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Apr
6
comment Algorithm for fast factorization of polynomial over $\mathbb Z$ or over $\mathbb F_p$
Sage uses FLINT flintlib.org to do factorisation in $\mathbb{Z}[X]$ and $\mathbb{Z}/n\mathbb{Z}[X]$. For an introduction into the very basic factorisation algorithms, maybe Cohen's "A course in computational number theory" pp. 124 could help.
Apr
6
answered Is Ш a good parameter for the failure of Global-Local principle for abelian varieties?
Apr
6
awarded  Fanatic
Apr
4
comment Is Ш a good parameter for the failure of Global-Local principle for abelian varieties?
Sorry for the chain of comments, but the question is not (yet ?) open for an answer.
Apr
4
comment Is Ш a good parameter for the failure of Global-Local principle for abelian varieties?
There is some evidence that the fine Sha is much smaller. For instance even over infinite extension considered in Iwasawa theory the fine Sha should be finite (all the time ?) while the full Sha can get very large.
Apr
4
comment Is Ш a good parameter for the failure of Global-Local principle for abelian varieties?
As to the kernel of your map: This is what I would call the "fine Tate-Shafarevich group". There are examples of when it is trivial, half or all of the Tate-Shafarevich group when the latter has 4 elements. In general I would think the kernel could just be anything. Proc. Camb. Soc. 142 (2007), no. 1, p. 1-12.
Apr
3
comment Is Ш a good parameter for the failure of Global-Local principle for abelian varieties?
(After edit): Take $\ell$-primary parts everywhere. Then the local product of the first term is just $E(\mathbb{Q}_{\ell})\otimes\mathbb{Q}_{\ell}/\mathbb{Z}_{\ell}$ which is cofree of rank $1$. So $\operatorname{coker}(a)[\ell^{\infty}]$ is finite if the rank of $E(\mathbb{Q})$ is positive and is cofree of corank $1$ otherwise. That does not look analogous to your huge local product in $r$.
Apr
1
comment Is Ш a good parameter for the failure of Global-Local principle for abelian varieties?
And the title of the question als odoes not seem to have much relation to the question itself.
Apr
1
comment Is Ш a good parameter for the failure of Global-Local principle for abelian varieties?
What is your definition of $Sel(E/\mathbb{Q})$ ? My first guess would be the inductive limit of $n$-Selmer groups. But then I can't see how you defined the map you want to be injective. Did you mean the target to be $E(\mathbb{Q}_p)\otimes \mathbb{Q}/\mathbb{Z}$ ? I think this question needs some improvement to be understandable.
Mar
30
awarded  Yearling
Mar
11
comment integral basis for the Lie algebra of the Neron model of an abelian variety
Lie$(A^{\vee})$ is defined to be the $O_K$-dual of the differentials on the Néron model. So if that is not free, neither will Lie$(A^{\vee})$. But a basis of Lie${}_K(A^{\vee})$ is all that you need.
Mar
10
answered integral basis for the Lie algebra of the Neron model of an abelian variety
Mar
3
comment If $E(K)=E(L)$ for an elliptic curve $E$ and an algebraic extension $L/K$, what can we say about $Sel(E/K), Sel(E/L), L/K$?
I assume that Sel$(E/K)$ is the group that fits into a short exact sequence between $E(K)$ and Sha$(E/K)$. Otherwise you need to fix your $n$ in $n$-Selmer group (or your isogeny). So your question is equivalent to asking about the growth of the Tate-Shafarevich group in the tower. Basically anything can happen. Iwasaw theory gives examples of exploding growth as well as frequent stabilisation.
Feb
13
comment Is it possible the division polynomials evaluated at fixed point to be perfect powers unbounded number of times?
Sure, sorry, if $x(P)$ and $y(P)$ are integers and the equation has integer coefficients then $a_n$ is an integer. - Non-square-free, yes, that is very easy, but perfect squares or even perfect powers are hard. But why do you want them perfect powers ?
Feb
13
comment Is it possible the division polynomials evaluated at fixed point to be perfect powers unbounded number of times?
For a given $(E,P)$ it will only finitely many times give an integer, I would think. Why are you interested in this question ? It seems hard that one can say anything interesting about it other than that it is very unlikely that there could be infinitely many perfect powers in any such sequence.
Feb
12
comment Is it possible the division polynomials evaluated at fixed point to be perfect powers unbounded number of times?
So $a_n$ is a rational number.
Feb
12
comment Is it possible the division polynomials evaluated at fixed point to be perfect powers unbounded number of times?
Do you assume $E/\mathbb{Q}$ ? Do you assume $P$ to have integer coordinates in a given Weierstrass equation ? Do you assume that $P$ has good reduction at all primes ?
Feb
8
comment Why does inconstructibility of $\sqrt[3]{2}$ imply impossibility of cube doubling?
No. But you can place your cube such that it appears as two unit squares and the cube of twice the volume appears as a larger square with sides $\sqrt[3]{2}$. So in the two planes of descriptive geometry you face the same problem of constructing that number, no ?
Feb
8
comment Why does inconstructibility of $\sqrt[3]{2}$ imply impossibility of cube doubling?
In descriptive geometry it is quite clear that doubling the cube is the same as constructing $\sqrt[3]{2}$. But it may well be that your students did not see any descriptive geometry.
Feb
5
comment Heegner points on elliptic curves
I might be wrong, but my guess is that we do not expect anything particular to happen when the elliptic curve you map the Heegner points to has complex multiplication.