It is the point $((\lambda,0),{\cal{O}},\ldots)\in \prod E^{\sigma}$ that generates $L$.
Dec
3
comment Adjoining torsion points from abelian varieties
@LaurentMoret-Bailly I guess part of the problem is that the $\mathbb{Q}$-structure of the product of the conjugates is ambiguous. It's only after fixing this structure that one can talk about the field of definition of an algebraic point. In the case of a variety defined over $\mathbb{Q}$, the structure over $\mathbb{Q}$ of the restriction of scalars is not that of the $n$-th power of the original variety.
Dec
3
comment Adjoining torsion points from abelian varieties
@LaurentMoret-Bailly Not in general. They are twists of each other. Maybe, in Filippo's case, it doesn't matter as he is base-changing to $L$ but then he needs to use a different point, since Ulrich is definitely correct and that point corresponds to a rational point on the Weil restriction.
Dec
3
reviewed Close Well-known or prolific mathematicians that have never written a sole-author article?