%% alf@mpce.mq.edu.au

%% Copyright 1993
%% Alf van der Poorten
%% ceNTRe for Number Theory Research
%% Macquarie University
%% NSW 2109 AUSTRALIA


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{\it Again in 1850 the Acad\'emie des Sciences de Paris offered a
golden medal and a prize of $3000$ Francs to the mathematician who
would solve Fermat's problem. 

In 1856 it determined to withdraw the question from competition but
to award the medal to Kummer `for his beautiful researches on the
complex numbers composed of roots of unity and integers'.}\smallskip
\rightline{P. Ribenboim, in part translating
Cauchy.\hskip1cm}\bigskip

All we really will need about ideal numbers is that, according to
their purpose, there is unique factorisation into ideals in every
number field. But there is a bonus. An element $\alpha$ corresponds
to the {\it principal\/} ideal $(\alpha)$, and so do its {\it
associates\/} $\eta\alpha$, where $\eta$ is any unit. Thus ideals
hide units, as it were, which is great. Of course, {\it vice
versa\/} the ideal $(\alpha)$ corresponds to one or other of the
{\it associates\/}
$\eta\alpha$ of $\alpha$, where $\eta$ is some unit; that's not so
good. But Kummer copes. 


{\bf The regular case:}\  An odd prime $p$ is {\it regular\/} if,
roughly speaking, one can do common sense arithmetic --- as regards
Fermat's Last Theorem --- in the field
$\Q(\zeta)$, where $\zeta=\zeta_p$ is a primitive $p$-th root of
unity. In his first arguments of 1847 Kummer works subject to the
assumptions that the $p$-th power of an ideal can be treated as a
number in the field, and the truth of a lemma concerning $p$-th
powers of units that I mention below. A little later he found that
both of these conditions follow from the regularity of $p$.

We recall that in the
$p$-th cyclotomic field the integers are
$a_0+a_1\zeta+\cdots+a_{p-2}\zeta^{p-2}$, with the $a_i\in\Z$. 

Our object is to demonstrate the impossibility of
$$x^p+y^p+z^p=0$$ in nonzero rational integers $x$, $y$ and $z$.

{\bf Case I} $p$ regular (Kummer).\ \ We begin with the first case
of the FLT, thus when
$p\ndiv xyz$.

Then 
$$x^p+y^p=\prod_{j=0}^{p-1}(x+\zeta^jy)=-z^p\,,$$ and the factors
$x+\zeta^jy$ are relatively prime, since their only possible common
divisor is a factor of $p$, whilst $p\ndiv z$. Hence each is a
$p$-th power --- technically, each {\it principal ideal\/}
$(x+\zeta^jy)$ is the $p$-th power of an ideal of
$\Q(\zeta)$. When $p$ is regular it follows that we have
$$x+\zeta y=\eta\alpha^p\,,$$ where $\alpha$ is an integer, and
$\eta$ a unit of $\Q(\zeta)$ [compare
$36=-4\cdot-9$, and note that $-4$ is a square times a unit $-1$,
not just a square].

The difficulty is that $\Q(\zeta)$ has nontrivial units $\eta$. But
Kummer succeeded in proving that $\eta/\overline\eta$ is always a
$p$-th root of unity,
$\zeta^{2a}$, say.

Now if
$$\alpha=a_0+a_1\zeta+\cdots+a_{p-2}\zeta^{p-2}
\quad\hbox{then}\quad
\alpha^p\equiv a_0+a_1+\cdots+a_{p-2}\pmod p\,,$$ so
$\alpha^p\equiv\overline\alpha^p\pmod p$, and we have
$$x+\zeta y\equiv(x+\overline\zeta y)\zeta^{2a}\pmod p
\quad\hbox{or}\quad x(1-\zeta^{2a})+y(\zeta-\zeta^{2a-1})\equiv0\pmod
p\,.$$ But the latter is impossible if $p\ndiv xy$, unless
$2a\equiv1\pmod p$, in which case
$p\div x-y$. But then, by symmetry, we can also establish $p\div
y-z$ and $p\div z-x$, so 
$$0=x^p+y^p+z^p\equiv3x^p\equiv3x\pmod p\,,$$ whence $p\div x$ if
$p\gt 3$, and that is a contradiction.

Finally, for completeness, FLT I with $p=3$ is impossible by Sophie
Germain. One need only note that all cubes not divisible by
$7=2\cdot3+1$ are
$\equiv\pm1\pmod7$.

