User françois brunault - MathOverflow

Answer by François Brunault for link to a paper by Ramanujan

2013-05-15T11:15:40Z

It's freely available on Google Books

http://books.google.fr/books?id=oSioAM4wORMC&printsec=frontcover&hl=fr#v=onepage&q&f=false

(on page 23).

Answer by François Brunault for Is there an algebraic curve over Q which is not modular?

2013-05-11T23:51:51Z

It is conjectured that there are only finitely many curves over $\mathbf{Q}$ of given genus $g \geq 2$ which are covered by a modular curve, see Conjecture 1.1 in the following paper :

Baker, M. H. ; González-Jiménez, E. ; González, J. ; Poonen, B. . Finiteness results for modular curves of genus at least 2. Amer. J. Math. 127 (2005), no. 6, 1325--1387.

In fact, the authors prove a strong result towards this conjecture, namely that there are only finitely such curves which are new (in an appropriate sense).

Answer by François Brunault for Verifying the correctness of a Sudoku solution

2013-05-04T02:09:17Z

The consequence relation $\models$ defined in Emil Jeřábek's answer is a matroid. In fact, it is a linear matroid.

Let $X=\{r_1,\ldots,r_9,c_1,\ldots,c_9,b_1,\ldots,b_9\}$ be the set of possible checks. Recall that given $S \subset X$ and $x \in X$, the notation $S \models x$ means that every Sudoku grid which is valid on $S$ is also valid on $x$.

We may embed $X$ into the free abelian group $V$ generated by the 81 cells of the Sudoku grid, by mapping a check $x \in X$ to the formal sum of the cells contained in $x$. The span of $X$ has rank $21$, and the kernel of the natural map $\mathbf{Z}X \to V$ is generated by the six relations of the form $r_1+r_2+r_3-b_1-b_2-b_3$.

Proposition. We have $S \models x$ if and only if $x \in \operatorname{Vect}(S)$.

Proof. By Proposition 2 from Emil's answer, the consequence relations $\models$ and $\vdash$ coincide, so we may work with $\vdash$. Let us prove that $S \vdash x$ implies $x \in \operatorname{Vect}(S)$. By transitivity, we may assume $S=D \backslash \{x\}$ for some $D \in \mathcal{D}$. It is straightforward to check that $x \in \operatorname{Vect}(D \backslash \{x\})$ in each case (i)-(iv).

Conversely, let us assume $x=\sum_{s \in S} \lambda_s s$ for some $\lambda_s \in \mathbf{Z}$. Since the elements of $X$ have degree 9, we have $\sum_{s \in S} \lambda_s = 1$. Any Sudoku grid provides a linear map $\phi : V \to E$, where $E$ is the free abelian group with basis $\{1,\ldots,9\}$ (map each cell to the digit it contains). If the grid is valid on $S$ then $\phi(s)=[1]+\cdots+[9]$ for every $s \in S$, and thus $\phi(x)=[1]+\cdots+[9]$, which means that the grid is valid on $x$. QED

Note that a set of checks $S$ is complete if and only if $\operatorname{Vect}(S)=\operatorname{Vect}(X)$. In particular, the minimal complete sets are those which form a basis of $\operatorname{Vect}(X)$, and it is now clear that every such set has cardinality $21$.

We also obtain a description of the independent sets : these are exactly the sets which are linearly independent when considered in $V$. Any independent set may be extended to a minimal complete set (we may have worked with $\mathbf{Q}$-coefficients instead of $\mathbf{Z}$-coefficients above).

Answer by François Brunault for elliptic curve with a degree 2 isogeny to itself?

2013-05-03T10:34:33Z

Yes, but the elliptic curve needs to have complex multiplication, since the multiplication-by-$n$ map has degree $n^2$. For an explicit example, you can take $E=\mathbf{C}/(\mathbf{Z}+i\mathbf{Z})$ with the isogeny being multiplication by $1+i$.

Answer by François Brunault for Modular Forms on $\Gamma_0(4)$ with Nebentypus

2013-04-26T10:50:46Z

EDIT : In what follows the weight $k$ is assumed to be an integer.

The matrix $\begin{pmatrix} 1 & 1/2 \\ 0 & 1 \end{pmatrix}$ normalizes the group $\Gamma_0(4)$, so in your case $g$ is still a modular form on $\Gamma_0(4)$ with trivial Nebentypus.

More generally if you start with $f$ on $\Gamma_0(N)$ and twist it by $1/m$ then you get a form on $\Gamma_0(N') \cap \Gamma_1(m)$ with $N'=\operatorname{lcm}(N,m^2)$.

In general, when you consider $g(z)=f(z+1/m)$, you are twisting $f$ by the additive character $\alpha(n)=\exp(2\pi i n/m)$. You can always write $\alpha$ as a linear combination of (not necessarily primitive) Dirichlet characters $\chi$ of level dividing $m$. Thus you can write $g$ as a linear combination of twists $f \otimes \chi$ for such $\chi$ (up to bad Euler factors). If you want to do this completely explicitly then the formulas are quite complicated in general, see Merel, Symboles de Manin et valeurs de fonctions L, Section 2.5. At some point, it may be useful to switch to the adelic language and work with the automorphic representation associated to the newform $f$.

