23

Takagi's goal is the following:
show the existence of sufficiently many cyclic extensions defined
by division values of sn $u$ (the lemniscatic sine);
prove that each abelian extension of ${\mathbb Q}(i)$ is contained in
the compositum of these fields.
Step 1 is analogous to the construction of the fields of $p^n$-th roots
of unity (in particular, ...

answered Jul 17 '13 at 18:58

Franz Lemmermeyer

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17

Yes, let's use the fact that the moduli stack of elliptic curves is etale-locally a scheme. We could also use the formal deformation space of the elliptic curve mod $\pi$ (of course we may assume $E_1 \cong E_2 \mod \pi$). We pick an etale-local model of the moduli stack of pairs of elliptic curves that includes $(E_1,E_2)$ as an $R$-point. (This many ...

15

No such newform exists.
If $\psi$ is a Groessencharacter of an imaginary quadratic field $K$, the level of the associated newform is $N_{K/\mathbf{Q}}(\mathfrak{f}) \cdot \operatorname{disc}(K/\mathbf{Q})$, where $\mathfrak{f}$ is the conductor of $\psi$. So a CM-type newform of prime level would have to come from an imaginary quadratic field $K$ of prime ...

answered Oct 11 '13 at 6:59

David Loeffler

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13

Here is the standard argument. Obviously, there are details to fill in. I'm glad to fill in the ones you have trouble with, but I want to get the overview down first.
Isomorphism over $\overline{\mathbb{Q}}$ is the same as isomorphism over $\mathbb{C}$, by the usual nonsense that allows us to exchange any two algebraically closed fields of the same ...

answered Jul 17 '15 at 15:13

David E Speyer

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13

There is a more down-to-earth definition. A newform $f=\sum_{n=1}^\infty a_n q^n$ of level $N$ and weight $k$ has CM
if there is a quadratic imaginary field $K$ such that $a_p=0$ as soon as $p$ is a prime which is inert in $K$. The field $K$ is then unique (if the weight $k \geq 2$), and one says that $f$ has CM by K.
A quick way to see the uniqueness of $K$...

13

No. If $\alpha$ is an endomorphism over $\bar L$ such that $n\alpha$
is defined over $L$, then so is $\alpha$. Proof: suppose some element
of Gal$(\bar L / L)$ takes $\alpha$ to an endomorphism $\beta$.
We shall prove that $\beta=\alpha$. Indeed by hypothesis $n\beta = n\alpha$.
But then $n(\beta-\alpha) = 0$, so the endomorphism $\beta-\alpha$
takes $E$ ...

answered Nov 16 '16 at 21:42

Noam D. Elkies

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11

What's usually meant when phrased this way is that within the $\overline K$-isomorphism class of $E$, there is an elliptic curve defined over $K(j(E))$. An indeed, for any elliptic curve $E$ defined over $\overline{\mathbb Q}$, there is an elliptic curve $E'$ defined over $\mathbb Q(j(E))$ that is $\overline{\mathbb Q}$-isomorphic to $E$. So that's the sort ...

answered Jan 14 '18 at 22:42

Joe Silverman

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10

The Answer to Question 1 is YES. It follows from Serre's variant of Hilbert's irreducibility theorem (for infinite Galois extensions) combined with the Tate conjecture on homomorphisms of abelian varieties in char 0 (proven by Faltings); see Prop. 1.3 and Cor. 1.5 of 1995 Compositio paper by Rutger Noot (vol 97, Oort Festschrift) for details.
The main ...

10

Your conjecture is true and follows from a theorem in Cox's book ''Primes of the form $x^{2} + ny^{2}$.'' (Theorem 14.16 on page 317, although this is from the first edition.) In particular, if $\mathcal{O}$ is an order in an imaginary quadratic field, $L$ is the ring class field associated to $\mathcal{O}$ and $E/L$ is an elliptic curve, with good reduction ...