{\bf Case II} $p$ regular (Kummer).\ \ Suppose, without loss of
generality, that
$p\div z$, and put
$\lambda=1-\zeta$. Recall that the significance of $\lambda$ is
that the principal ideal $(\lambda)$ satisfies
$(\lambda)^{p-1}=(p)$.

Suppose also that we already have the Case I result for integers
$\alpha$, $\beta$, $\gamma$ of $\Q(\zeta)$, thus that there is no
nontrivial solution of
$$\alpha^p+\beta^p+\gamma^p=0 \quad\hbox{if}\quad
(p,\alpha\beta\gamma)=1\,.$$ We consider the apparently more
general equation
$$\alpha^p+\beta^p=\varepsilon\lambda^{np}\gamma^p\,,$$ with
$\varepsilon$ a unit, and $\alpha$, $\beta$, $\gamma$ nonzero
integers of
$\Q(\zeta)$. We will use descent on $n$, and so may suppose that
$\alpha$ and
$\beta$ are prime to $\lambda$.

Firstly, one remarks that we may further suppose that $\alpha$ and
$\beta$ each have the property of being congruent to a rational
integer modulo $\lambda^2$. The idea is that, since, for example,
we are only given $\alpha^p$, there is no loss in replacing
$\alpha$ by $\zeta^k\alpha$.

[Since $\zeta=1-\lambda$, we have
$\zeta^k\equiv1-k\lambda\pmod{\lambda^2}$. So if
$\alpha\equiv l+m\lambda\pmod{\lambda^2}$ then 
$$\zeta^k\alpha\equiv (1-k\lambda)(l+m\lambda)\equiv
l+(m-lk)\lambda\pmod{\lambda^2},$$ and it suffices to choose $k$ so
that
$lk\equiv m\pmod p$.]

A major Hilfsatz (Kummer's lemma) that Kummer will need is the
following: {\sl Let
$\eta$ be a unit of
$\Q(\zeta)$ and suppose there is a rational integer $a$    so that
$$\eta\equiv a \mod \lambda^p\,.$$ Then $\eta$ is the $p$-th power
of a unit of
$\Q(\zeta)$.\/}

We can rewrite our equation in terms of ideals as 
$$\prod_{r=0}^{p-1}(\alpha+\zeta^r\beta)=(\lambda)^{np}(\gamma)^p\,.$$
One notices that the ideals on the left do each have a common
factor $(\lambda)$, and that $\lambda\div \alpha+\beta$ implies
$\lambda^2\div\alpha+\beta$, so $n\gt1$. Any other common factor
$\frak d$, say, must be a common factor of
$(\alpha)$ and 
$(\beta)$. For the rest, the ideals on the left must be $p$-th
powers. So we have
$$(\alpha+\zeta^r\beta)=\frak d(\lambda)\frak a_r^p
\quad\hbox{and}\quad (\alpha+\beta)=\frak
d(\lambda)^{p(n-1)+1}\frak a_0^p\,.$$ The point is that the ideals
$\frak a^p$ may be viewed just as $p$-th powers of integers of
$\Q(\zeta)$. Now consider the identity (known, somehow, in the
transcendence trade as `Siegel's identity')
$$(\zeta-\overline\zeta)(\alpha+\beta)+
(\overline\zeta-1)(\alpha+\zeta\beta)
+(1-\zeta)(\alpha+\overline\zeta\beta)=0\,.$$ After dividing by
$\lambda$ we have integers $\alpha''$, $\beta'$ and $\gamma'$ so
that
$$\varepsilon'\lambda^{p(n-1)}{\gamma'}^p=
{\beta'}^p+\varepsilon''{\alpha''}^p$$
for units $\varepsilon'$ and $\varepsilon''$. But, in effect by
Fermat's (little) theorem, a $p$-th power of an integer of
$\Q(\zeta)$ is a rational integer mod $p$, so
$\varepsilon''$ is a rational integer mod $p$. Thus, and here we
invoke Kummer's lemma,
$\varepsilon''$ is the $p$-th power of some unit. Hence we have an
integer $\alpha'$ in $\Q(\zeta)$ so that
$$\varepsilon'\lambda^{p(n-1)}{\gamma'}^p={\beta'}^p+{\alpha'}^p\,,$$
contradicting the minimality of $n$.

{\bf About cyclotomic fields.}\ The class number $h$ of a number
field measures the extent to which unique factorisation fails;
technically $h$ is the {\it order\/} of the {\it group\/} of ideal
{\it classes\/}, that is, of ideals modulo the class of {\it
principal\/} ideals. A cyclotomic field
$\Q(\zeta_p)$ is said to be {\it regular\/} if
$p\ndiv h$.