Embeddings of finite groups into GL(n,Q_p)

2013-04-22T07:10:40Z

This question is inspired by some interesting comments on this recent question.

Fix an integer $n \geq 1$ and a finite subgroup $G$ of $\mathrm{GL}_n(\mathbf{C})$. It is known that there are infinitely many primes $p$ such that $G$ embeds into $\mathrm{GL}_n(\mathbf{Q}_p)$. This follows from the Cassels Embedding Theorem : every finitely generated field of characteristic 0 embeds into $\mathbf{Q}_p$ for infinitely many primes $p$ (see Cassels's book Local Fields or Cassels's article An embedding theorem for fields. Bull. Austral. Math. Soc. 14 (1976), 193-198; see also Chapter 4 of Pete L. Clark's notes).

Representation theoretic arguments actually show that $G$ occurs in $\mathrm{GL}_n(\overline{\mathbf{Q}})$ and thus in $\mathrm{GL}_n(K)$ for some number field $K$. From this and Chebotarev's density theorem, we deduce that the set $S(G)$ of primes $p$ such that $G \hookrightarrow \mathrm{GL}_n(\mathbf{Q}_p)$ has positive density.

Does the set $S(G)$ have a natural density?

Is the set $S(G)$ a Galoisian set of prime numbers in the sense of this question?

Answer by François Brunault for A $\mathbb{Q}$-rational canonical model for $X(N)$?

2013-04-23T12:55:09Z

This is done by Shimura in Introduction to the Arithmetic Theory of Automorphic Functions, Chapter 6.

By definition, the field $\mathbf{Q}(X(N))$ of rational functions with respect to this model is the field of modular functions for the congruence subgroup $\Gamma(N)$ whose Fourier expansions belong to $\mathbf{Q}(\zeta_N)((q^{1/N}))$.

Answer by François Brunault for cube + cube + cube = cube

2013-04-16T15:11:55Z

The solution mentioned in the question consists in fact of $9$ pieces, as shown in the following picture I took yesterday. I don't know whether a solution with $8$ pieces exists.

Answer by François Brunault for Lower bounds for Petersson inner products of cuspforms with integral Fourier coefficients

2013-04-10T17:57:23Z

It's not hard to see that the answer to (a) is yes. There is a basis of $S_2^{\textrm{new}}(\Gamma_0(N))$ consisting of newforms. These newforms come into Galois orbits $\{f^\sigma\}_{\sigma}$ . Here $\sigma$ runs through the embeddings of $K_f$ into $\mathbf{C}$, where $K_f$ is the field generated by the Fourier coefficients of $f$. A basic fact is that the $\mathbf{C}$-span of a given Galois orbit is generated by cusp forms with integral coefficients. This follows from considering the forms $\sum_{\sigma} \sigma(a) f^\sigma$ where $a$ runs through a $\mathbf{Z}$-basis of the ring of integers of $K_f$.

It follows that if the newform $f$ has integral coefficients, then its orthogonal complement $f^\perp$ is generated by cusp forms with integral coefficients. Now consider the linear map $\lambda_f : S_2(\Gamma_0(N),\mathbf{Z}) \to \mathbf{R}$ given by $\lambda_f(g)=(f,g)$. By the previous remark, the kernel of $\lambda_f$ has rank one less than the rank of $S_2(\Gamma_0(N),\mathbf{Z})$, which implies that the image of $\lambda_f$ is of the form $c_f \mathbf{Z}$ for some $c_f >0$. Thus we can take for $c$ the minimum of all the $c_f$. In fact $c_f = (f,f)/m_f$ for some integer $m_f$ measuring the congruences of $f$ with other cusp forms.

Answer by François Brunault for Status of Beilinson conjectures?

2013-04-07T08:57:19Z

In addition to Andreas's excellent answer, we should also mention the Tamagawa number conjecture of Bloch and Kato, which predicts the undetermined rational factor arising in Beilison's conjectural description of the $L$-value. The Bloch-Kato conjecture was later reformulated and generalized by Fontaine and Perrin-Riou to the case of motives with coefficients in an arbitrary number field. Here are some references :

Bloch, Kato, L-functions and Tamagawa numbers of motives.

Fontaine, Perrin-Riou, Autour des conjectures de Bloch et Kato: cohomologie galoisienne et valeurs de fonctions L.

Colmez, Fonctions L p-adiques.

Kings, The Bloch-Kato conjecture on special values of L-functions. A survey of known results.

Flach, The equivariant Tamagawa number conjecture : A survey.

Gealy, On the Tamagawa Number Conjecture for Motives Attached to Modular Forms.

Bellaïche, An introduction to the conjecture of Bloch and Kato.

