\lecture{22 : Towards the AKS Primality Test }{CS681}{Piyush P Kurur}{Ramprasad Saptharishi}

\section*{Overview}

Our next goal is to do the deterministic polynomial time primality
test that was found by Manindra Agrawal, Neeraj Kayal and Nitin Saxena
in 2002. It was a remarkable achievement and it also gave an example
of how certain problems that are open for a long time can have such
beautiful, elegant solutions. 

\section{Derandomization Approaches and the Riemann Hypothesis}

We had very good randomized algorithms like the Solovay-Strassen test
which we discussed last time, or the Rabin-Miller
test.\footnote{another randomized algorithm for primality testing. We
  will see this as an assignment.} But people wanted a deterministic
algorithm for primality testing that runs in polynomial time. 

However, it was known that under some strong assumptions, the above
algorithms can be derandomized and made into a deterministic
algorithm. If the {\em Extended Riemann Hypothesis} (ERH) is true,
then we can completely remove randomness from the Solovay-Strassen
test (or the Miller-Rabin test) and make it a deterministic polynomial
time algorithm.

This conjecture is widely believe to be true and has been a very
important long-standing open problem for a long time. Infact, Clay
Mathematical Institute has a prize of a million dollars for anyone who
solves it. Of course, if someone proves the riemann hypothesis, we
already have a deterministic primality test. But trying to prove the ERH
for a primality test is like trying to uproot a whole tree to get a
fruit on top of it; completely avoiding the tree and instead using
some stick would be better.

Let us have a small discussion on what the ERH, or the Riemann
Hypothesis is. 



\subsection{The Zeta Function}
Riemann introduced a function called the {\em Riemann Zeta Function.}
The Riemann Hypothesis is to do with the roots of this function. Some
of you might have seen the equalities like 
\bea{
1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2}\cdots & =
&\frac{\pi^2}{6}\\
1 + \frac{1}{2^4} + \frac{1}{3^4} + \frac{1}{4^4}\cdots & = &
\frac{\pi^4}{90}
}

One question is what happens when we vary the exponent of the
summation. What can we say about
$$
\zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s}
$$

Note that there are diverging series like $\zeta(s)$ since 
$$
1+\frac{1}{2} + \frac{1}{3} + \cdots+ \frac{1}{n} \approx \log n
$$
and therefore if $n$ goes to infinity, the sum goes to infinity as
well; the infinite some diverges. Infact, the series $\zeta(s)$
converges for all real $s$ if and only if $|s|>1.$ 

Now why not $s$ when it is complex? Why do we restrict ourselves to
just real exponents? Then we can define $\zeta(s)$ to be the function
when $s$ is any complex number. There again, the infinite sum issues
props up. It can be shown that $\sum \frac{1}{n^s}$ converges for all
$s$ such that the real part of $s$, denoted by $\Re(s)$ is greater
than $1.$ 

And hence $\zeta(s)$ is well defined for all numbers $s$ such that
$\Re(s) > 1$. A pictorial way to look at it is if you consider the
complex plane, it is well-defined for all point to the right of the
line $x = 1.$\\

Suppose we consider functions over real values, we have this notion of
continuity. A function being continuous at a point $x_0$ essentially
means that irrespective of whether we approach $x_0$ from the left (also
denoted by left hand limit) or the right (right hand limit), the value
should coincide with the functional value at $x_0$. This is sometimes
also written as 
$$
\lim_{x\rightarrow x_0^-}f(x) = \lim_{x\rightarrow x_0^+}f(x) = f(x_0)
$$

In the case of complex functions, we have a function over a plane. So
it is not just a line and hence just left or right approaches. In the
complex case, it is said to be continuous if no matter what part you
take to approach $x_0$, the limits should match $f(x_0).$

Similarly for derivatives in the real case, we want the derivative on
the left hand side to match the derivative on the right hand side. In
the complex case, the derivative must be the same on all directions. 

A complex function that satisfies these conditions is called an {\em
  analytic function.} We say  function is analytic over a region $D$
if the above properties hold for every point in $D.$

Analytic functions have very strong properties like not just the first
derivative but all higher order derivatives exist, a whole bunch of
properties. It imposes a lot of restriction on the function. \\

Riemann showed that the zeta function is analytic in the region
$\Re(s) > 1.$ \\

Suppose we have a function $f$ that we define over a small domain $D$
over the compex plane. And over this domain, suppose the function $f$
is analytic. We haven't defined the function outside this domain at
all. A process known as {\em analytic continuation} can be used to
extend the domain of this function. 