It has been known for a long while (Jensen 1915) that there are
infinitely many irregular primes. Curiously, since heuristically
the majority of primes are regular, it is not known that there are
infinitely many regular primes.

The class number $h^+$ of the {\it real\/} cyclotomic field
$\Q(\zeta_p+\zeta_p^{-1})$ is a divisor of $h$. It also happens that
$h^+=[E:E_0]$ is the {\it index\/} of the subgroup of the units of
$\Q(\zeta_p)$ generated by the {\it real cyclotomic units\/}
$$\left|\frac{1-\zeta^r}{1-\zeta}\right|
=\frac{\sin
r\pi/p}{\sin \pi/p}\qquad r=2, 3,\ldots, \tfrac12(p-1)$$ in the
group $E$ of positive real units of $\Q(\zeta)$.

Kummer showed that if $p\div h^+$ then $p\div h_*=h/h^+$. Thus
whenever $p\div h$ certainly $p\div h_*$. Kummer found a criterion
for $p\div h_*$ that could be reasonably conveniently checked, at
least for small $p$, say less than
$100$. It is, I believe, still not known whether in fact ever
$p\div h^+$.

To be precise, Kummer showed that if $p\div h_*$ then $p$ divides
(the numerator of) a Bernoulli number $B_{2k}$ with $2\le
2k\le p-3$. The irregular primes less than $100$ are $37$, $59$ and
$67$ (for example, $37\div B_{32}$ and $59\div B_{44}$).

By 1857 he had completed an analysis of the irregular case to an
extent that allowed him to deal with the cases for which $p$
divides at most one of the  numbers $B_{2}$,
$B_4$, $\ldots$, $B_{p-3}$, thereby proving the FLT beyond
exponent $n=100$.

I don't see much point here in struggling to explain the 
calculations which Kummer has to carry out, other than to say that in 1850
Kummer shows that $p\div h_*$ if and only if there is some integer $k$ with
$1\le k\le (p-3)/2$, so that $p^2$ divides the sum $\sum_{j=1}^{p-1}j^{2k}$. 
This explains the appearance of 
the Bernoulli numbers.
 
However, by shying away
from technicalities and gory details, I do threaten to miss the point of
Kummer's contributions to mathematics. Sure, the ancients are a little hard
to read, because of poor notation and conventions now strange to us, 
because they find some things, now easy to us, difficult; and, I guess,
because they gloss over some things we now find difficult. And to top that
off, they often fail to write in American English. It was no great 
surprise to hear John Coates say, at the time of the Coates--Wiles work on
 the Birch--Swinnerton-Dyer conjectures in the complex multiplication
case, that core ideas had come from reading Kummer (the context was, I think,
remarks in praise of Kurt Mahler --- who would admonish us to read the
masters). Some years later, Thaine's contributions could even be recognised
by me as an ingenious transfer of Kummer's ideas from the cyclotomic to the
elliptic case. Indeed,  Kummer introduced $p$-adic
analysis some sixty years before Hensel.\footnote""{Most of this material
comes from lectures I prepared c. 1977, when I was probably aided by Le
Veque and by Ribenboim {\it op. cit.\/}; I looked at Neal Koblitz, $p$-{\it
adic Numbers, $p$-adic analysis, and zeta-functions\/}, GTM 58,
Springer-Verlag 1977, for my concluding remarks.}


Specifically, it had been known since 1840 by work of Clausen and of von
Staudt that if $k\ge1$ then
$B_{2k}+\sum_{(p-1)\mid2k}1/p$
is a rational integer. Here and in the sequel $p$ denotes only prime
numbers. It follows that if
$p-1\div 2k$ then
$pB_{2k}\equiv-1\pmod p$. Kummer proves that if $p-1\ndiv k$ then $p$ does
not divide the denominator of $B_k/k$, and that if, further, $k\equiv
k'\pmod {p^n}$ then $(1-p^{k-1})B_k/k\equiv (1-p^{k'-1})B_{k'}/k' \pmod
{p^{n+1}}$. These congruences were just mysterious oddities until Kubota and
Leopoldt interpreted them in 1964 in terms of $p$-adic interpolation of the
Riemann $\zeta$-function. And so, on to $p$-adic $L$-functions. Thus, after
all, in the work of Kummer we find the genesis of the
eventual solution of Fermat's Last Theorem.

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