bio | website | ma.utexas.edu/users/voloch |
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visits | member for | 5 years |
seen | 43 mins ago | |
stats | profile views | 10,894 |
<-------- 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 |
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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As for why 2) is no longer valid, I suspect it's because MacWilliams died in 1990.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 |
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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 |
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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 |
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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 |
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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 |
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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 |
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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 |
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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 |
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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 |
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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? |