Answer by François Brunault for Algebraic independence of $E_2$, $E_4$ and $E_6$

2013-03-28T09:32:38Z

Here is a reference for the algebraic independance of $E_2,E_4,E_6$ over $\mathbf{C}$ :

MR2186573 (2007a:11065) Martin, François ; Royer, Emmanuel . Formes modulaires et périodes. (French) [Modular forms and periods] Formes modulaires et transcendance, 1--117, Sémin. Congr., 12, Soc. Math. France, Paris, 2005.

See Lemme 117 p. 80. The proof is along the lines suggested by Emerton.

Answer by François Brunault for computing the order of the image of 0 under the modular parametrization map for an elliptic curve

2013-03-27T08:33:04Z

Here is an expanded version of g6hq's answer. Your question is indeed sensitive to the Manin constant of the modular parametrization $X_0(N) \to E$. This in turns depends on whether the elliptic curve $E$ is the so-called "strong Weil curve" in its isogeny class, which by definition means that the kernel of $\phi_* : J_0(N) \to E$ is connected.

For example when $E=11a1=X_0(11)$, then $\phi$ is an isomorphism so everything is fine. But when $E=11a3=X_1(11)$, the modular parametrization $\phi : X_0(11) \to X_1(11)$ of least degree has degree 5 and in fact satisfies $\phi^* \omega_E = c_E \omega_f$ where $f=f_E \in S_2(\Gamma_0(11))$ is the newform associated to $E$, with a non-trivial Manin constant $c_E=5$. Thus the torsion point $I(0)$ has, in fact, order $c/c_E=5$.

In general, given any elliptic curve $E/\mathbf{Q}$, there is an optimal parametrization $\phi : X_1(N) \to E$. Such a parametrization conjecturally satisfies $\phi^* \omega_E = \omega_f$ (Stevens' conjecture). But the class of the divisor $[0]-[\infty]$ in the Jacobian of $X_1(N)$ is defined only over $\mathbf{Q}(\mu_N)^+$, not over $\mathbf{Q}$.

Answer by François Brunault for critical values of motives

2013-03-21T15:50:35Z

The factor $L_\infty(M,s)$ is holomorphic and non-vanishing for $\operatorname{Re}(s)$ large enough, so it is definitely necessary to also ask that $L_\infty(\hat{M},1-s)$ has no pole at the given integer. As an example, for the Riemann zeta function $\zeta(s)$, only the even integers $n \geq 2$ are critical.

I haven't done the computation of critical integers for $L$-functions of $K3$ surfaces, but there should be no difficulty in this computation as the answer depends only on the behaviour of the motive at infinity, which in this case just means (the cohomology of) the complex variety.

Answer by François Brunault for Does split L-function imply split jacobian

2013-03-13T11:47:46Z

Faltings proved that two abelian varieties $A$ and $B$ defined over a number field are isogenous if and only if they have the same $L$-function. See Korollar 2 p. 361 in

Faltings, G. Endlichkeitssätze für abelsche Varietäten über Zahlkörpern. Invent. Math. 73 (1983), no. 3, 349--366.

So if you assume $L(A,s)=L(B,s) L(C,s)$ for some abelian varieties $A,B,C$ defined over a number field $K$, you can deduce that $A$ is $K$-isogenous to $B \times C$.

Answer by François Brunault for An expression for the function $f_e$ that appears in the Weil Pairing

2013-02-18T08:05:04Z

To expand on wccanard's comment : see my answer to this question for a general method to compute a rational function with given (principal) divisor.

This will give you $f_e$ as a rational function of $x$ and $y$. You then just need to expand this in terms of the standard formal coordinate $z=-x/y$.

Answer by François Brunault for The value $\pm 1$ for the square root of Wilson's theorem, ((p-1)/2)! mod p

2013-02-13T11:20:52Z

Regarding your question (1), here are two articles I know dealing with the problem of computing factorials :

Crandall, Dilcher, Pomerance. A search for Wieferich and Wilson primes. Math. Comp. 66 (1997), no. 217, 433--449. (MR1372002)

Costa, Gerbicz, Harvey. A search for Wilson primes.

(Recall that Wilson primes are those primes $p$ such that $(p-1)! \equiv -1 \mod{p^2}$.)

It turns out that there exists an algorithm to compute the product of $N$ terms in arithmetic progression in no more than $O(N^{\alpha+\epsilon})$ multiplications, with $\alpha = \frac{\sqrt{5}-1}{2} = 0.618...$ (see p. 441 in the first article). They also mention an algorithm to compute $(p-1)! \mod{p^2}$ in time $O(p^{1/2+\epsilon})$.

The second article seems to use a somewhat different method and achieves an average polynomial time algorithm to compute this quantity, making use of the fact that there are redundant products when considering many values of $p$.

Both articles always work mod $p^2$. It is certainly faster to work only mod $p$, but I'm not sure whether this has an impact on the theoretical running time of the algorithms.

