User tommaso centeleghe - MathOverflow

Power series whose partial sums attain only finitely many values

2013-05-16T00:41:33Z

I am wondering if there is a general explanation for the following phenomenon. The partial sums of the geometric series $\sum_{n\geq 0} x^n$ evaluated at a root of unity $\zeta\neq 1$ in the complex plane attain only finitely many values (since $1+\zeta+\ldots+\zeta^{n-1}=0$, where $\zeta^n=1$). The average of these values is $1/(1-\zeta)$, which is the value at $\zeta$ of the meromorphic function that analytically continues the geometric series. Out of curiosity, I would like to ask if this is a special case of a general theorem?

Decomposition of primes in Galois closures of number fields

2013-02-13T11:03:16Z

Let $L/K$ be an extension of number fields, and $M/K$ the Galois closure of $L/K$ (everything happens inside a suitably large characteristic zero field $\Omega$). Let $p$ be a discrete prime of $K$.

Given the group $G={\rm Gal}(M/K)$, its subgroup $H={\rm Gal}(M/L)$, and the splitting of $p$ in $L/K$ can one find the splitting of $p$ in $M/K$? That is, can one find the ramification index $e$ and inertial degree $f$ of a prime $P$ of $M$ lying above $p$?

Turning the above question into a group-theoretic one, I got the following: let $D$ be a finite group, and $X$ a finite set on which $D$ acts faithfully on the right. Can we obtain the order $d$ of $D$ knowing the sizes of all the $D$-orbits in $X$?

(To switch to this question from the original one, look at the natural right action of a decomposition group (resp. an inertia group) $D$ at $P$ on the coset space $X=H\backslash G$. The fact that $M$ is the Galois closure of $L$ ensures that $D$ acts on $X$ faithfully.)

Answer by Tommaso Centeleghe for which integers take the form x^2 + xy + y^2 ?

2011-10-19T14:07:50Z

One can start showing the following:

The integers $n$ which are of the form $x^2+xy+y^2$, for two relatively prime integers $x,y$ are precisely those positive integers occurring as divisors of $m^2+m+1$, for some integer $m$.

In other words, the polynomial $f(x)=x^2+x+1$ has the property that the positive divisors of the integers it represents are precisely those integers that can be properly represented by its homogenization $F(x,y)=x^2+xy+y^2$ ("properly" here refers to the condition $(x,y)=1$).

The proof uses the fact that the imaginary quadratic order $\mathbf{Z}[x]/f(x)$ has class number one, as anticipated by Elkies. It goes as follows:

Let $n$ be a positive divisor of $m^2+m+1$, for some integer $m$. Consider the quadratic form in $x,y$ given by:

$Q(x,y)=\frac{m^2+m+1}{n}x^2-(2m+1)xy+ny^2$;

it has integer coefficients, positive definite, and has discriminant equal to $-3$ (in particular it is primitive).

Since $h(-3)=1$, there is only one reduced, positive definite quadratic form of discriminant $-3$. This is $E(x,y)=x^2+xy+y^2$. Therefore $Q$ and $E$ are properly equivalent (that is there is a determinant-one change of variables taking one into the other), and since $Q$ certainly properly represents $n$, so does $E$.

The converse is similar and uses the fact that if a quadratic form $Q$ properly represents an integer $n$, then $Q$ is properly equivalent to a form of the type $nx^2+bxy+cy^2$ (this is lemma 2.3 of Cox's wonderful book "Primes of the form x^2+ny^2").

Once you understand the positive integers $n$ that are properly represented by $E$, then you can get them all, after scaling by squares.

The original problem is then reduced to understanding those integers $n$ for which $x^2+x+1$ has a zero in $\mathbf{Z}/(n)$. Using the Chinese Remainder Theorem, this can be reduced to the case where $n$ is a prime power $p^s$. Then for $p\neq 3$ Hensel's Lemma tells you that your equation has a solution mod $p^s$ if and only if it has a solution mod $p$. With quadratic reciprocity you can conclude that any prime divisor of $n$ has to be congruent to $1$ mod $3$. I am at the moment missing how you can solve the equation $x^2+x+1=0$ in $\mathbf{Z}/3^s$, but I think that a version of Hensel's Lemma applies. I'll think about it.

[EDIT: The equation $x^2+x+1=0$ has a solution in $\mathbf{Z}/(3^s)$, with $s\geq 1$, if and only if $s=1$. (Therefore $x^2+xy+y^2$ represents properly only $3$ and $1$ as powers of $3$.) One can see this by checking that there is no solution for $s=2$, and therefore for $s>2$. More generally, if $p>2$ then $x^{p-1}+x^{p-2}+\ldots+x+1=0$ has no solutions in $\mathbf{Z}/(p^s)$ with $s>1$. The $p$--adic valuation of an integer of the form $x^{p-1}+x^{p-2}+\ldots+x+1=(x^p-1)/(x-1)$ is either zero or one.]

(Suggested reading. Cox's book quoted above and Serre's paper: $\Delta=b^2-4ac$)

Ordinary vs Non-ordinary for GL(2)-type Abelian Surfaces over Q

2012-11-17T14:51:00Z

Let $A_f$ be an abelian surface over $\mathbf{Q}$ of $\mathbf{GL}_2$-type arising from a weight $2$ cuspidal eigenform $f\in S_2(\Gamma_0(N))$. What is known (or expected to be true) for the size of the set of primes $p$ such that $A_f$ mod $p$ is ordinary (resp. non-ordinary)? Do you know of a reference where this is discussed?

How to construct Weil numbers in a given CM quartic field?

2012-11-02T01:42:48Z

Let $L$ be a CM field of degree $4$ over the rationals, and let $p$ be a prime number. If $q$ is a power of $p$, I would like to know if it is possible to characterize (in some way) all Weil ${\bf F}_q$-numbers inside $L$.

I was inspired by the corresponding question when $L$ has absolute degree $2$ (i.e., it is an imaginary quadratic field), which has a simple answer: Weil $q$-numbers $\pi$, up to roots of unity in $L$, correspond to principal ideals of norm $q$. The answer in this case is especially simple to get because $L$ has only one archimedean place.

How harder is the problem when $L$ is quartic? Thanks.

On simple factors of modular jacobians: endomorphism ring and simplicity of mod p reduction

2012-10-09T18:46:28Z

Let $A_f$ be the abelian variety over $\mathbf{Q}$ arising as a $\mathbf{Q}$-simple factor of the Jacobian $J_0(N)$ of the modular curve associated to a normalized newform $f$ of weight $2$ on the congruence subgroup $\Gamma_0(N)$. There is a piece $\mathbf{T}_f$ of the Hecke ring acting on $A_f$. Such a ring is an order in the number field generated by the Fourier coefficients of $f$, whose degree equals the dimension of $A$. It is a result of Ribet (I think) that the natural map $\mathbf{T}_f\otimes\mathbf{Q}\rightarrow\textrm{End}(A)\otimes\mathbf{Q}$ is an isomorphism, where $\textrm{End}(A)$ is the ring of $\mathbf{Q}$-enndomorphisms of $A$ (I couldn't attach the index to End..).

