# E. Kowalski's blog

## Update on a bijective challenge

A long time ago, I presented in a post a fun property of the family of curves given by the equations
$x+1/x+y+1/y=t,$
where $t$ is the parameter: if we consider the number (say $N_p(t)$) of solutions over a finite field with $p\geq 3$ elements, then we have
$N_p(t)=N_p(16/t),$
provided $t$ is not in the set $\{0, 4, -4\}$. This was a fact that Fouvry, Michel and myself found out when working on our paper on algebraic twists of modular forms.

At the time, I knew one proof, based on computations with Magma: the curves above “are” elliptic curves, and Magma found an isogeny between the curves with parameters $t$ and $16/t$, which implies that they have the same number of points by elementary properties of elliptic curves over finite fields.

By now, however, I am aware of three other proofs:

• The most elementary, which motivates this post, was just recently put on arXiv by T. Budzinski and G. Lucchini Arteche; it is based on the methods of Chevalley’s proof of Warning’s theorem: it computes $N_p(t)$ modulo $p$, proving the desired identity modulo $p$, and then concludes using upper bounds for the number of solutions, and for its parity, to show that this is sufficient to have the equality as integers.  Interestingly, this proof was found with the help of high-school students participating in the Tournoi Français des Jeunes Mathématiciennes et Mathématiciens. This is a French mathematical contest for high-school students, created by D. Zmiaikou in 2009, which is devised to look much more like a “real” research experience than the Olympiads: groups of students work together with mentors on quite “open-ended” questions, where sometimes the answer is not clear (see for instance the 2016 problems here).
• A proof using modular functions was found by D. Zywina, who sent it to me shortly after the first post (I have to look for it in my archives…)
• Maybe the most elegant argument comes by applying a more general result of Deligne and Flicker on local systems of rank $2$ on the projective line minus four points with unipotent tame ramification at the four missing points (the first cohomology of the curves provide such a local system, the missing points being $0, 4, -4, \infty$). Deligne and Flicker prove (Section 7 of their paper, esp. Prop. 7.6), using a very cute game with products of matrices and invariant theory, that if $\mathcal{V}$ is such a local system and $\sigma$ is any automorphism of the projective line that permutes the four points, then $\sigma^*\mathcal{V}$ is isomorphic to $\mathcal{V}$.

Not too bad a track record for such a simple-looking question… Whether there is a bijective proof remains open, however!

September 24th, 2016 at 10:47 am

## Kummer extensions, Hilbert’s Theorem 90 and judicious expansion

This semester, I am teaching “Algebra II” for the first time. After “Algebra I” which covers standard “Groups, rings and fields”, this follow-up is largely Galois theory. In particular, I have to classify cyclic extensions.

In the simplest case where $L/K$ is a cyclic extension of degree $n\geq 1$ and $K$ contains all $n$-th roots of unity (and $n$ is coprime to the characteristic of $K$), this essentially means proving that if $L/K$ has cyclic Galois group of order $n$, then there is some $b\in L$ with $L=K(b)$ and $b^n=a$ belongs to $K^{\times}$.

Indeed, the converse is relatively simple (in the technical sense that I can do it on paper or on the blackboard without having to think about it in advance, by just following the general principles that I remember).

I had however the memory that the second step is trickier, and didn’t remember exactly how it was done. The texts I use (the notes of M. Reid, Lang’s “Algebra” and Chambert-Loir’s delightful “Algèbre corporelle”, or rather its English translation) all give “the formula” for the element $b$ but they do not really motivate it. This is certainly rather quick, but since I can’t remember it, and yet I would like to motivate as much as possible all steps in this construction, I looked at the question a bit more carefully.

As it turns out, a judicious expansion and lengthening of the argument makes it (to me) more memorable and understandable.

The first step (which is standard and motivated by the converse) is to recognize that it is enough to find some element $x$ in $L^{\times}$ such that $\sigma(x)=\xi x$, where $\sigma$ is a generator of the Galois group $G=\mathrm{Gal}(L/K)$ and $\xi$ is a primitive $n$-th root of unity in $L$. This is a statement about the $K$-linear action of $G$ on $L$, or in other words about the representation of $G$ on the $K$-vector space $L$. So, as usual, the first question is to see what we know about this representation.

And we know quite a bit! Indeed, the normal basis theorem states that $L$ is isomorphic to the left-regular representation of $G$ on the vector space $V$ of $K$-valued functions $\varphi\,:\, G\longrightarrow K$, which is given by
$(\sigma\cdot \varphi)(\tau)=\varphi(\sigma^{-1}\tau)$.
(It is more usual to use the group algebra $K[G]$, but both are isomorphic).