Answer by François Brunault for calculate function from its divizor

2013-02-11T07:53:38Z

Every principal divisor is a finite ${\bf Z}$-linear combination of divisors of the form $D=[P]+[Q]-[P+Q]-[O]$ with $P,Q \in C$. For such $D$, let $L:\lambda=0$ be the line through $P$ and $Q$.

If $P+Q=O$ then $L$ is vertical and $D={\rm div}(\lambda)={\rm div}(x-x_P)$.

If $P+Q \neq 0$ then ${\rm div}(\lambda)=[P]+[Q]+[-P-Q]-3[O]$ so that $D={\rm div}\bigl(\frac{\lambda}{x-x_{P+Q}}\bigr)$.

This method is ok if the coefficients of your divisor are reasonably small. Otherwise, you might consider using some kind of fast exponentiation. The idea is that a divisor $2N[P]$ is equivalent to $N[2P]+N[O]$ modulo the principal divisors. So you just need to compute a function $f$ such that ${\rm div} f=2[P]-[2P]-[O]$ and then compute $f^N$ by fast exponentiation. At each step the coefficient in your divisor has been divided by $2$, thus giving a logarithmic (instead of linear) number of steps.

I'm currently writing a PARI/GP script which takes as input a divisor on elliptic curve, checks if it's principal and if so outputs a rational function with this divisor. If you or other people are interested I could put the link here.

Finally a precision on the decomposition $f=f_1(x)+yf_2(x)$ I mentioned in my comment. Since the function $x$ has degree $2$ on $C$, the degree of $f_1$ as a usual rational function is in fact $\leq \deg(f)$. A more careful analysis also shows $\deg(f_2) \leq \deg(f)$ since there is compensation when you divide by $y$.

Answer by François Brunault for Are there noncongruence subgroups (of finite index) of the modular group generated only by 2 or 3 elements?

2013-01-10T08:30:24Z

A. J. Scholl has written a series of articles on modular forms on noncongruence subgroups. In the article

On the Hecke algebra of a noncongruence subgroup, Bull. London Math. Soc. 29 (1997), no. 4, 395–399

he gives two examples of index 7 subgroups of $\mathrm{SL}_2(\mathbf{Z})$ generated by 3 elements :

$$\Gamma_{4,3} = \Bigl\langle \begin{pmatrix} 1 & 4 \\ 0 & 1 \end{pmatrix},\begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix},\begin{pmatrix} 1 & -1 \\ 2 & -1 \end{pmatrix} \Bigr\rangle$$

$$\Gamma_{5,2} = \Bigl\langle \begin{pmatrix} 1 & 5 \\ 0 & 1 \end{pmatrix},\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix},\begin{pmatrix} 2 & 3 \\ 1 & 2 \end{pmatrix} \Bigr\rangle$$

The subscripts refer to the widths of the cusps (both groups have exactly two cusps).

Constructive proof of "Projective implies proper"

2012-12-19T16:41:45Z

For every ring $A$, the structural morphism of schemes $\pi_A : {\bf P}^n_{A} \to {\rm Spec}{A}$ is a closed map. The usual proof of this fact is not constructive : given equations of a closed subset $Z$ of ${\bf P}^n_{A}$, it doesn't produce equations for $\pi_A(Z)$.

In the case $A$ is a polynomial ring over an algebraically closed field $k$, this result is none other than the fundamental theorem of elimination theory : the image of a Zariski-closed subset of ${\bf P}^n(k) \times k^m$ under the second projection is a Zariski-closed subset of $k^m$. The first proofs of this theorem (Cayley, Kronecker, Sylvester) used resultants and thus were constructive.

In fact, the proof using elimination theory is universal in the following sense. Given integers $n,r \geq 1$, $d_1,\ldots,d_r \geq 1$, consider the universal homogenous polynomials $P_1,\ldots,P_r$ of degree $d_1,\ldots,d_r$ in the indeterminates $T_0,\ldots,T_n$, having coefficients in the polynomial ring $\widetilde{A} = \mathbf{Z}[Y_{i,\alpha} : 1 \leq i \leq r]$, where the indeterminates $Y_{i,\alpha}$ are the coefficients of $P_i$. Then there exists an explicit "resultant system" $R_1,\ldots,R_s \in \widetilde{A}$ such that $\pi_{\widetilde{A}}(V_+(P_1,\ldots,P_r))=V(R_1,\ldots,R_s)$. This means that specializations of $P_1,\ldots,P_r$ in some algebraically closed field $k$ have a common root in ${\bf P}^n(k)$ if and only if the corresponding specializations of $R_1,\ldots,R_s$ all vanish. Of course $s$ has to depend on $n,r,d_i$, but everything is explicit (at least from a theoretical point of view).