I have two questions, please:

1) Can we find an abelian variety $A_f'$ over $\mathbf{Q}$, which is $\mathbf{Q}$-isogenous to $A_f$, and such that $\textrm{End}(A_f')$ is the maximal order of $\textrm{End}(A_f')\otimes\mathbf{Q}=\textrm{End}(A_f)\otimes\mathbf{Q}$? (Notice that in the one dimensional case $\textrm{End}(A_f)$ is already maximal, being it the ring of integers)

2) Does the fact that $A_f$ has such a large endomorphism ring imply that the mod $p$ reduction of $A_f$ be simple over the prime field with $p$ elements? Here $p$ is a prime of good reduction for $A_f$.

Thanks!

Answer by Tommaso Centeleghe for Conjugation in GL(n) (p-adic setting)

2012-03-06T21:21:18Z

The problem can be reduced to that of classifying $GL(n,\mathbf{Z}_p)$ conjugacy classes in $M(n,\mathbf{Z}_p)$. The situation for general $n$ is complicated, but for $n=2$ the problem is settled by the following.

Let $F\in M(2,\mathbf{Z}_p)$ be any matrix, let $f(x)$ be its characteristic polynomial, and let

$n(F)=\sup_i(i\in\mathbf{Z}_{\ge 0},$ $F$ mod $p^i$ is multiplication by a scalar $)$.

Then the $GL(2,\mathbf{Z}_p)$ conjugacy class of $F$ is uniquely determined by $f(x)$ and $n(F)$.

If $n(F)$ is infinite, then $F$ is a scalar matrix and therefore central.

If $n(F)$ is finite then there exists a unique integer $\lambda\in\mathbf{Z}$, with $0\le\lambda\le p^{n(F)}-1$ such that $F$ is conjugate to $\begin{pmatrix} \lambda&0\\ 0&\lambda \end{pmatrix}+p^{n(F)}\begin{pmatrix} 0&-a_0\\ 1&-a_1 \end{pmatrix}$ .

Here $a_0$ and $a_1$ are the constant and linear term of the polynomial $f_0(x):=p^{-2n(F)}f(p^{n(F)}x+\lambda)$, which has coefficients in $\mathbf{Z}_p$.

We have that $p^{n(F)}$ is the index of the ring $\mathbf{Z}_p[F]$ inside the ring $R_F:=\mathbf{Q}_p[F]\cap M(2, \mathbf{Z}_p)$.

All rings are viewed as subrings of $M(2,\mathbf{Q}_p)$. We will sometime think of $F$ and of elements of $R_F$ as endomorphisms of the standard lattice $\mathbf{Z}_p^2$ inside $\mathbf{Q}_p^2$.

Proof: if $n(F)$ is infinite then there is not much to prove, therefore we assume $n(F)$ finite, and $F$ not central. The ring $R_F$ as defined above contains $\mathbf{Z}_p[F]$ with finite index, since they both are finite free $\mathbf{Z}_p$-modules of rank two. This is clear for $\mathbf{Z}_p[F]$, since $f(x)$ is the minimal polynomial of $F$, since $F$ is not central. For $R_F$ it follows from the fact that $\mathbf{Q}_p[F]\cap M(2, \mathbf{Z}_p)$ is open, compact, and non-empty in $\mathbf{Q}_p[F]$, which has rank two over $\mathbf{Q}_p$, since $F$ is not central.

The ring $R_F$ has a $\mathbf{Z}_p$-basis of the form $(1, F')$ where $F'=(a+bF)/p^{h}$, for some $a,b\in\mathbf{Z}_p$ not both divisible by $p$, and where $p^h$, with $h\ge 0$, is the index of $\mathbf{Z}_p[F]$ in $R_F$. The $p$-adic integer $b$ is a unit, for otherwise $p^{h-1}F'-b'F=a/p$, with $b'=b/p\in\mathbf{Z}_p$, would belong to $R_F$, which is not possible since $a/p$ is not a $p$-adic integer. This shows that the natural action of $F$ on $\mathbf{Z}_p^2/(p^h)$ is multiplication by $-ab^{-1}$ mod $p^h$. Therefore $h\leq n(F)$.

On the other hand, if $\lambda$ is any integer such that $F-\lambda$ is zero mod $p^{n(F)}$, then $F'':=(F-\lambda)/p^{n(F)}$ is an element of $R_F$, since $F-\lambda$ commutes with $F$ and it is divisible by $p^{n(F)}$ in $M(2,\mathbf{Z}_p)$. Therefore $n(F)\leq h$. Thus $n(F)=p^h$ and $(1, F'')$ is a $\mathbf{Z}_p$-basis of $R_F$ (since it spans a lattice of the correct index).

Now, by the maximality of $n(F)$ we see that $F''$ does not act via scalar multiplication on $\mathbf{Z}_p^2/p$. This implies that there is a $\mathbf{Z}_p$-basis $(e_1, e_2)$ of $\mathbf{Z}_p^2$ such that $F''(\mathbf{Z}_p\cdot e_1)\not\equiv \mathbf{Z}_p e_1$ mod $p$. It follows that $(e_1, F''(e_1))$ is also a $\mathbf{Z}_p$-basis of $\mathbf{Z}_p^2$.

With respect to this basis the action of $F''$ is given by a matrix of the form $\begin{pmatrix} 0&-a_0\\ 1&-a_1 \end{pmatrix}$ , where $a_0$ and $a_1$ are the constant and the linear term of the characteristic polynomial of $F''$, which is $f_0(x):=p^{-2n(F)}f(p^{n(F)}x+\lambda)$ and has coefficients in $\mathbf{Z}_p$. By picking $\lambda$ in the range $0,\ldots,p^{n(F)}-1$, we see that the action of $F$ with respect to the basis $(e_1, F''(e_1))$ is that given by the statement.

Notice that this shows that $\mathbf{Z}_p^2$ is a free $R_F$-module of rank one, and classifying the action of $F$ on $\mathbf{Z}_p^2$ is roughly equivalent to finding $R_F$. I was interested exactly in this in the context of Tate modules of elliptic curves over finite fields ($F=$Frobenius). Probably there is a more conceptual/simpler proof. I would be interested to hear what you get in higher dimension. It won't be that easy, I expect. What makes this case simple is that orders of $\mathbf{Q}_p[F]$ containing $F$ are classified by the index with which $\mathbf{Z}_p[F]$ sits in them.

When do the sizes of conjugacy classes and squares of degrees of irreps give the same partition for a finite group?

2010-04-05T09:05:51Z

I should admit the question below does not have a serious motivation. But still I found it somehow natural.

Let $G$ be a finite group of order $n$ with $h$ conjugacy classes. If $c_1,\ldots,c_h$ are the orders of the conjugacy classes of $G$, then clearly

$n=c_1+c_2+\ldots+c_h$.

Let now $\pi_1,\ldots,\pi_h$ be the pairwise non-isomorphic, irreducible complex representations of $G$. It is well known that another partition of $n$ of length $h$ is given by the squares of the degrees $d_i$'s of the $\pi_i$'s:

$n=d_1^2+d_2^2+\ldots+d_h^2$.

Question: Assume that, up to reordering, the two partitions of $n$ described above are the same. Then what can we say about $G$? Is $G$ forced to be abelian?

Answer by Tommaso Centeleghe for about Kummer Theory

2012-03-11T09:55:06Z

http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.jmsj/1261734945

Chevalley shows the following related statement (Remarque p. 39): let $p$ be a prime, $K$ a field of characteristic different from $p$, and $\zeta$ a $p^e$-th primitive root of $1$ in some algebraic extension of $K$, where $e\geq 1$ is any integer. Assume moreover that $-1$ is a square in $K$ if $p=2$. Then if an element $x\in K(\zeta)$ is a $p^e$-th power, then it is already a $p^e$-th power in $K$.

