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

%% alf@mpce.mq.edu.au


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


%%If you don't have a Mac with Textures comment out the
%%blatantly false claim that \MacTexturestrue below. 
%%Now DVIPS will produce the pictures, if you have the file
%%logo.art in your directory, that is.
%%If you don't have either TeXtures or DVIPS, despair. 
%%Tell TeX that it can do without the file epsf 
%%when it asks for it,
%%and remove the comment just above \Part,
%%so as to undefine the picture. 
%%Also deny \MacTexturestrue of course.
\newif\ifMacTextures 
\MacTexturestrue
%%Remember, you've got to have logo.art

\magnification=\magstep1
%\input amstex.tex	%Make sure that it is v2.1  
%Update from e-math, if not.
%\documentstyle{amsppt}
\vsize 9.4truein

\input AlfPreamble  %%I do hope you downloaded this vital file
%\def\centrelogo{\relax} \def\smallcentrelogo{\relax}

\Part=VIII

\input FermatTopmatter %%and this one

It seems perfectly reasonable to say that if something is true
locally everywhere then it is true globally, and conversely. The
only question is what on earth one might mean in saying such a
thing in the context of diophantine problems. Let me make some
suggestions, starting with the Riemann $\zeta$-function
$\zeta(s)=\sum n^{-s}$. By the way, here $s$ is a complex variable,
traditionally given as $s=\sigma+it$. Of course the series
converges absolutely only in the right plane $\sigma\gt1$. We owe
to Euler the `factorisation'
$$\sum_{n=1}^{\infty}\frac1{n^s}=
\prod_p\Bigl(1-\frac1{p^s}\Bigr)^{-1}\,.$$
Indeed, the typical Euler factor is
$1+p^{-s}+p^{-2s}+p^{-3s}+\cdots\ $,
and the product representation amounts to the unique
factorisation theorem whereby each positive integer has a unique
representation as a product of primes. In this case we might hope
to obtain local information, that is, about the primes, from
global knowledge, of the positive integers. For example, because
$1+\frac12+\frac13+\cdots+\frac1n\sim\log n$, we find, after taking a
logarithm, that $\sum_{p\lt x}p^{-1}\sim\log\log x$.
Since $\log\log x=\int_e^{x}(t\log t)^{-1}dt$, we might deduce
that the $m$-th prime is $\sim m\log m$ and that, therefore, the
number of primes less than $x$ is $\sim x/\log x$.

But I have something more down-to-earth in mind. Suppose I am
wondering whether there are rational points on the elliptic curve
$x^4-17=2y^2$. I will probably set $x=X/Z$, $y=Y/Z$ and multiply by
$Z^4$ to obtain a homogeneous \poly\ in the {\it projective\/}
co-ordinates $X$, $Y$ and $Z$ --- this also has the advantage of
giving a glimpse of $\infty$: just consider the line
$Z=0$. But, whatever, after failing to notice a solution we might
well choose to consider the problem mod $p$ --- with the intention of
employing the  evident principle that if there is no solution mod $m$
there is certainly none {\it globally\/}; that is, there then cannot be
a rational solution. However, in the present example there is a solution
mod $p$ for all
$p$. That doesn't require an infinite amount of work; in fact it
suffices to check by trial and error at $p=2$ and $17$, the other {\it
places\/} being quite automatic. Of course we do want to deal with an
arbitrary modulus
$m$, so we also need the powers of the primes. However, once one has a
solution mod $p$ it is easy to see whether or not there is a solution
mod ${p^n}$ for all $n$. Let me illustrate that by showing that
$x^2\equiv2\pmod {7^n}$ has a solution for all positive integers $n$,
given that $3^2\equiv2\pmod7$. It will give a feeling of generality if
I set $f(x)=x^2-2$ and $p=7$.