Now let $A$ be any ring and let $I=(f_1,\ldots,f_r)$ be an homogenous ideal of finite type of $A[T_0,\ldots,T_n]$. Then the resultant system above specialized at $f_1,\ldots,f_r$ provides explicit equations for $\pi_A(V_+(I))$ (this can be seen by studying the geometric fibers of $\pi_A$). In particular if $A$ is noetherian, then the map $\pi_A$ is closed, and we have a constructive proof for that. But in general, a closed subset of ${\bf P}^n_A$ need not be defined by finitely many equations. This raises the following questions :

Is there a way to prove that the map $\pi_A$ is closed for every ring $A$, by some clever reduction to the noetherian case?
If $Z$ is a closed subset of ${\bf P}^n_A$, given to us by infinitely many explicit equations $(f_i)_{i \in I}$, is there a way to produce explicit equations for $\pi_A(Z)$? In other words, is there a constructive proof of the fact that $\pi_A$ is closed?
Regarding question 2, an obvious thing to do is to look at all finite subfamilies $(f_i)_{i \in J}$, where $J$ is a finite subset of $I$, and to consider the associated resultant systems. Are all these equations sufficient to define $\pi_A(Z)$?

EDIT. Will Sawin has proved that the answers to all these questions is yes. Following Daniel Litt's comment, we can also consider $\pi_A(Z)$ as a closed subscheme of $\operatorname{Spec} A$, namely the closed subscheme defined by the kernel of the morphism $A \to \mathcal{O}_Z(Z)$. Do the resultant systems generate this ideal of $A$?

Answer by François Brunault for How do you compute the primes of bad reduction?

2012-12-13T17:57:23Z

Assume for simplicity that $Y$ has pure relative dimension $d$. By considering the standard affine cover of $\mathbf{P}^n_{\mathbf{Z}}$, one easily reduces to the case where $Y=V(F_1,\ldots,F_k)$ is a closed subscheme of $\mathbf{A}^n_{\mathbf{Z}}$. Then the special fiber $Y_p = V(F_{1,p},\ldots,F_{k,p}) \subset \mathbf{A}^n_{\mathbf{F}_p}$ is smooth if and only if for every $x \in Y_p(\overline{\mathbf{F}}_p)$ , the Jacobian matrix $(\frac{\partial F_{i,p}}{\partial X_j}(x))$ has rank $n-d$. Note that $Y_p$ is $d$-dimensional by the flatness assumption, so all geometric tangent spaces have dimension $\geq d$, which implies that the rank of the Jacobian matrix is everywhere $\leq n-d$. Thus $Y_p$ is smooth if and only if the ideal generated by $F_{1,p},\ldots,F_{k,p}$ together with all $(n-d)$-minors of the Jacobian matrix is equal to $\mathbf{F}_p[X_1,\ldots,X_n]$.

Now consider the ideal $I$ of $R=\mathbf{Z}[X_1,\ldots,X_n]$ generated by $F_1,\ldots,F_k$ together with all $(n-d)$-minors of the Jacobian matrix. Since the generic fiber of $Y$ is smooth, we have $I \cap \mathbf{Z}=N\mathbf{Z}$ for some integer $N \geq 1$, and the prime factors of $N$ are precisely the primes of bad reduction of $Y$.

Proof. If $p$ doesn't divide $N$ then $(I,p)$ contains $N$ and $p$, so $(I,p)=R$ and $Y_p$ is smooth. If $p$ divides $N$, write $N=p^k M$ with $p$ not dividing $M$. If we had $(I,p)=R$, then we would also have $(I,p^k)=R$. Then $M \in (MI,p^k M)=(MI,N) \subset I$, a contradiction. Thus $Y_p$ is not smooth.

You can compute the integer $N$ using Gröbner bases for polynomials over $\mathbf{Z}$, see e.g. http://magma.maths.usyd.edu.au/magma/handbook/text/1112#12189 for the Magma implementation. I'm not expert in Gröbner bases, so I would appreciate if someone could confirm whether it is always possible to find $I \cap \mathbf{Z}$ when a set of generators of $I$ is given, and whether the Magma implementation works in all cases.

Answer by François Brunault for Dimension of polynomial algebras

2012-12-05T08:28:06Z

The following reference should be of interest :

Brewer, Montgomery, Rutter, Heinzer, Krull dimension of polynomial rings.

For example, they prove, see Corollary 2 p. 30, that any semi-hereditary ring (all finitely generated ideals are projective) satisfies the dimension formula above. This generalizes Seidenberg's result since a Prüfer ring is a semi-hereditary integral domain.

It might be interesting to study the class of rings $R$ satisfying the following condition : for every prime ideal $P$ of $R$ and every $n \geq 1$, we have $\textrm{height}(P[X_1,\ldots,X_n]) = \textrm{height}(P)$. This condition implies $\dim R[X_1,\ldots,X_n] = \dim R +n$ for every $n$ (this can be deduced from Thm 1 of this paper), but I don't know about the converse.