Endomorphisms rings of elliptic curves and congruences of $j$

2012-03-02T23:27:09Z

Let $p$ be a prime number, $K/\mathbf{Q}_p$ a finite extension, with integers $O_K$, valuation ideal $\mathfrak{p}$, and residue field $k_\mathfrak{p}$. Let $E$ be an elliptic curve over $K$ with good reduction $E_\mathfrak{p}$ over $k_\mathfrak{p}$.

If $\ell$ is a prime $\neq p$, then $T_\ell(E)$ is identified with $T_\ell(E_\mathfrak{p})$ in a natural way, by the good reduction of $E$. As it turns out such a Galois representation is determined, up to isomorphism, by the characteristic polynomial $f_{E_\mathfrak{p}}(x)=x^2-a_{E_\mathfrak{p}}x+|k_\mathfrak{p}|$ associated to $E_\mathfrak{p}$ and by $j_E$ mod $\mathfrak{p}=j_{E_\mathfrak{p}}$ UNLESS we are in the following (very) ${\it special}$ case:

$p\equiv 3$ mod $4$; $|k_\mathfrak{p}|=p^{2m+1}$; $a_{E_\mathfrak{p}}=0$; $\ell=2$; and $j_E\equiv 1728$ mod $\mathfrak{p}$.

If the first three conditions hold, then $E_\mathfrak{p}$ is supersingular and its endomorphisms ring over $k_\mathfrak{p}$ is "only" isomorphic to an order in $\mathbf{Q}(\sqrt{-p})$ containing $\sqrt{-p}$, and thus isomorphic to either $\mathbf{Z}[\sqrt{-p}]$ or to $\mathbf{Z}[(1+\sqrt{-p})/2]$. The second case occurs precisely when all the two torsion is defined over $k_\mathfrak{p}$, the first case when $E_\mathfrak{p}[2]$ has only two $k_\mathfrak{p}$-points. Both cases do arise and give rise to non-isomorphic $T_2(E_\mathfrak{p})$.

Essentially by Deuring's Lifting Lemma one can decide which of the two possibilities occurs by looking at the $j$-invariant of $E_\mathfrak{p}$ UNLESS this is equal to 1728. The point is that if $j_{E_\mathfrak{p}}\neq 1728$ then the two $k_\mathfrak{p}$-forms of $E_\mathfrak{p}\otimes_{k_\mathfrak{p}}\bar k_\mathfrak{p}$ lying in the $k_\mathfrak{p}$-isogeny class $a_{E_\mathfrak{p}}=0$ have the same ring of endomorphisms over $k_\mathfrak{p}$, out of the two possibilities listed above. The opposite being true when $j_{E_\mathfrak{p}}=1728$ (this fact is very related to the analysis of the mod $p$ reduction of Hilbert Class Polynomials associated to discriminants $-p$ and $-4p$ done by Gross and Elkies (cf. $\S 2$, Proposition, in Elkies' "The existence of infinitely many supersingular primes for every elliptic curve over $\mathbf{Q}$", Inventiones 89 (1987))).

In other words, in the special case the pair $(f_{E_{\mathfrak{p}}}(x), j_E$ mod $\mathfrak{p})$ does ${\it not}$ determine $T_2(E_\mathfrak{p})$.

Here is the question then: in the special case can we determine what is the endomorphism ring of $E_\mathfrak{p}$ (and hence $T_2(E)$) from congruences of $j_E$ mod a higher power of $\mathfrak{p}$ (or of $p$)?

It is not even clear to me whether this should be possible, let alone what power of $p$ we would need to tell one case from the other. The hope behind this is that the $j$-invariant of $E$ be "close" to that of the CM lift of $E_\mathfrak{p}$ and of its endomorphisms ring over $k_\mathfrak{p}$. Thanks.

PS: I do not know if the question above has anything to do with http://mathoverflow.net/questions/45638/is-there-a-classical-proof-of-this-j-value-congruence

[EDIT: I realize that for clarity of exposition I should have probably recalled that $T_\ell(E_\mathfrak{p})$ for $\ell\neq p$, in the above notation, is a free ${\rm End}(E_\mathfrak{p})\otimes\mathbf{Z}_\ell$-module of rank one. Therefore, roughly, the knowledge of either of the two is equivalent to that of the other]

Generalization of singular moduli

2012-02-09T21:33:31Z

$j$-invariants of CM curves $E$ over (say) the complex numbers are known as singular moduli. As the theory of complex multiplication explains, singular moduli are algebraic integers of great arithmetical significance, for they generate the maximal abelian extensions of imaginary quadratic fields.

I would be interested in reading about generalizations of this, where $E$ is replaced by an abelian variety of higher dimension with a big endomorphism ring. What are the singular moduli in this case, and what are their features? Can you recommend some references please? Or better address this question? Thanks!

[ERRATUM: singular moduli only generate the maximal abelian extensions of imaginary quadratic fields K over which ${\rm Gal}(K/\mathbf{Q})$ acts by inversion.]

Answer by Tommaso Centeleghe for reduction of CM elliptic curves

2012-02-09T17:28:53Z

Let $k$ be a finite field of char. $p$ and $E$ an elliptic curve over $k$. Denote by $R_E$ the endomorphism ring of $E$ over $k$. This ring has a distinguished element given by the purely inseparable $k$-isogeny $\pi_E:E\rightarrow E$ given by the Frobenius endomorphism relative to $k$. Its degree is $|k|$. We know that $\pi_E$ satisfies the Weil polynomial $f(x)=x^2-a_Ex+|k|$, where $a_E$ is the error term $|k|+1-|E(k)|$ (it is not hard to deduce this from the fact that $|E(k)|$ is equal to the degree of the separable $k$-isogeny $1-\pi_E$).

Assume now that $F$ is an imaginary quadratic field that embeds inside $R_E\otimes\mathbf{Q}$. If $E$ is supersingular, then there is a finite extension $k'$ of $k$ such that $R_E':={\rm End}_{k'}(E\otimes_k k')$ becomes a maximal order of "the" quaternion over $\mathbf{Q}$ ramified precisely at $p$ and infinity (in fact $k'$ of degree $2$ suffices "most times"). Considering the embedding $F\rightarrow R_E\otimes\mathbf{Q}\subset R_E'\otimes\mathbf{Q}$ you see that $F$ cannot be split at $p$ (and at infinity) for otherwise it would not embed in such a quaternion algebra (Vigneras' book has all of this basics).

If $E$ is ordinary then $R_E\otimes\mathbf{Q}$ is isomorphic to the quadratic field obtained by joining to $\mathbf{Q}$ a root of $f(x)$, therefore so is $F$ and since $p\nmid a_E$ we see easily that the discriminant of $f(x)$ is not divisible by $p$ and it is a square mod $p$, thus $p$ splits in $F$.

Back to your original question, the reduction mod $p$ of a CM ellipitic curve induce a nice, injective reduction map at the endomorphisms level. Therefore you should get what you wanted!