The desired equation implies (because $G$ is generated by $\sigma$) that $Kx$ is a sub-representation of $L$. In $V$, we have an explicit decomposition in direct sum
$V=\bigoplus_{\chi} K\chi,$
where $\chi$ runs over all characters $\chi\,:\, G\longrightarrow K$ (these really run over all characters of $G$ over an algebraic closure of $K$, because $K$ contains all $n$-th roots of unity and $G$ has exponent $n$). So $x$ (if it is to exist) must correspond to some character. The only thing to check now is whether we can find one with the right $\sigma$ eigenvalue.

So we just see what happens (or we remember that it works). For a character $\chi\in V$ such that $\chi(\sigma) = \omega$, and $x\in L^{\times}$ the element corresponding to $\chi$ under the $G$-isomorphism $L\simeq V$, we obtain $\sigma(x)=\omega^{-1}x$. But by easy character theory (recall that $G$ is cyclic of order $n$) we can find $\chi$ with $\chi(\sigma)=\xi^{-1}$, and we are done.

I noticed that Lang hides the formula in Hilbert’s Theorem 90: an element of norm $1$ in a cyclic extension, with $\sigma$ a generator of the Galois group, is of the form $\sigma(x)/x$ for some non-zero $x$; this is applied to the $n$-th root of unity in $L$. The proof of Hilbert’s Theorem 90 uses something with the same flavor as the representation theory argument: Artin’s Lemma to the effect that the elements of $G$ are linearly independent as linear maps on $L$. I haven’t completely elucidated the parallel however.

(P.S. Chambert-Loir’s blog has some recent very interesting posts on elementary Galois theory, which are highly recommended.)

March 26th, 2015 at 5:45 pm

## A parity lemma of A. Irving

one comment

In his recent work on the divisor function in arithmetic progressions to smooth moduli, A. Irving proves the following rather amusing lemma (see Lemma 4.5 in his paper):

Lemma Let $p$ be an odd prime number, let $k\geq 1$ be an integer and let $h=(h_1,\ldots,h_k)$ be a $k$-tuple of elements of $\mathbf{F}_p$. For any subset $I$ of $\{1,\ldots, k\}$, denote
$h_I=\sum_{i\in I}{h_i},$
and for any $x\in\mathbf{F}_p$, let
$\nu_h(x)=|\{I\subset \{1,\ldots, k\}\,\mid\, h_I=x\}|$
denote the multiplicity of $x$ among the $(h_I)$.
Then if none of the $h_i$ is zero, there exists some $x$ for which $\nu_h(x)$ is odd.

I will explain two proofs of this result, first Irving’s, and then one that I came up with. I’m tempted to guess that there is also a proof using some graph theory, but I didn’t succeed in crafting one yet.

Irving’s proof. This is very elegant. Let $\xi$ be a primitive $p$-th root of unity. We proceed by contraposition, hence assume that all multiplicities $\nu_h(x)$ are even. Now consider the element
$N=\prod_{i=1}^k(1+\xi^{h_i})$
of the cyclotomic field $K_p=\mathbf{Q}(\xi)$. By expanding and using the assumption we see that
$N=\sum_{x\in\mathbf{F}_p} \nu_h(x)\xi^{x}\in 2\mathbf{Z}[\xi].$
In particular, the norm (from $K_p$ to $\mathbf{Q}$) of $N$ is an even integer, but because $p$ is odd, the norm of $1+\xi^{h_i}$ is known to be odd for all $h_i\not=0$. Hence some factor must have $h_i=0$, as desired.

A second proof. When I heard of Irving’s Lemma, I didn’t have his paper at hand (or internet), so I tried to come up with a proof. Here’s the one I found, which is a bit longer but maybe easier to find by trial and error.

First we note that
$\sum_{x\in \mathbf{F}_p} \nu_h(x)=2^k$
is even. In particular, since $p$ is odd, there is at least some $x$ with $\nu_h(x)$ even.

Now we argue by induction on $k\geq 1$. For $k=1$, the result is immediate: there are two potential sums $0$ and $h_1$, and so if $h_1\not=0$, there is some odd multiplicity.

Now assume that $k\geq 2$ and that the result holds for all $(k-1)$-tuples. Let $h$ be a $k$-tuple, with no $h_i$ equal to zero, and which has all multiplicities $\nu_h(x)$ even. We wish to derive a contradiction. For this, let $j=(h_1,\ldots,h_{k-1})$. For any $x\in\mathbf{F}_p$, we have
$\nu_h(x)=\nu_j(x)+\nu_j(x-h_k),$
by counting separately those $I$ with sum $x$ which contain $k$ or not.

Now take $x$ such that $\nu_j(x)$ is odd, which exists by induction. Our assumptions imply that $\nu_j(x-h_k)$ is also odd. Then, iterating, we deduce that $\nu_j(x-nh_k)$ is odd for all integers $n\geq 0$. But the map $n\mapsto x-nh_k$ is surjective onto $\mathbf{F}_p$, since $h_k$ is non-zero. Hence our assumption would imply that all multiplicities $\nu_j(y)$ are odd, which we have seen is not the case… Hence we have a contradiction.