The authors also discuss the class of strong $S$-rings introduced by Kaplansky (see the paper for the definition). This class contains the Noetherian rings and the Prüfer rings, and is stable by localizations and quotients. Kaplansky proved that a strong $S$-ring $R$ satisfies $\mathrm{height}(P[X]) = \textrm{height}(P)$ for every prime ideal $P$ of $R$, and thus $\textrm{dim}(R[X])=\textrm{dim}(R)+1$. But a strong $S$-ring doesn't necessarily satisfy the dimension formula for every $n$. In the other direction, the authors give an example of a ring which satisfies the height formula $\textrm{height}(P[X_1,\ldots,X_n]) = \textrm{height}(P)$ for every prime ideal $P$, but which is not a strong $S$-ring.

Answer by François Brunault for Is this height-transcendence-degree inequality true without AC ?

2012-11-30T13:38:58Z

Here is a proof of (**) by induction on the height of $P$.

If $P=0$, the inequality (**) is obvious. Let $P$ be a prime ideal of $R$ of height $d \geq 1$, and consider a chain of prime ideals $0=P_0 \subset P_1 \subset \ldots \subset P_d = P$ of length $d$ in $R$. The domain $R/P_1$ has finite Krull dimension, and the prime ideal $P/P_1$ has height $d-1$ so by the induction hypothesis \begin{equation*} {\rm height}(P/P_1) + {\rm trdeg}_k (R/P) \leq {\rm trdeg}_k(R/P_1). \end{equation*} It remains to prove that ${\rm trdeg}_k(R/P_1) \leq{\rm trdeg}_k(R)-1$. Let $x_1,\ldots,x_e$ be elements of $R$ whose images in $R/P_1$ are algebraically independent over $k$. Let $y$ be any element of $P_1 \backslash \{0\}$. Consider the map $\phi : k[X_1,\ldots,X_e,Y] \to R$ sending $X_i$ to $x_i$ and $Y$ to $y$. If $Q = \sum_j Q_j(X_1,\ldots,X_e) Y^j$ lies in the kernel of $\phi$, then reducing modulo $P_1$ gives $Q_0(x_1,\ldots,x_e)=0$, so that $Q_0=0$. Since we are working in a domain and $y \neq 0$, an easy induction gives $Q=0$. Thus $x_1,\ldots,x_e,y$ are algebraically independent over $k$, which concludes the proof.

Answer by François Brunault for Is a domain all of whose localizations are noetherian itself noetherian ?

2012-11-28T10:02:48Z

The ring of integers $\mathcal{O}_{\mathbf{C}_p}$ of $\mathbf{C}_p$ is not noetherian, but its only nontrivial localization is $\mathbf{C}_p$, which is noetherian.

EDIT This doesn't answer the question : the ring $\mathcal{O}$ is local, so its localization at the maximal ideal is $\mathcal{O}$ itself, which isn't noetherian.

The nonzero ideals of $\mathcal{O}$ are of the form $I_{\geq \alpha} = \{x \in \mathcal{O} : v(x) \geq \alpha\}$ with $\alpha \in \mathbf{Q}_{>0}$ , and $I_{> \alpha} = \{x \in \mathcal{O} : v(x) > \alpha\}$ with $\alpha \in {\bf R}_{\geq 0}$ . Here $v$ is the $p$-adic valuation on $\mathbf{C}_p$ . The ring $\mathcal{O}$ is one-dimensional : its only prime ideals are $(0)$ and the maximal ideal $I_{>0}$.

Answer by François Brunault for Existence of an $R$-basis with at least one unit in it?

2012-11-28T08:00:51Z

First some general remarks. You're asking whether $S/R$ is free as a $R$-module. If $S$ is a $R$-algebra which is free of finite rank, then the map $R \to S$ splits as a map of $R$-modules. This fact was already noticed by Florian Eisele in his answer to the other MO question.

Now for an explicit counter-example to your question. Consider the ring $R={\bf Z}[x,y,z]/(x^2+y^2+z^2-1)$. It is an integral domain. It is known that there exists a $R$-module $M$ which is not free such that $R \oplus M \cong R^3$. For a nice construction, see e.g. Keith Conrad's notes. Explicitly we can take $M=\{(f,g,h) \in R^3 : xf+yg+zh=0\}$. Note that we can embed $M$ in $R^2$ by $(f,g,h) \mapsto (f,g)$, and the cokernel $R^2/M$ is a torsion module, so there exists $F \in R \backslash \{0\}$ such that $F \cdot R^2 \subset M$.

Now, we would like to construct a $R$-algebra structure on $R \oplus M$. We can do this by considering the $R$-algbera $S_0 = R \otimes_{\mathbf{Z}} \mathcal{O}$ where $\mathcal{O}$ is an order of a cubic field $K$. It is an integral domain, since the polynomial $x^2+y^2+z^2-1$ is irreducible over any field of characteristic not $2$. Let $(1,\alpha,\beta)$ be a $\mathbf{Z}$-basis of $\mathcal{O}$. Embed $R \oplus M$ in $S_0$ by $(f,(g,h)) \mapsto f+g\alpha+h\beta$. This won't be a subring of $S_0$ in general, but $S=R \oplus FM$ is a subring of $S_0$ since $(FM) \cdot (FM) \subset F^2 S_0 \subset R \oplus FM$. So we have constructed an integral domain $S$ over $R$ such that $S/R \cong M$ is not free over $R$.