(Waterhouse thesis "Abelian varieties over finite fields" is a great source to learn these things. The Bourbaki talk by Tate on abelian varieties over finite fields is a must to learn these things)

How does one get that for elliptic curves $A$, $B$ over a number field, $a_\mathfrak{l}=b_\mathfrak{l}$ implies that the curves are isogenous?

2011-11-26T12:29:22Z

Let $K$ be a number field, and let $A$ and $B$ be two elliptic curves over $K$. For a nonzero prime ideal $\mathfrak{l}$ of $K$, outside a finite set of primes, let $a_\mathfrak{l}$ and $b_\mathfrak{l}$ be the usual error terms $N(\mathfrak{l})+1-\bar A_\mathfrak{l}(\mathcal{O}/\mathfrak{l})$ and $N(\mathfrak{l})+1-\bar B_\mathfrak{l}(\mathcal{O}/\mathfrak{l})$ respectively (here $N$ is the absolute norm and $\bar X_\mathfrak{l}$ denotes the reduction of the curve $X$ at a good place $\mathfrak{l}$).

It is well known that the collection of these error terms determines the $K$-isogeny class of the curve considered. I am at the moment missing how the "correct" proof of this fact goes.

What I see so far is that, thanks to Faltings' Isogeny Theorem, it is enough to show that ${\rm Hom}_{G_K}(T_p(A),T_p(B))$ is non-zero, for some prime $p$. Now, our assumption that the packages of error terms of $A$ and $B$ coincide for almost all primes of $K$ ensures that the $G_K$-representations $T_p(A)\otimes\mathbf{Q}_p$ and $T_p(B)\otimes\mathbf{Q}_p$ are ${\it locally}$ isomorphic for almost all primes of $K$. How do we get then that they are ${\it globally}$ isomorphic, and hence conclude

${\rm Hom}_{G_K}(T_p(A),T_p(B))\neq 0$? If they came from an automorphic form, I see that one can use the strong multiplicity one result by J-L. But otherwise how can one complete the argument?

What's known about complete split primes in Q(E[p])?

2011-10-10T02:20:44Z

Let E be an elliptic curve over $\mathbf{Q}$, and p be a prime of good reduction for E such that the Galois representation $\bar\rho_p$ of $\mathbf{Q}$ on the p-torsion of E surjects onto Aut(E[p]). Let now K be the number field cut out by the kernel of $\bar\rho_p$. What is known about the primes q that are completely split in K? Have they been characterized?

Answer by Tommaso Centeleghe for Which math paper maximizes the ratio (importance)/(length)?

2011-11-08T00:17:48Z

What about Ribet's great inventiones paper from the 70s $\textit{ A modular construction of unramified }p-\textit{extensions of }\mathbf{Q}(\mu_p)$? I think it should be mentioned!

Reference request for projective representations of finite groups over a non-problematic field

2011-05-10T14:21:52Z

I would like to get a reference where I can learn about the theory of projective representations of finite groups over the complex numbers (or over any field K such that the order of the given group under study is invertible in K). And how one can relate this to character theory for linear representations (with which I am more familiar). Thanks.

Can the Galois representation on the $p$-adic Tate module of $E/\mathbf{Q}_p$ be recovered from the $p$-divisible group associated to the mod $p$ good reduction of $E$?

2011-10-11T22:26:33Z

Let $p$ be a prime number, and $E/\mathbf{Q}_p$ and elliptic curve with good reduction $\bar E_p$. Can the Galois representation of $\mathbf{Q}_p$ given by the $p$-adic Tate module of $E$ be recovered from the Dieudonn\'e module associated to the $p$-divisible group $\bar E_p[p^\infty]$ of the reduction $\bar E_p$?

In case of ordinary reduction, can the [EDIT: possible] splitting of the $p$-torsion $E[p]$, as Galois module, be understood in terms of the Dieudonn\'e module of the $p$-torsion $\bar E_p[p]$?

Thanks!

Finite dimensional automorphic representations of a definite quaternion with prime discriminant and Hecke action

2010-09-26T12:58:30Z

Before stating the questions that I have, which are very specific and probably not so interesting to someone who has never thought about these things, I need to introduce some notation.

Let $p$ be any prime, and let $D$ be the quaternion algebra over $Q$ ramified precisely at $p$ and infinity. Choose a maximal order $R$ inside $D$. For any prime $\ell\neq p$, fix an isomorphism of $D_\ell:=D\otimes Q_\ell$ with the algebra $M_2(Q_\ell)$ in such a way that the maximal compact subring $R_\ell:=R\otimes Z_\ell$ corresponds to $M_2(Z_\ell)$.

If $N$ is an integer $>0$ not divisible by $p$, let $U_\ell(N)$ be the subgroup of $R_\ell^\star\simeq GL_2(Z_\ell)$ given by those matrices whose bottom row is congruent to $(0$ $1)$ modulo $N$ (equivalently, modulo the highest power of $\ell$ dividing $N$).

The ring $R_p:=R\otimes Z_p$ has a unique maximal, two-sided principal ideal $(\pi)$ generated by any uniformizer $\pi$, the residue field $R/(\pi)$ is a finite field with $p^2$ elements. We let $R_p^\star(1)$ denote the subgroup of $R_p^\star$ given by the units that are congruent to $1$ modulo $(\pi)$.

Let now $D^\star$ be the multiplicative group of $D$, viewed as an algebraic group over $Q$. For any integer $N>0$ not divisible by $p$, we are going to define an open subgroup $U(1,N)$ of the group $D^\star_A$ of points of $D^\star$ valued in $A$, the adele ring of $Q$. Namely $U(1,N)$ is the product of all the $U_\ell(N)$, for $\ell\neq p$; of $R_p^\star(1)$; and of the full (connected) component at infinity $D^\star_\infty$.

Consider the space $S(1,N)$ of complex valued functions on the double coset $D^\star\backslash D^\star_A/U(1,N)$, which is known to be finite. For any prime $\ell\neq p$ there is an Hecke operator $T_\ell$ acting on $S(1,N)$ that can be defined in terms of double cosets in the usual way.

(Let $\alpha_\ell$ be the matrix whose top row is $(\ell$ $0)$ and whose bottom is $(0$ $1)$; decompose $U_\ell(N)\alpha_\ell U_\ell(N)$ as a finite union of left cosets $\gamma_i U_\ell(N)$; for $f\in S(1,N)$ define $T_\ell(f)(x)=\sum f(x\gamma_i)$).

Let $V$ be the vector space of locally constant, complex valued functions on $D^\star_A$ that are left invariant by $D^\star$. Observe that $S(1,N)$ can be viewed as a finite dimensional subspace of $V$. Right translation defines an admissible representation of $D^\star_A$ on $V$ which is known to be completely decomposable into a discrete direct sum of irreducible admissible representations of $D^*_A$. If $f\in S(1,N)$, then denote by $V_f$ the smallest subspace of $V$ that is stable by $D^\star_A$.

Questions:

1) Let $f\in S(1,N)$, for some $N$. Is it true that the space $V_f$ is finite dimensional if and only if $f:D^\star_A\rightarrow C$ factors through the reduced norm map $Nr:D^\star_A\rightarrow A^\star$?

2) Does the subspace of $S(1,N)$ given by those functions that factor through the reduced Norm admit an Hecke stable complement?

3) Is the action of $T_\ell$ on $S(1,N)$, for $\ell\nmid pN$, semisimple?