March 16th, 2015 at 12:31 pm

Posted in Exercise,Mathematics

## 0.00023814967230605090687395214144185337601

Yesterday my younger son was playing dice; the game involved throwing 6 dices simultaneously, and he threw a complete set 1, 2, 3, 4, 5, 6, twice in a row!

Is that a millenial-style coincidence worth cosmic pronouncements? Actually, not that much: since the dices are indistinguishable, the probability of a single throw of this type is

$\frac{6!}{6^6}\simeq 0.015432098765432098765432098765432098765,$

so about one and a half percent. And for two, assuming independence, we get a probability

$\frac{(6!)^2}{6^{12}}\simeq 0.00023814967230605090687395214144185337601,$

or a bit more than one chance in five throusand. This is small, but not extraordinarily so.

(The dices are thrown from a cup, so the independence assumption is quite reliable here.)

October 27th, 2014 at 7:38 pm

## Leo’s first theorem

I learnt the following from my son Léo: the teacher asks to compute $9+9$; that’s easy
$9\ +\ 9\ =\ 18.$
But no! The actual question is to compute $9$ times $9$! We must correct this! But it’s just as easy without starting from scratch: we turn the “plus” cross a quarter turn on the left-hand side:
$9\ \times\ 9$
and then switch the digits on the right-hand side:
$9\ \times\ 9\ =\ 81.$

This is a fun little random fact about integers and decimal expansions, certainly.

But there’s a bit more to it than that: it is in fact independent of the choice of base $10$, in the sense that if we pick any other integer $b\geq 2$, and consider base $b$ expansions, then we also have

$(b-1)\ +\ (b-1)\ =\ 2b-2\ =\ b+(b-2)= \underline{1}\,\underline{b-2}$

as well as

$(b-1)\ \times\ (b-1)\ =\ (b-1)^2=b(b-2)+1= \underline{b-2}\,\underline{1},$

(where we underline individual digits in base $b$ expansion.)

At this point it is natural to ask if there are any other Léo-pairs $(x,y)$ to base $b$, i.e., pairs of digits in base $b$ such that the base $b$ expansions of the sum and the product of $x$ and $y$ are related by switching the two digits (where we always get two digits in the result by viewing a one-digit result $z$ as $\underline{0}\, \underline{z}$).

It turns out that, whatever the base $b$, the only such pairs are $(b-1,b-1)$ and the “degenerate” case $(0,0)$.

To see this, there are two cases: either the addition $x+y$ leads to a carry, or not.

If it does, this means that $y=b-z$ where $x>z$. The sum is then

$x+y=b+(x-z)=\underline{1}\,\underline{x-z}.$

So this is a Léo-pair if and only if

$xy=\underline{x-z}\,\underline{1}.$

This equation, in terms of $x$ and $z$, becomes

$x(b-z)=b(x-z)+1,$

which holds if and only if $z(b-x)=1$. Since the factors are integers and non-negative, this is only possible if $z=b-x=1$, which means $x=y=b-1$, the solution found by Léo.

Now suppose there is no carry. This means that we have $0\leq x,y\leq b-1$ and $x+y\leq b-1$. Then
$x+y=\underline{0}\,\underline{x+y},$
and we have a Léo-pair if and only if
$xy=\underline{x+y}\,\underline{0},$
i.e., if and only if $xy=b(x+y)$.

This is not an uninteresting little equation! For a fixed $b$ (which could now be any non-zero rational), this defines a simple quadratic curve. Without the restrictions on the size of the solution $(x,y)$, there is always a point on this curve, namely
$(x_0,y_0)=(2b,2b).$
This does not fit our conditions, of course. But we can use it to find all other integral solutions, as usual for quadratic curves. First, any line through $(x_0,y_0)$ intersects the curve in a a second point, which has rational coordinates if the line is also defined by rational coefficients, and conversely.

Doing this, some re-arranging and checking leads to the parameterization

$\begin{cases} x=b+k\\ y=b+\frac{b^2}{k}\end{cases}$

of the rational solutions to $xy=b(x+y)$, where $k$ is an arbitrary non-zero rational number. In this case, this can also be found more easily by simply writing the equation in the form
$0=xy-bx-by=(x-b)(y-b)-b^2\ldots$

Now assume that $b\geq 2$ is an integer, and we want $(x,y)$ to be integers. This holds if and only if $k$ is an integer such that $k\mid b^2$.

Such solutions certainly exist, but do they satisfy the digit condition? The answer is yes if and only if $k=-b$, which means $x=y=0$, giving the expected degenerate pair. Indeed, to have $x, the parameter $k$ must be a negative divisor of $k^2$. We write $k=-d$ with $d\mid k^2$ positive. Then to have non-negative digits, we must have
$\begin{cases} x=b-d\geq 0\\ y=b-\frac{b^2}{d}\geq 0\end{cases},$
the first one of these inequalities means $b\geq d$, while the second means that $b\leq d$

June 11th, 2014 at 11:46 am

Posted in Exercise,Mathematics