Suppose we already know that $f(x_{n-1})\equiv0\pmod{p^n}$. We want
$x_n=x_{n-1}+xp^n$ so that $f(x_{n-1}+xp^n)\equiv0\pmod{p^{n+1}}$. But
the Taylor expansion
$$f(x_n)=f(x_{n-1}+xp^n)=f(x_{n-1})+xp^nf'(x_{n-1})+
\cdots\ ,$$ where the $\cdots$ indicate terms in yet higher powers of
$p$, tells us that
$$0\equiv p^{-n}{f(x_{n-1})}+xf'(x_{n-1})\equiv 
p^{-n}{f(x_{n-1})}+xf'(x_{0})\pmod p\,;$$ recall that
$p^n\div f(x_{n-1})$ and $x_{n-1}\equiv x_{0}\pmod p$. So we have only
to solve a simple linear congruence in $x$ to obtain $x_n$. This 
argument is known as Hensel's lemma. Mind you, there is a little more
work to be done if $x_0$ happens to be a singular point, that is if
$f'(x_0)\equiv0$, but ultimately things settle down to just a
sequence of linear problems as above. So it's almost painless to decide
whether there is a solution mod $p^{\infty}$, so to speak. In our
example we obtain
$$x_{\infty}=3+1\cdot7+2\cdot7^2+6\cdot7^3+1\cdot7^4+2\cdot7^5
+1\cdot7^6+2\cdot7^7+4\cdot7^8+\cdots\
.$$ 
It is a well known principle both of magic and of mathematics that
once one knows the true name of an object one controls it. Thus we say
that $x_\infty$ is a $7$-{\it adic integer\/} and observe that we can do
arithmetic with  objects such as it much as if they were absolutely
convergent power series in powers of $7$, provided we remember to
`carry' to cope with coefficients outside the range
$\{0,1,2,3,4,5,6\}$.

To see what is actually going on here it is best first to remember
that we barely understand the real numbers $\R$. We would be
satisfied to say that the reals are the set of all decimals and proud to
pop out with the statement that
$\R$ {\sl is the field of limit points of convergent sequences of
rationals\/}; which it is. If we were asked to clarify this claim we'd
remark that a sequence
$(a_h)=\{a_0,a_1,a_2,\ldots\}$ of rationals is convergent if it is
Cauchy, that is, if $|a_n-a_m|$ is arbitrarily small as soon as $m$ and
$n$ are sufficiently large; we'd show that the termwise difference and
product of Cauchy sequences is again Cauchy; and we'd agree to identify
such sequences if they have the same limit --- that is, if they differ
by a {\it null sequence\/}: one for which $|a_n|$ is as small as we like
whenever $n$ is  large enough. In a trice we would have taken a
 leap in sophistication and would hear ourselves saying that $\R$
is just {\sl the ring of Cauchy sequences of rationals modulo its
maximal ideal of null sequences\/}.

The joker in this explanation is that we could give it without telling
anyone what we meant by `small'; that is, what we mean by the absolute
value
$|\phantom{a_n}|$. Well, yes, we'd have used that $|\phantom{a_n}|$ is
a positive definite map on the rationals preserving multiplication --- 
$|a|\ge0$ and 
$=0$ if and only if
$a=0$; and $|a\cdot b|=|a|\cdot|b|$. And we will have needed the
{\it triangle inequality\/}: $|a+b|\le|a|+|b|$. But that's it.

So a listener could say, ``Aha! If I were to pick my favourite prime
$p$ then I could take a rational $r/s$ and apply unique factorisation to
write $r/s=\prod_p p^{v_p}$ with integers $v_p$, mostly zero but
otherwise positive or negative,  and set $|r/s|=p^{-v_p}$.'' On
overcoming our surprise we'd have to admit that there is indeed
nothing wrong with believing that `small' means `divisible by a
high power of $p$'. After making her write
$|\phantom{a_n}|_p$ instead of just
$|\phantom{a_n}|$ to avoid confusion (we might agree to write
$|\phantom{a_n}|_\infty$ for the absolute value we thought we had
meant, so as not to make too special a claim for it) we'd find
that all was well. She has indeed nominated an absolute value
and her {\it completion\/} $\Q_p$ --- the $p$-adic rationals --- has as
much right to respect as does the completion $\R$.