4) Assuming that 2) holds, and letting $S_0(1,N)$ be such complement, how do we relate the C-subalgebra $T_0(N)$ of End($S_0(1,N)$) generated by the Hecke operators $T_\ell$, with $\ell\nmid pN$, to a C-algebra of Hecke operators acting on weight 2 cusps forms of a certain level? What I mean is: out of the J-L correspondence, can we read off an isomorphism between $T_0(N)$ and some Hecke algebra coming from classical modular forms?

Thanks.

[EDIT: In the 2nd and 3rd lines above "Questions:" I should have probably have said "discrete Hilbert direct sum", the "direct sum" being only dense in V]

Elliptic curves over $\mathbf{Q}$ with isogenous mod $\ell$ reductions, for several $\ell$

2011-10-07T13:31:55Z

Characteristic polynomials of Hecke operators $T_\ell$, with $\ell$ prime, acting on cusp forms $S_k$ of level one and weight $k$ "appear to be" squarefree (even irreducible!).

This can be interpreted as follows: if $p$ is any prime, then the $p$-adic Galois representations $\rho_i$, where $1\leq i\leq \dim(S_k)$, attached to eigenforms in $S_k$ "appear to be" pairwise non isomorphic ${\it locally}$ at primes $\ell\neq p$.

This is completely false for other levels. For example in the two dimensional space $S_2(\Gamma_0(37))$, I learn from Magma and Cremona's tables that there are exactly two normalized eigenforms, $f_1$ and $f_2$, with rational coefficients, corresponding to two elliptic curves $E_1$ and $E_2$ defined over $\mathbf{Q}$, of conductor $37$, and uniquely determined up to $Q$-isogeny.

Looking at some Hecke operators on this space, one easily finds examples of $T_\ell$ acting diagonally on $S_2(\Gamma_0(37))$, i.e., examples of primes $\ell\neq 37$ for which the two elliptic curves have $p$-adic Tate modules isomorphic as local Galois modules at $\ell$ (some of the $\ell$'s for which this happens are $7$, $31$, $41$, $101$, $137$, $173$, $179$,..$39769$).

$Q1$: Is it reasonable to suspect that $E_1$ and $E_2$ become isogenous over an extension $F$ of $Q$? If this were the case, then one should see the phenomenon described above for primes $\ell$ that are split in $F$, right?

$Q2$: On the other hand, given an elliptic curve $E$ over $\mathbf{Q}$, what are the known ways to construct more elliptic curves $A$, defined over $\mathbf{Q}$, possibly of the same conductor as $E$, which are not $\mathbf{Q}$-isogenous to $E$ but such that they become so over a non-trivial extension of $Q$?

$Q3$: Can we say why we do not see the above phenomenon in level one?

Why a certain "Hecke polynomial" is equal to a certain "Galois polynomial"?

2011-09-20T14:17:39Z

Let $f$ be a modular form of weight $k>0$ on $\Gamma_1(N)$ that is an eigenvector for all the Hecke operators $T_\ell$, for $\ell$ prime, and with Nebentype $\epsilon$. If $\ell$ is a prime not dividing $N$, set $\iota_\ell(f)(z):=f(\ell z)$. As it is well known, $f$ and $\iota_\ell(f)$ are distinct modular forms of weight $k$ on $\Gamma_1(N\ell)$ that span a two-dimensional space denoted by $V_f$. It is also well known that any Hecke operator $T_p$, for $p$ a prime other than $\ell$, acts on $V_f$ as a scalar, and that the characteristic polynomial of the linear action of $T_\ell$ (often denoted by $U_\ell$ in this context) on $V_f$ is $X^2-a_\ell X +\epsilon(\ell)\ell^{k-1}$, where $a_\ell$ is the $\ell$--th eigenvalue of our original $f\in M_k(\Gamma_1(N))$. A possible proof of this fact is a simple computation based on the explicit formulas for the effect of Hecke operators on $q$--expansions.

On the other hand, it is also well known that if $\lambda$ is a finite prime of the number field attached to $f$, and $\rho_\lambda$ is the two dimensional, $\lambda$--adic Galois representation of $Q$ attached to $f$, then the "Hecke polynomial" mentioned in the previous paragraph coincide with the characteristic ("Galois") polynomial of $\rho_\lambda$ at a Frobenius element at $\ell$, when $\lambda$ does not divide $\ell$.

Question: is there a conceptual proof of the equality between these two polynomials? Can you point a reference where this fact is explained? thanks.

Answer by Tommaso Centeleghe for Why is there a weight 2 modular form congruent to any modular form

2011-07-09T12:20:01Z

If $N\geq 1$ is an integer not divisible by $p$, one can see that any system of Hecke eigenvalues $(a_\ell)$ arising from $S_k(\Gamma_1(N))$ is congruent mod $p$ to a system $(b_\ell)$ arising from $S_2(\Gamma_1(Np^n))$, for some $n$, using an interplay between a theorem of Serre (describing a purely mod $p$ Jacquet-Langlands correspondence), and the more classical, characteristic zero J-L between ${\rm GL}_2$ and the multiplicative group $G$ of the $\mathbf{Q}$-quaternion algebra ramified at $p$ and infinity.

I know that what I describe here is perhaps not the right way of proving the result you are asking, but it seems to me worth to mention.

In his '87 letter to Tate Serre proves:

${\rm Theorem:}$ Systems of mod $p$ Hecke eigenvalues arising from $M_k(\Gamma_1(N))$ are the same as those arising from locally constant function $f:G(A)\rightarrow\overline{F}_p$ that are left invariant under $G(\mathbf{Q})$ and right invariant under a certain open subgroup $K_N$.

Here $G(A)$ is the adelic group associated to $G$. Notice that the functions considered on the quaternion side are independent of the archimedean variable. Moreover, the double coset $G(\mathbf{Q})\backslash G(A)/K_N$ is finite and any mod $p$ system of eigenvalues arising from it can be lifted to characteristic zero.

Therefore applying the theorem and then lifting, we see that for any (char. zero) eigensystem $A=(a_\ell)$ arising from $M_k(\Gamma_1(N))$ there is a (char. zero) eigensystem $B=(b_\ell)$ arising from the space of locally constant functions $f:G(A)\rightarrow\mathbf{C}$ such that $A\equiv B$ mod $P$, where $P$ is a fixed prime of $\overline{\mathbf{Z}}$ lying over $p$.

Assuming that the automorphic form $\Pi_B$ on $G$ associated to $B$ is infinite dimensional, by the J-L correspondence we have that there is a cuspidal automorphic form $\Pi'_B$ on ${\rm GL}_2$ associated to the same eigensystem $B$. The type of $\Pi'_B$ at any finite place other than $p$ is the same as that of $\Pi_B$, while at infinity $\Pi'_B$ is the discrete series of lowest weight $2$. This basically says that there is a cusp form in $S_2(Np^n)$ whose associated system of eigenvalues is $B=(b_\ell)$.

We are only left with deciding when $\Pi_B$ is infinite dimensional, or can be chosen as such. This happens only for systems of eigenvalues of the form $B=(\chi(\ell)(1+\ell))_{\ell\nmid pN}$, where $\chi:\mathbf{Z}/p\rightarrow\mathbf{C}^*$ is any character (in order to show this one has to consider the particular shape of $K_N$, which I did not even define..). The reduction mod $P$ of such eigensystems are all of the form $(\ell^k+\ell^{k+1})_{\ell\nmid pN}$.

