# E. Kowalski's blog

## 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

Today’s Google doodle celebrates Emmy Noether.

Noether

Algebra would look very different without her (“successive sets of symbols with the same second suffix“).

March 23rd, 2015 at 11:35 am

Posted in Mathematics

## 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

## Who proved the Peter-Weyl theorem for compact groups?

Tamas Hausel just asked me (because of my previous post on the paper of Peter and Weyl) how could Peter and Weyl have proved the “Peter-Weyl Theorem” for compact groups in 1926, not having Haar measure at their disposal? Indeed, Haar’s work is from 1933! The answer is easy to find, although I had completely overlooked the point when reading the paper: Peter and Weyl assume that their compact group is a compact Lie group, which allows them to discuss Haar measure using differential forms!

Peter-Weyl

So the question is: who first proved the full “Peter-Weyl” Theorem for all compact groups? Pontryaguin, in 1936, certainly does, without remarking that Peter-Weyl didn’t, possibly because it was clear to anyone that the argument would work as soon as an invariant measure was known to exist. But since there are “easier” proofs of the existence of Haar measure for compact groups than the general one for all locally-compact groups (using some kind of fixed-point argument), it is not inconceivable that someone (e.g., von Neumann) might have made the connection before.

In fact, there is an amusing mystery in connection with Pontryaguin’s paper and von Neumann: concerning Haar measure, he refers to a paper of von Neumann entitled Zum Haarschen Mass in topologischen Gruppen, and gives the helpful reference Compositio Math., Vol I, 1934. So we should be able to read this paper on Numdam? But no! The first volume of Compositio Mathematica there is from 1935; it is identified as Volume I, and there is no paper of von Neumann to be found…

[Update: as many people pointed out, the paper of von Neumann is indeed on Numdam, but appeared in 1935; I was tricked by the absence of 1934 on the Compositio archive and the author’s name being written J.V. Neumann (I had searched Numdam with “von Neumann” as author…)]

March 15th, 2015 at 8:03 pm

Posted in Mathematics