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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.
Jan
21
awarded  Nice Answer
Dec
24
comment $\zeta$ function for ambiguous class group
It seems to me this is just the Riemann zeta function at $[K:\mathbb{Q}]s$ with finitely many factors modified, no ?
Dec
21
comment Can the pre-image of the real points in the complex upper-half plane of a modular elliptic curve under the modular parametrization be identified?
Coordinates on the affine model over $\mathbb{Q}$ are $j(\tau)$ and $j(N\tau)$. These are power-series in $q=e^{2\pi i \tau}$ with real coefficients. Now $\bar{q} = e^{2\pi i(-\bar\tau)}$ confirms that complex conjugation is the reflection on the imaginary axis.
Dec
21
comment Can the pre-image of the real points in the complex upper-half plane of a modular elliptic curve under the modular parametrization be identified?
I guess you answer what the preimage of $X_0(N)(\mathbb{R})$ in the upper half plane looks like. Surely the imaginary axis belongs to it as complex conjugation on the curve comes from the reflection through the imaginary axis $\tau\mapsto -\bar\tau$.
Nov
25
comment Fermat's last theorem over larger fields
Could a Russian speaker add a summary of the results in this paper to this answer, please ?
Nov
25
answered On a minimal algebraic number field which satisfies the principal ideal theorem
Nov
22
awarded  Nice Question
Nov
20
comment Is elliptic curve point division defined over the field of real numbers?
If you are looking for "procedures" to compute it for general fields, then look at sage's function P.division_points(n). It starts by finding roots of the division polynomial in the field.
Nov
20
comment How to explicitly compute lifting of points from an elliptic curve to a modular curve?
If increasing the conductor is computationally hard, maybe you could enlarge the base field to get to rank 2 examples. Find a small conductor curve with large rank over a small quadratic field.
Nov
17
comment Does this equation has a closed-form solution for $t$? ($(1-p)\sum_{i=0}^{n}t^i = p\sum_{i=0}^{n}(1-t)^i)$)
The Galois group of the equation for $p=3/5$ and $n=7$ is $S_7$, which is not soluble. So don't expect a solution by radicals for $n=7$.