Concluding: Let $A=(a_\ell)$ be a sytstem of char. zero eigenvalues arising from $M_k(\Gamma_1(N))$, with $p\nmid N$. Assume that the mod $P$ reduction of $A$ is not of the form $(\ell^k+\ell^{k+1})_{\ell\nmid pN}$. Then, there exists a cusp form in $S_2(\Gamma_1(Np^n))$ such that its associated system of eigenvalues $B$ is congruent to $A$ mod $P$.

Answer by Tommaso Centeleghe for Which 'well-known' algebraic geometric results do not hold in characteristic 2?

2011-07-08T09:01:40Z

If $E$ is an elliptic curve over a finite field $k$ of characteristic $p>3$, then the group homomorphism ${\rm Aut}_k(E)\rightarrow k^*$ sending an automorphism of $E$ to its effect on the tangent space at $O_E$ is injective.

This can be seen (quite easily) using that the degree map $d:{\rm End}_k(E)\rightarrow \mathbf{Z}$ is a positive definite quadratic form (for a precise definition see Silverman p. 88), thus satisfies $d(a+b)\leq d(a)+d(b)+2\sqrt{d(a)d(b)}$, for all $a,b\in {\rm End}_k(E)$.

If now $u\in {\rm Aut}_k(E)$ acts as the identity on the tangent space at $O_E$, then the endomorphism $1-u$ is either zero or it is an inseparable isogeny. In both cases $p$ divides $d(1-u)$. Setting $a=1$, $b=u$ in the previous inequality we see that $d(1-u)\leq 4$ and hence $d(1-u)=0$, and $u=1$ (since $p>3$).

In char $2$ and $3$ there are elliptic curves with automorphism groups that are not cyclic (even not abelian), thus they cannot be embedded in $k^*$. This different behavior of the primes $2$ and $3$ is perhaps due to the fact that they are "archimedeanly" small.

Answer by Tommaso Centeleghe for Cool problems to impress students with group theory

2011-01-29T02:55:18Z

This problem might be related to some of the ones suggested above, I did not check carefully.

Assume that you have to prepare the calendar of a basketball league to which $2n$ teams take part, where $n$ is an integer $\geq 1$. The full season consists of $2n-1$ dates where in each date each of the team has to play against another one (different than itself :-)). This has to be done in such a way that, overall, every team meets every other team exactly once (we all know what the calendar of a basketball league look like).

A solution is given in the comment below. The solution I had first suggested was wrong.

Answer by Tommaso Centeleghe for Hecke algebra generated by a single element

2011-06-12T01:01:58Z

In certain cases a reason that causes the subring generated by a single Hecke operator $T_\ell$ in $\mathbb{T}_\mathbb{Z}$ to have index greater than one is the existence of exceptional primes for the normalised eigenforms $f\in S_k(\Gamma,\mathbb{C})$ (we shall say that $p$ is exceptional for $f$ if (one of) the $p$-adic Galois representation $\rho_{f,\lambda}$ associated to $f$ has small residual image, i.e., the image of $\bar\rho_{f,\lambda}$ does not contain $SL_2(F)$, for any finite extension $F/F_p$).

The special cases that I have in mind are instances of the inner twist phenomena that Kevin Buzzard has already mentioned in one of his comments, and you can see this also in level one, if I am right. Let me elaborate on that.

Let $p$ be a prime $\equiv 3$ mod $4$, and set $k=(p+1)/2$. Let $h$ be the class number of $Q(\sqrt{-p})$. The space $S_k(\Gamma(1))$ gives rise to $n=(h-1)/2$ distinct systems of mod $p$ Hecke eigenvalues so that the associated mod $p$ Galois representations $\rho_1,\ldots,\rho_n$ (which are unramified outside $p$) have dihedral image. This implies that if $\ell$ is a prime that is not a quadratic residue mod $p$, then the trace of $\rho_i$ at a Frobenius element at $\ell$ is zero.

In particular, if $n>1$ (i.e. $h>3$) then there are at least two DISTINCT systems of mod $p$ Hecke eigenvalues $(a_q)$ and $(b_q)$ such that their $\ell$-th members are equal. This implies that the integral Hecke ring $\mathbb{T}$ has more ring homomorphisms valued in $\bar{F_p}$ than the subring generated by $T_\ell$ alone does. In this case one can show that $p$ divides the index of the latter in the former.

With the previous argument you see that for almost all primes $p\equiv 3$ mod $4$ in weight $(p+1)/2$ half of the primes $\ell$ are such that $T_\ell$ does not generate $\mathbb{T}$. But what about the other half of the primes $\ell$?

By relating the traces of $\rho_i$ at Frobenius elements over primes $\ell$ that are split in $Q(\sqrt{-p})$ to the characters of the class group of $Q(\sqrt{-p})$ valued in $\bar{F_p}$ we get the following:

PROPOSITION: Let $A$ be the class group of $Q(\sqrt{-p})$. Assume that for every $a\in A$ there exists a pair of non trivial characters $\chi_i:A\rightarrow\bar{F_p}^*$, with $\chi_1\neq\chi_2$ and $\chi_1\neq \chi_2^{-1}$, such that $\chi_1(a)+\chi_1(a^{-1}) =\chi_2(a)+\chi_2(a^{-1})$. Then the integral Hecke ring $\mathbb{T}$ in weight $(p+1)/2$ and level one cannot be generated by a single $T_\ell$, for $\ell$ prime.

Few remarks: 1) The above does not say anything about the possibility of having a $T_n$ (with $n$ not prime) generating $\mathbb{T}$. In fact I think we cannot rule this out a priori.

2) Probably a little more group theoretic work can be done to reformulate the condition on the mod $p$ characters of $A$, and turn it into something nicer. I think if $A$ is not cyclic and $3$ does not divide its order then the assumption is satisfied.

3) Notice that we do not need to worry about $T_p$ generating $\mathbb{T}$ when $n>1$. In fact, since $P$ splits in $H/Q({\sqrt{-p}})$, where $H$ is the Hilbert Class Field of $Q(\sqrt{-p})$, all the $\rho_i$'s are the same locally at $p$ and one can show that the eigenvalue $a_p$ is $1$ for all of them.

An example. The cuspidal, integral Hecke ring of level one and weight $k=246=(491+1)/2$ is so that $p=491$ does not divide its discriminant. However every Hecke operators $T_n$, for $2\leq n\leq 153$ generates a subring of $\mathbb{T}$ of index divisible by $p$.

May be it could be nice to see whether the class group of $Q(\sqrt{491})$ satisfies the assumption of the proposition. I suspect it does!

[EDIT: The assumption of the proposition is NOT satisfied for $p=491$, unless I am wrong to an infinite amount. Therefore my suspicion is not confirmed. The proposition above, as far as I can tell, remains valid, although I have no example of a prime $p\equiv 3$ mod $4$ such that the class group $A$ of $Q(\sqrt{-p})$ satisfies the assumption of the proposition.