I don't know whether it's possible to find a counterexample where $R \to S$ splits as a map of rings, in other words where $S=R \oplus I$ where $I$ is an ideal of $S$.

Answer by François Brunault for What are some consequences of the Mumford-Tate conjecture?

2012-11-21T08:16:35Z

Here is a result in a somewhat different direction. Assuming MTC, Hindry and Ratazzi recently proved a bound for the torsion subgroup of abelian varieties of ${\rm GSp}$-type over arbitrary finite extensions.

Answer by François Brunault for Models of the modular curve $Y_1(N)$

2012-11-16T21:07:08Z

In the model you describe, the cusp $\infty$ of $X_1(N)$ is not defined over ${\bf Q}$ (but the cusp $0$ is). A way to see this is that the marked elliptic curve $({\bf C}/({\bf Z}+\tau{\bf Z}),1/N)$ is isomorphic to the marked Tate curve $E_q=({\bf C}^\times/q^{\bf Z},e^{2\pi i/N})$ with $q=e^{2\pi i\tau}$. When you let $\tau \to \infty$, you get $q \to 0$ so that $E_q \to ({\bf G}_m,e^{2i\pi/N})$, which is not defined over ${\bf Q}$. This fact is explained in Diamond-Im, Modular forms and modular curves, see 9.3.5 and 9.3.6.

There is an alternative model $Y_\mu(N)$ classifying elliptic curves $E$ together with a closed immersion $\mu_N \hookrightarrow E$ (see loc. cit. 8.2.2). In this model the cusp $\infty$ is defined over ${\bf Q}$, so it gives an affirmative answer to your second question.

You can switch from one model to another with the Atkin-Lehner involution $W_N$, which becomes an isomorphism defined over ${\bf Q}$ — it is only defined over ${\bf Q}(\mu_N)$ when considered as an involution of either $X_1(N)$ or $X_{\mu}(N)$. But I don't see a nice way to characterize those functions which are rational for the canonical model in terms of the $q$-expansion at $\infty$.

On a theorem of Galois

2012-10-26T06:19:03Z

I am currently teaching Galois theory and this week, I mentioned the following theorem of Galois :

Let $P(x) \in \mathbf{Q}[x]$ be an irreducible polynomial of prime degree. Then $P$ is solvable by radicals if and only if the splitting field of $P$ is generated by any two roots of $P$.

I was asked by a student whether this theorem can be generalized to polynomials whose degree is composite, maybe allowing the splitting field to be generated by more than two roots. I know that the proof of Galois's theorem relies on determining the solvable subgroups of $\mathfrak{S}_p$, but I don't know enough group theory to tell what can be proved in the case where the degree of the polynomial is composite, say $pq$ where $p$ and $q$ are (possibly equal) primes.

Does such a generalization of Galois's theorem exist? Or is there a conceptual reason why such a generalization cannot hold? In the latter case, do there already exist generalizations of Galois's theorem, possibly in different directions?

Answer by François Brunault for Can we ascertain that there exist an epimorphism $G\rightarrow H?$

2012-11-01T07:10:41Z

As explained in the comments, the result is true if $H$ is abelian.

Here is an argument which shows that the result is true in the somewhat orthogonal case where $H$ has trivial center [EDIT] and is indecomposable [/EDIT].

Write the epimorphism $G \times G \to H \times H$ as $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ with $a,b,c,d : G \to H$. Let $A,B,C,D$ be the respective images of $a,b,c,d$ in $H$.

The groups $A$ and $B$ commute elementwise in the sense that $xy=yx$ for every $x \in A$ and $y \in B$. Moreover, they generate $H$ by assumption. So we have an exact sequence

\begin{equation*} 1 \to A \cap B \to A \times B \to H \to 1 \end{equation*}

and similarly for $C,D$. Note that $A \cap B$ commutes with $A$ and $B$, so it must lie in the center of $H$, thus it should be trivial. Therefore $H=A \times B = C \times D$. It follows that $A=\{e\}$ or $B=\{e\}$, thus $a$ or $b$ is surjective.

The same argument also works in some cases where the center of $H$ is not trivial, for example when $H$ is a group of order $p^3$ with $p$ prime.

Over which fields does the Mordell-Weil theorem hold?

2012-10-01T13:01:11Z

According to a well-known theorem of Mordell, the group of rational points $E(\mathbf{Q})$ of an elliptic curve $E/\mathbf{Q}$ is finitely generated. Weil generalized this theorem to abelian varieties over number fields.

Less well-known is the following generalization, due I believe to Néron : if $K$ is a field of finite type (that is, finitely generated over its prime field) and $A$ is an abelian variety over $K$, then $A(K)$ is finitely generated. There is an even more general statement, the Lang-Néron theorem, for relative field extensions which are finitely generated (see Brian Conrad's article for the precise statement and a proof of this theorem).