Is $\R$ special? Well. physical reality, as we,
and Archimedes, believe it, has  no infinitesimals. In
$\R$, given
$a\ne0$ no matter how close to $0$, and $b$ no matter how large, there
is always an integer $n$ so that $|na|_\infty\gt |b|_\infty$. We say
that $\R$, and thence the absolute value $|\phantom{a_n}|_\infty$, are
{\it archimedean\/}. On the other hand, the $p$-adic absolute values
satisfy the {\it ultrametric\/} inequality
$|a+b|_p\le\max\(|a|_p,|b|_p\)$, a stronger form of the triangle
inequality. That entails $|na|_p\le |a|_p$ for all integers $n$.
We say that the fields $\Q_p$, and thence the absolute values
$|\phantom{a_n}|_p$, are {\it nonarchimedean\/}. By the way, {\it
pace\/} Archimedes, in recent years the theoretical physicists have
learned about the $p$-adic fields and have begun to wonder whether
they may not help in modelling just what happens in the nuclei of
atoms, for instance. So much for reality.

The $x_\infty$ of our example is an element of the field $\Q_7$. It is
in fact a
$7$-adic {\it integer\/}, that is, it satisfies $|x_\infty|_7\le1$.
Just as $\Q$ is the field of quotients of elements of $\Z$, so the
fields $\Q_p$ are the respective quotient fields of the domains
$\Z_p$. The elements of $\Q_p$ can be represented as sums
$\sum_{n=-m}a_np^n$ just as the elements of $\R$ are often represented
as decimals $\sum_{n=-m}b_n10^{-n}$. All's well, because just as
$10^{-n}$ gets small at exponential rate in $\R$, so $p^n$ becomes
small in $\Q_p$.

The bottom line is that checking diophantine equations mod $m$ can
usefully be viewed as asking for solutions in the fields $\Q_p$
of $p$-adic rationals. That's true, by the Chinese Remainder Theorem,
whereby any congruence modulo
$m=\prod p^{\ord_pm}$ is the same thing as a collection of
congruences modulo the respective $p^{\ord_pm}$. One talks about looking
for solutions {\it at\/}
$p$; these are the solutions {\it locally\/}. But we'd better not forget
about $\infty$. The diophantine equation $x^2+y^2=-1$ has solutions at
all $p$, but it really has no rational solutions, because it has no
solutions in $\R$. Thus, when we speak about {\it locally
everywhere\/} we mean at all $p$ {\it and\/} $\infty$. The absolute
values are not totally unrelated. For nonzero $x$, unique factorisation
gives the `product formula' $\sum_p\log|x|_p=-\log|x|_\infty$.

With all that said, is it true that if a diophantine equation has
solutions locally everywhere --- thus in all $\Q_p$ and in $\R$ ---
then it has {\it global\/} solutions --- thus in $\Q$? Sadly, no. There
is a theorem of Hasse-Minkowski showing that a {\it quadratic form\/}
--- a homogeneous \poly\ expression of degree $2$ --- takes a given
rational value if and only if it has a solution locally everywhere. But
this {\it Hasse principle\/} does not apply generally. In fact I
obnoxiously chose the example equation $x^4-17=2y^2$ to display an
equation with solutions locally everywhere but with no rational
solutions. Nonetheless, our view of diophantine equations is that they
would like to obey the Hasse principle. We will want to study the
obstruction to their doing so.

{\bf An elementary aside.}\ Suppose that $r/s$ is a rational zero of
the
\poly\ $f(x)=a_0x^n+a_1x^{n-1}+\cdots+a_{n-1}x+a_n\in\Z[x]$. If $p\div
r$ then this zero is $0\pmod p$ so $0$ must be a zero of $f(x) \pmod p$.
That is, $a_n\equiv0\pmod p$, or $p\div a_n$. This beginning of a
local argument of course eventually entails that $r\div a_n$. Now 
consider the so-called {\it reciprocal\/} \poly\ $x^nf(x^{-1})$. We
see that it has $s/r$ as a zero and conclude that $s\div a_0$.