There still remains to give an explanation of the fact that $p=491$ seems to divide the index of the subring generated by $T_\ell$ in the integral, cuspidal Hecke ring of level $1$ and weight $246$. The arguments above explain such divisibility for primes $\ell$ that are non-quadratic residue mod $p$. For what concerns the other primes $\ell$, I am tempted to say that there should be a mod $491$ Galois representation arising from weight $246$ and level $1$ that is tamely ramified, reducible at $p$, and non-dihedral (may be with small image?): this in fact would cause the existence of a system of mod $p$ eigenvalues $(a_\ell)$, with $\ell\neq p$, arising from $S_{246} (\Gamma(1))$ such that its quadratic twist $(a_\ell \ell^{(p-1)/2})$ also arises from the same space.

Summarising

PROPOSITION: Let $p\equiv 3$ mod $4$. Assume that

1) the class number of $Q(\sqrt{-p})$ is > 3;

2) there exists a mod $p$ representation $\rho$ of $G_Q$ arising from $S_{(p+1)/2}(\Gamma(1))$ that is tamely ramified and reducible at $p$ and it is non-dihedral;

then for every prime $\ell$, $p$ divides the index of the subring generated by $T_\ell$ in the integral, cuspidal Hecke ring of weight $(p+1)/2$ and level $1$.

When $p=2083$ I learnt from some tables that there exists an odd, $A_5$ extension of $Q$ unramified outside $p$ that gives rise to a representation of the type we want in 2). Since condition 1) is also satisfied we see that the Hecke ring in weight $1044$ and level $1$ cannot be generated by a single $T_\ell$, for $\ell$ prime.]

Elliptic curves over finite fields and 2x2 matrices

2011-06-03T08:28:29Z

Let $k$ be a finite field of order $p^a$ and characteristic $p$, and $\mathcal{C}$ a $k$ isogeny class of elliptic curves over $k$. Let $w$ be the (Galois conjugacy class of the) Weil $k$-number attached to $\mathcal{C}$, and let $f(x)\in\mathbf{Z}[x]$ be its minimal, monic polynomial.

Assume that $\mathcal{C}$ is such that $f(x)$ has degree $>1$ (this amounts to exclude the case $a=2a'$ even and $f(x)=x\pm p^{a'}$). Equivalently, assume that any object $E$ in $\mathcal{C}$ has a commutative ring of $k$-endomorphisms, which is an order in an imaginary quadratic field (notice that this is weaker than to require $\mathcal{C}$ be given by ordinary elliptic curves).

Question: what is the number $R(f)$ of isomorphisms classes of objects in $\mathcal{C}$?

Candidate answer: $R(f)$ is the same as the number $S(f)$ of $SL_2(\mathbf{Z})$-orbits of 2x2 matrices $m\in M_2(\mathbf{Z})$ with integral coefficients with characteristic polynomial equal to $f(x)$, and with bottow left entry, say, positive (the $SL_2(\mathbf{Z})$-action is via conjugation, the sign of the bottom left entry is indeed a conjugacy invariant thanks to the fact that the discriminant of $f(x)$ is $<0$).

It seems that in the literature people tend to considered the weighted number $R'(f)$ of isomorphism classes of elliptic curves. This, at least when $k$ is the prime field, is known to be equal to a certain Kronecker class number $H(D)$ (cf. Lenstra, $\textit{Factoring integers with elliptic curves}$, Annals of Math., 126 (1987), 649-673), depending on the discriminant $D$ of $f(x)$. The Kronecker number $H(D)$ parametrizes weighted isomorphisms classes of positive definite, binary quadratic forms of discriminant $D$ (not necessarily primitive!).

The link with conjugacy classes of $2x2$ matrices proposed above comes from the fact that such number $S(f)$ is equal to the number of isomorphism classes of positive definite, binary quadratic forms of discriminant $D$ (not necessarily primitive!).

[One way to see this is by noticing that multiplication to the left by $(0, 1; -1, 0)$ gives a map from $M_2(\mathbf{Z})$ to itself which turns the $SL_2(\mathbf{Z})$ conjugation action into the $SL_2(\mathbf{Z})$ similitude action, i.e., $m\rightarrow g m g^t$, where $g^t$ is the transpose of $g$]

Basically, I would like a procedure that given an elliptic curve E in $\mathcal{C}$ produces a 2x2 matrix with characteristic polynomial equal to $f(x)$, whose conjugacy class determines that of $E$ (one might need to make some choices at the beginning, but that's fine). The standard idea (may be?) would be to fix an "origin" $E_0$ of $\mathcal{C}$ and then to attach to each isogeny $\varphi:E_0\rightarrow E$ the ideal $I_E$ of $\textrm{End}_k(E_0)$ given by those morphisms factoring through $\varphi$. Then one could try to show that the free rank $2$, $\mathbf{Z}$-module $I_E$ equipped with the multiplication action by $\pi_{E_0}\in\textrm{End}_k(E_0)$ (the Frobenius of $E_0$ relative to $k$) determines completely the isomorphism class of $E$. (I learnt of this approach from Waterhouse, $\textit{Abelian varieties over finite fields}$. However he tends to consider isomorphism classes of objects lying in a given isogeny class and with a fixed endomorphism ring that is a maximal order). Any comments, hints, or the answer itself would be appreciated. Thanks.

Reference for a theorem of Tate on the endomorphism rings of AVs over finite fields

2011-05-24T08:54:57Z

Let $k$ be a finite field, and $A$, $B$ abelian varieties over $k$. Let $T_p(A)$, $T_p(B)$ be the (contravariant) Dieudonn\'e modules associated to the p-divisible groups attached to $A$ and $B$, respectively. The theorem of Tate in the question is that the natural map

${\rm Hom}_k(A,B)\otimes\mathbf{Z}_p\longrightarrow{\rm Hom}(T_p(B),T_p(A))$

is an isomorphism.

It seems to me that the standard reference for the proof of this theorem is a paper of Tate titled Endomorphisms of abelian varieties over finite fields. II that should have appeared 50 years ago in Inventiones but, I am right, never did.

Is there another reference in the literature for the proof of this theorem?

Thanks.

What are the zeroes of $E_{p+1}$ on the modular curve $X_1(N)_{\overline{\mathbf{F}}_p}$?

2011-04-25T18:57:46Z

If $p$ is prime $>3$, then the $(p+1)$-st Eisenstein series

$E_{p+1}=-\frac{B_k}{2k}+\Sigma_{n\geq 1}\sigma_{p}(n)q^n$

is the $q$-expansion of a modular form of level one and weight $p+1$ ($B_k$ is the $k$-th Bernoulli number, $\sigma_p(n)$ is the sum of the $p$-th powers $d^p$ of all the positive divisors $d$ of $n$).

It can be viewed as a (meromorphic) section of a certain line bundle on the modular curve $X_1(N)$, where $N\geq 1$ is an integer prime to $p$.

Consider the base change $X_1(N)_{\overline{\mathbf{F}}_p}$ to an algebraic closure of the finite field with $p$ elements ($X_1(N)$ can be constructed over $\mathbf{Z}[1/N]$). I would be interested in computing the divisor of $X_1(N)_{\overline{\mathbf{F}}_p}$

defined by the base change of $E_{p+1}$. (Notice that the analogous question for $E_{p-1}$ has a classical interpretation as the Hasse invariant).

Answer by Tommaso Centeleghe for bad reduction for elliptic curves

2011-04-07T03:57:58Z

I would like to stress something that, I think, has not been point out very explicitly in any of the nice answers above.

The question that you are asking, if I understand it correctly, is a local question. So let's start with, say, a $p$-adic field $K$, and with the equation $f(x,y)=0$ of (the affine piece of) an elliptic curve $E$ with coefficients in the ring of integers $R$ of $K$.