9

For the elliptic curve $E$ in the original post, we have two periods $\lambda_i = \int_{\gamma_i} \frac{dx}{y}$, $i=1,2$, and quasi-periods $\eta_i = \int_{\gamma_i} \frac{x\,dx}{y}$, $i=1,2$.
Theorem (Schneider 1936): Assume that $E$ is defined over $\overline{\mathbb{Q}}$. I.e., $g_2$, $g_3 \in \overline{\mathbb{Q}}$.
Each of $\lambda_1$, $\lambda_2$, $\...

9

I recently learned a fact which seems to answer part of my question: the $CM$ extension $\mathbb{Q}^{CM}$ of $\mathbb{Q}$ is exactly the extension of $\mathbb{Q}$ generated by the Weil numbers (see for example Appendix D of this paper by Drinfeld: http://arxiv.org/abs/1007.4004 ).
Recall that $\alpha \in \overline{\mathbb{Q}}$ is a Weil number if is a $q$-...

9

The answer to your Question 3 is YES with the ground field $\mathbb{Q}$.
Here is a sketch of the proof. For each positive integer $q$ and a "parameter" $t$ (in char 0) consider the smooth projective model $C_{q,t}$ of an affine curve $y^q=x^3-x-t$. Let $P_{8,t}$ be the Prym variety of the double cover
$$C_{8,t}\to C_{4,t}, (x,y)\mapsto (x,y^2).$$
Then $P_{8,...

8

The term "$X_{(p)}$ has Hasse invariant 0" means that "$X_{(p)}$ is supersingular". See the definition in Hartshorne's book 2 pages before the quoted remark. So the quoted remark says that there are infinitely many supersingular primes for an elliptic curve over $\mathbb{Q}$, i.e., for any finite set $S$ of primes there is a supersingular prime outside $S$.

8

There is such an elliptic curve in char $p$ if and only if $p$ splits in $K$. The elliptic curve in char. zero with $K$ as endomorphisms is CM so has potential good reduction everywhere and the reduction modulo a prime above $p$ is an elliptic curve in char $p$ with $K$ contained in the (field of fractions of) the endomorphism ring. If the reduction is ...

answered Nov 30 '14 at 14:26

Felipe Voloch

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8

For $|q|<1$ we consider the null Jacobi theta functions
$$
\theta_2(q):=\sum^{\infty}_{n=-\infty}q^{(n+1/2)^2}\textrm{, }
\theta_3(q):=\sum^{\infty}_{n=-\infty}q^{n^2}\textrm{, }
\theta_4(q):=\sum^{\infty}_{n=-\infty}(-1)^nq^{n^2}.
$$
For $q=e^{-\pi \sqrt{r}}$, $r>0$ the elliptic singular modulus $k=k_r$ is given by
$$
k_r=\left(\frac{\theta_2(q)}{\...

nt.number-theory analytic-number-theory algebraic-number-theory modular-forms complex-multiplication

7

The action of $L$ on global 1-forms would give an embedding of $L$ into the algebra $M$ of $g$-by-$g$ matrices over $\mathbb{Q}$ (since we're in characteristic 0). But any maximal commutative $\mathbb{Q}$-subalgebra of $M$ is $g$-dimensional.

7

I am sure others (magma, etc) can do the same. With sage the following lines will produce the rational functions $\bigl(f(x,y), g(x,y)\bigr)$ representing the multiplication by $(1\pm\sqrt{-7})/2$ on $E$ in the Weierstrass equation given at the link. Once you have that you can get all endomorphisms.
sage: E = EllipticCurve([1,-1,0,-2,-1])
sage: K.<t> =...

7

The answer to your Question 1 is NO. See Th. 5.1 of my 2015 JPAA paper.
The answer to your Question 2 is YES. See 1963 Annals paper of Shimura "On analytic families ..".