Q1. Are there other fields $K$ for which the group of $K$-rational points of an abelian variety over $K$ is always finitely generated?

In the other direction, there exist fields $K$ for which $A(K)$ is clearly never finitely generated whenever $\operatorname{dim}(A) \geq 1$. For example $K=\mathbf{C}$, in which case it follows from the description af abelian varieties as complex tori. If $K$ is a finite extension of $\mathbf{Q}_p$, then $A(K)$ contains a finite-index subgroup isomorphic to $\mathcal{O}_K^{\operatorname{dim} A}$, so $A(K)$ is again never finitely generated. Other examples I can think of are complete discretely valued fields and algebraically closed fields. Note that we often have the stronger result that $A(K) \otimes \mathbf{Q}$ is infinite-dimensional (except when $K=\overline{\mathbf{F}}_p$, in which case $A(K)$ is a torsion group).

Q2. Are there other fields $K$ for which the group of $K$-rational points of a non-trivial abelian variety over $K$ is never finitely generated?

Answer by François Brunault for Conjugating the Lyness 5-cycle into a rotation of the plane

2012-10-27T17:51:09Z

Yes, this map is conjugate to an automorphism of $\mathbf{P}^2$. See

A. Beauville, J. Blanc, On Cremona transformations of prime order, C.R. Acad. Sci. Paris 339 (2004), no4, 257-259.

See also T. de Fernex, On planar Cremona maps of prime order, Nagoya Math. Journal, Vol. 174 (2004), 1–28, which contains a classification of planar Cremona maps of prime order up to conjugation. According to Remark 1.3.4 therein, the fact above was already known to Iskovskikh.

Comment by François Brunault

2013-05-17T07:43:08Z

@Deane : See also Terence Tao's blog post terrytao.wordpress.com/2007/09/14/…

Comment by François Brunault

2013-05-15T09:42:21Z

For those interested in putting an intersection of two quadrics into cubic and then Weierstrass form, you can have a look at the first 2 pages of the following note I wrote perso.ens-lyon.fr/francois.brunault/recherche/… The equations are not precisely the same, but you can easily adapt and find the right change change of variables.

Comment by François Brunault

2013-05-15T09:32:43Z

The conclusion is that 4/9 is not the ratio of any two Pythagorean fractions.

Comment by François Brunault

2013-05-15T09:30:45Z

Sorry, this is 120a2 and not 120a1.

Comment by François Brunault

2013-05-15T09:30:20Z

This elliptic curve is 120a1 in Cremona's tables, it has rank 0 and its Mordell-Weil group is isomorphic to $\mathbf{Z}/4\mathbf{Z} \times \mathbf{Z}/2\mathbf{Z}$. So it seems that the only rational points on this intersection of 2 quadrics are those with $x=0$ or $y=0$.

Comment by François Brunault

2013-05-14T11:52:28Z

@tkluck: You should be careful when speaking of the Zariski topology on $\mathbf{R}^n$. Usually we only consider Zariski topology on (the underlying set of) schemes like $\mathbf{A}^n_{\mathbf{R}$, which is bigger than $\mathbf{R}^n$ set-theoretically. Of course, you can consider the topology induced on the set of real points, but some care is needed. In the examples above, the set $X(\mathbf{R})$ is dense in $X$ so is still irreducible by general topology. So the isolated singularities provide counterexamples to your question as stated.

Comment by François Brunault

2013-05-14T07:39:46Z

For the first question, I would simply read the introduction of Helfgott's paper.

Comment by François Brunault

2013-05-12T18:18:14Z

I don't have access to Lascar's article neither, so I cannot check this.

Comment by François Brunault

2013-05-12T18:11:55Z

A MathSciNet search gives the article Automorphism groups of fields, and their representations by Rovinskii, according to which Lascar actually proves the result for any extension of algebraically closed fields of uncountable transcendence degree.

Comment by François Brunault

2013-05-10T16:10:57Z

Use the functional equation and the existence of Euler product for s>1.

Comment by François Brunault

2013-05-10T08:39:17Z

I meant $2^{(3^d-1)/2}$ in the last comment

Comment by François Brunault

2013-05-10T08:37:38Z

If I'm not mistaken, the number of disjoint unions $P_0 \cup \cdots \cup P_{d-1}$ satisfying your condition is $(3^d-1) (3^{d-1}-1) \cdots (3-1)$, and the number of possible $X$'s is $2{\^}((3^d-1)/2)$ which is larger for $d \geq 3$.

Comment by François Brunault

2013-05-10T08:22:52Z

Isn't it possible to just compare the number of such disjoint unions vs. the number of such sets?

Comment by François Brunault

2013-05-08T20:05:10Z

Another useful reference are K. Conrad's notes : math.uconn.edu/~kconrad/blurbs/grouptheory/…

Comment by François Brunault

2013-05-06T12:46:47Z

But these elliptic curves seem to have irreducible mod 13 representations.