To define the reduction of $E$ modulo the maximal ideal $\mathfrak{m}$ of $R$, one cannot follow the naive approach of simply considering the reduction mod $\mathfrak{m}$ of the equation $f(x,y)=0$.

Instead one should look at what Tate called a minimal Weierstrass equation $f_{\rm min}(x,y)=0$ for $E$ (LNM 476 pag. 39), and define the reduction $\bar E$ of $E$ as the object obtained by considering the reduction mod $\mathfrak{m}$ of $f_{\rm min}(x,y)=0$.

The point now is that the minimal equation is NOT stable by taking field extensions: if $L/K$ is a finite extension then $f_{\rm min}(x,y)$ need not still be a minimal equation for the base change of $E$ to $L$ (but it is if $L/K$ is unramified). Therefore it might very well happens that while $f_{\rm K, min}(x,y)=0$ has a singular reduction mod $\mathfrak{m}$ the corresponding thing does not hold for $f_{\rm L, min}(x,y)=0$ (the meaning of the subscripts is obvious I hope).

I tried to come up with an explicit example illustrating this phenomena. I failed to find one in a reasonable amount of time. But I am sure they can be found (easily?) in the literature.

Answer by Tommaso Centeleghe for Do isogenies between AVs over finite fields separate finite subgroups?

2011-03-02T16:28:44Z

As the comments and the answer above explain, the answer to my original question is NO. For completeness, I include the (very easy) proof of the statement in the second edit of my question (whose notation will be adopted) which is a weaker statement than that I was asking in the first place.

The group of homomorphisms Hom$_k(A/H(I),A)$ has the property that the intersection of all the kernels of its elements is the trivial subgroup of $A/H(I)$. To see this, one can just observe that the ideal $I$ can be identified with a subset of Hom$_k(A/H(I),A)$, and this subset already has the property that the intersection of all the kernels of its elements is trivial.

It now immediately follows that $H(I)=H(J)$.

I would like to thank the authors of the comments and answer above for their interest.

Do isogenies between AVs over finite fields separate finite subgroups?

2011-03-02T11:07:10Z

Waterhouse in his thesis (Abelian varieties over finite fields, Ann. scient. \'Ec. Norm. Sup., t. 2, 1969, p 521-560) seems to use without comments the following fact:

Let $k$ be a finite field, and let $A$, $B$ be two abelian varieties over $k$ that are $k$-isogenous. Consider the set $I(A,B)$ of all the $k$-isogenies from $A$ to $B$. Then for any finite, non-trivial, subgroup $H$ of $A$, there is $\varphi\in I(A,B)$ that does not vanish identically on $H$.

This fact is used implicitly in lines 8-9 page 533, right after the definition of kernel ideal.

Does anyone have an argument to see it? Also, I do not know what role the assumption that the base field $k$ is finite should play. Thanks.

[EDIT: Actually what Waterhouse uses is that, under the assumption of the second paragraph above, there is a ${\it morphism}$ $\varphi:A\rightarrow B$ that does not vanish identically on $H$]

[EDIT 2: I report here Waterhouse's statement. Let $A$ be an abelian variety over a finite field $k$, and let $R$ be its $k$-endomorphism ring. Let $I$ be a left ideal of $R$ that contains an isogeny of $A$. Define $H(I)$ to be the finite subgroup of $A$ given by the intersection of all ker($\varphi$), as $\varphi$ ranges in $I$.

By definition, $I$ is a kernel ideal if $I=$ { $r\in R: r\cdot H(I)=0$ }.

Here comes the line I can't verify:

"Every $I$ is contained in a kernel ideal $J$ with $H(J)=H(I)$, namely $J=$ { $r\in R: r\cdot H(I)=0$ }."

The question is "how do we know that $J$, as just defined, is a kernel ideal?" I think this question is just a reformulation of the main question I asked above.]

Comment by Tommaso Centeleghe

2013-05-16T12:23:04Z

thanks. you link gives only the definition of cesaro mean, however.

Comment by Tommaso Centeleghe

2013-05-16T12:22:32Z

Thanks, this explains exactly what I was asking!

Comment by Tommaso Centeleghe

2013-05-14T22:00:26Z

@Neil: I see, thanks.

Comment by Tommaso Centeleghe

2013-05-14T21:43:10Z

Wouldn't uniform convergence on your disk imply that your function admits its derivative at the branch point? (By a "switch-sum-and-differentiation"-type argument.. Or am I wrong?)

Comment by Tommaso Centeleghe

2013-02-27T10:23:04Z

What if we give ourself the knowledge of the j-invariant of E? Do you think that using j_E one can say more about the structure of $T_p(E)$?

Comment by Tommaso Centeleghe

2013-02-26T08:48:51Z

I believe that if you are working with a $K$-isogeny class of elliptic curves whose members have rings of $K$-endomorphisms larger than $\mathbf{Z}$, then the isogeny theorem works.

Comment by Tommaso Centeleghe

2013-02-15T13:29:02Z

Nice example, thanks! I guess the question remains of whether the original question has a positive answer in the case where p is known to be tamely ramified. As we already decided, since inertia groups are cyclic in this case, the ramification of p in M/K is lcm of the ramification indeces in L/K. But what about f?

Comment by Tommaso Centeleghe

2013-02-13T16:26:48Z

@Peter: Oh..I see. I actually wrote also (resp. inertia group) because I wanted to take this into account. But I agree, that explanation on how to go from the original question to the group theoretic one could have been more precise. Thanks again for your example.

Comment by Tommaso Centeleghe

2013-02-13T16:12:12Z

@Peter: thanks for the example, this settles the answer to the original question (which by the way I do not find slightly vague).

Comment by Tommaso Centeleghe

2013-02-13T14:53:00Z

Wait: in your example primes $p$ of $Q$ lying below a prime $P$ of $M$ whose associated decomposition group $D_P$ is $Gal(M/L)$ do not split completely in $L$, I think.

Comment by Tommaso Centeleghe

2013-02-13T14:01:07Z

@Michael: thanks. One way to prove the assertion in your first comment is by remarking that the group-theoretic question I asked has a positive answer in the cyclic case (i.e., $D$=cyclic in the notation above), right? I'll think about your dihedral suggestion - cheers.

Comment by Tommaso Centeleghe

2013-02-13T13:27:59Z

I guess for cyclic groups the property above holds. Do you know, more generally, to what kind of groups it applies?

Comment by Tommaso Centeleghe

2013-02-13T13:02:52Z

I definitely allow the knowledge of the $I$-orbits, too. Since I assume that the decomposition of $p$ in $L/K$ is given, i.e., we are given the $e_i$'s and $f_i$'s. Now, are you saying that the ramification index of $p$ in $M/K$ is the least common multiple of the $e_i$'s?

Comment by Tommaso Centeleghe

2013-02-13T12:51:27Z

Thanks for your answer. I think I see the way you want to argue. One comment: knowing how $p$ decomposes in $L/K$ is equivalent to knowing the sizes of both the $D$ and the $I$-orbits in $H\backslash G$. I think those for $I$ would be different for $P$ and $P′$ in your counterexample. Right?

Comment by Tommaso Centeleghe

2013-02-13T12:45:20Z

Thanks! I was clearly asking too much.