An instructive {\it degenerate\/} case of this discussion deals with
the case of a zero $0$ of $f(x)$. Since anything whatsoever
divides $0$, anything whatsoever divides $a_n$, showing that of
course $a_n=0$. Now the preceding argument would have us say that the
reciprocal \poly\ has $\infty$ as a zero. Indeed, by rights it should
be a \poly\ of degree $n$. It displays its zero $\infty$ by
actually being of lower degree.

A generalisation is Gau\ss's lemma which asserts that if a \poly\
$f(x)$ with integer coefficients has a factorisation as a product of
\poly s with rational coefficients then it already has a
factorisation into a product of \poly s with integer coefficients. To
see this we write $df(x)=g(x)h(x)$, where the
nonzero integer $d$ is a multiplier required so that
$g$ and $h$  have integer coefficients. If we fear that $p\div d$ we
look at the identity modulo $p$, obtaining $0\equiv g(x)h(x)\pmod p$.
But, with $p$ prime, this entails that either $g(x)\equiv0$ or
$h(x)\equiv0\pmod p$. So we may divide through by $p$, in
particular replacing $d$ by $d/p$. Then by descent we may conclude that
$d$ has no prime divisors, thus $d=1$ as we wished to show.

The terminology `global' and `local', and the talk of `at' $p$ is
sometimes jolting. Its genesis can be
readily illustrated. Suppose we think about the arithmetic of functions
$f(z)$ defined on
$\C$. It seems reasonable to consider the \poly s as the integers and
the {\it rational functions\/} --- quotients of \poly s --- as the
rationals. What are the irreducibles? That's easy, they're just the
linear \poly s $z-\alpha$, one for each point $\alpha\in\C$. So that's
why we say `at' a prime! To say that $z-\alpha$ divides some \poly\
$f$ exactly $m=\ord_\alpha f$ times, is to say that $f$ has a zero of
order $\ord_\alpha f$ at $\alpha$; or, if you prefer, that the Taylor
expansion of $f$ at $\alpha$ has
$(m!)^{-1}f^{(m)}(\alpha)(z-\alpha)^m$ as its first nonzero term.

And just as the numbers have a special
`prime' corresponding to the absolute value
$|\phantom{a_n}|_\infty$, so there is a place in $\C$, the
point\footnote"$^*$"{Plainly $\C$ is not a plane, but a sphere. Just
put a globe on the origin of the plane and draw a line from the
North Pole $N$ to each point in
$\C$. Now identify each point in $\C$ with the point on the globe cut by
its line. Clearly $N$ corresponds to $\infty$, and we may speak of
{\it the\/} point at $\infty$.} at $\infty$. The fundamental theorem of
algebra  that, in $\C$, a
\poly\ $f$ of degree $\deg f$ has exactly $\deg f$ zeros counted
according to multiplicity, becomes the formula
$\sum_{\alpha\in\C}\ord_\alpha f=-\ord_\infty f$.

Geometry with an ultrametric is great fun. Every triangle is isosceles
and each point inside a circle is its centre. However, we're
interested in arithmetic and what pleases us is that in $\Q_p$ the
integers $\Z$ no longer discretely keep their distance one from the
other. That allows one to look at difficult functions like the Riemann
$\zeta$-function in a new light. After all, we are a bit blas\'e to
take a function like $n^{-s}$ for granted. An honest integral power,
like
$n^{-2k}$, is fine; it just means $n^{-2}$ multiplied by itself $k$
times. But to suddenly generalise that to an analytic function
$e^{-s\log n}$ is a drastic step. In that spirit, one might well feel
that except for the perfectly decent rational numbers
$\zeta(2k)/\pi^{2k}$ all the other $\zeta(s)$ should be disregarded.
But, it was nice to have an analytic function. Now a
miracle comes to the rescue. Kummer's congruences: if
$k\equiv k'\pmod {p^n}$ then $(1-p^{k-1})B_k/k\equiv
(1-p^{k'-1})B_{k'}/k' \pmod {p^{n+1}}$, amount exactly to our being able
to view the numbers
$(-1)^k(1-p^{2k-1})(2k-1)!\zeta(2k)/2^{2k-1}\pi^{2k}$ as the values
of a $p$-adic analytic function of a $p$-adic variable $k$. This is the
$p$-adic
$\zeta$-function.

\end