7

Sorry for self-advertisement: the question of the "sign" that one must choose
(equivalently which $\pi$ or $\overline{\pi}$) is entirely answered in my
book GTM239, Section 8.5.2, some of the results being due to Mark Watkins
(unpublished I believe). The main idea is to show that any CM curve is
equivalent in an evident sense to a "basic" CM elliptic curve ...

6

``Given $(K,\Phi)$ , there always exists a $g$ -dimensional abelian variety $A$ such that $End(A)\otimes Q=K$ and that $K$ acts on $H^0(A,\Omega^1)$ through $\Phi$. One easily constructs as a complex torus, starting from the embedding $K\subset C^g$ given by $\Phi$."
Actually, this is not always the case. For example, if $K$ is a quartic CM-field ...

6

This is covered at the end of Serre's book "Abelian l-adic Representations and Elliptic Curves".
Basically, since $\mathrm{Gal}(\bar{K}/K)$ acts semi-simply on $V_p(E)$ (you might have to go to Lang's book "Abelian Varieties" for details on this) the sequence of $\mathrm{Gal}(\bar{K}/K)$-modules
$$0\to V_p(E(p)^\circ) \to V_p(E) \to Y \to 0$$
splits, ...

6

The usual method to do this is to use formulas due to Velu. There's an implementation in PARI given at http://perso.ens-lyon.fr/francois.brunault/parigp/velu.gp

answered Aug 16 '16 at 13:14

Joe Silverman

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6

The simplest generalisation to abelian surfaces is, I believe, the statement (which is a theorem, not a conjecture) that the Igusa invariants of an abelian surface with CM by $K$ generate an abelian unramified extension of the reflex field of $K$. Note that the reflex field is not, in general, equal to $K$, and that there is no claim that the full Hilbert ...

5

From the first page of Scholl's paper:
1.0.0. Consider integers $n \geq 3$, $k \geq 1$. (We do not treat here the case $k = 0$, which corresponds to cusp forms of weight $2$; the associated motives are then given by the Jacobians of modular curves, and are well understood.)
So (a) and (b) are indeed isomorphic, but this is basically just the modularity ...

5

See J. Silverman, Advanced Topics in the Arithmetic of Elliptic Curves (Springer GTM), Proposition 2.3.1. Note that since the class number of $\mathbb{Q}(\sqrt{-7})$ is 1, this is the only example defined over $\mathbb{Q}$.

5

Each cyclic subgroup is Galois invariant if and only if its $x$ coordinate is Galois invariant, which happens if and only if it lies in $\mathbb Q$. So the first one is Galois invariant and then, if $4B$ is a perfect cube, one of the other three is Galois invariant.
The action of $\mathcal O$ is generated by the action of the third roots of unity, which ...

5

The statement about $E_2^*(\tau)$ is Proposition 5.10.6 in Cohen-Strömberg, Modular forms: a classical approach.

answered Oct 4 '18 at 0:30

François Brunault

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5

I have seen this statement about $E_2^*$ tossed around off-handedly by experts a number of times, but never seen a complete proof referenced.
The tools to prove it are (mostly) in Masser's "Elliptic functions and transcendence", Appendix 1. There, Masser gives a formula for certain non-holomorphic modular functions. One of these fomrulas (Lemma A3) can be ...

5

Presumably, the author aimed only at showing that the order $\mathfrak{O}_t$ is a one-dimensional Noetherian domain, i.e., it is a Noetherian domain and every of its non-zero prime ideals is maximal. He succeeded in doing so.
The subsequent claim about the primary decomposition of a non-zero ideal in a one-dimensional Noetherian integral domain holds true ...

5

The statement that seems odd to you is indeed false, what (I believe) he wants to say is that $\mathfrak{D}_t$ is a Noetherian ring in which every prime ideal $\mathfrak{p}\neq 0$ is maximal.
I believe that all you need is in Atiyah-Macdonald's Introduction to Commutative Algebra, to which I will refer as [AM69].
Since $\mathfrak{D}_t$ is Noetherian, every ...

answered Nov 6 '19 at 1:44

Filippo Alberto Edoardo

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