Formally speaking, $g$ (whose domain is $D_g$) is an analytic
continuation of $f$ (whose domain is $D_f$) if the following
properties hold:
\begin{itemize}
\item $g$ is analytic over $D_g.$
\item $D_f \subseteq D_g.$
\item For all $z\in D_f$, $f(z) = g(z).$
\end{itemize}

In simple words, $g$ extends $f$ to a larger domain keeping in mind
that if $f$ was already defined at a point $z$, then $g$ shouldn't
change that value; $g$ should coincide with $f$ wherever it is defined
already. 

There is a remarkable results that if $g_1$ and $g_2$ both
analytically extend $f$ independently, then essentially $g_1 = g_2.$
Therefore, the analytic continuation of a function $f$ is uniquely
determined; we can hence talk of {\em the} analytic continuation of
$f.$\\

The zeta function is actually the analytic continuation of the
function $\sum \frac{1}{n^s}.$ It is however written as just 
$$
\zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s}
$$
The analytic continuation now extends the domain to the entire complex
plane except the point $z = 1.$ Thus our $\zeta(s)$ is a function
defined over the entire complex plane except $1$; there is a simple
pole at the point $z=1.$\\

\subsection{Why is this important?}

Now what is the importance of this function? What does it give us? The
answer is that it essentially captures the factorization of integers
in it. 

Every $n$ can be factorized as a product of primes
$p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k}$ and hence
$$
\frac{1}{n^s} = \frac{1}{(p_1^{\alpha_1})^s(p_2^{\alpha_2})^s\cdots (p_k^{\alpha_k})^s}
$$
And therefore, every term in the zeta function can be written as such
a term on the RHS. And therefore, 
$$
\sum_{n=1}^\infty \frac{1}{n^s} = \prod_{p\in
  \text{Primes}}\inparen{1+\frac{1}{p^s} + \frac{1}{p^{2s}} +\cdots }
= \prod_{p\in \text{Primes}}\inparen{\frac{1}{1 - \frac{1}{p^s}}}
$$
This is not a rigerous proof; mathematicians will stand on their head
and complain that such infinite sum/product manipulation is an
unforgivable sin. However, the final equality is true; can be made
rigerous. 

This relation between the zeta function and the prime product is one
of the many things that make it important. 

\subsection{The Riemann Hypothesis}

The analytically continued function has a lot of roots in $\C.$
Infact, it has a root at every negative integer. These are considered
as trivial roots since they provide us with no consequence. 
It has been shown that all the non-trivial zeroes lie in the critical
strip of $\setdef{s}{0<\Re(s)<1}$, the strip between the $y$-axis and
the line $x = 1.$ \\

The {\em Riemann Hypothesis} is the conjecture that all the
non-trivial zeroes of the zeta function lie on the line
$x=\frac{1}{2}.$ That is, any non-trivial root $s$ must satisfy
$\Re(s) = \frac{1}{2}.$ \\

As of now, all the roots that have been discovered lie on this
line. But we do not have a way of proving, or disproving, that all the
non-trivial roots lie on this line. 

\subsection{The Extended Riemann Hypothesis}

This is a slight generalization of the zeta function. A character
$\chi$ is a periodic\footnote{there exists a number $k$ such that
  $\chi(n+k) = \chi(n)$ for all $n$} function is
multiplicative. That is, $\chi(mn) = \chi(m)\chi(n).$ 

For example, $\chi = \legendre{a}{n}$ for a fixed $a$ is a character. 

The generalized zeta function is the analytic continuation of 
$$
L(\chi,s) = \sum_{n=1}^\infty \frac{\chi(n)}{n^s}
$$

Clearly, when $\chi(n) = 1$ for all $n$, this is just the zeta
function. \\

The Extended Riemann Hypothesis is the conjecture that for any
character $\chi$, all the non-trivial zeroes of the $L$ function lie
on the line $\Re(s) = \frac{1}{2}.$

\subsection{Derandomizing using ERH}

One of the major consequences of ERH, that is used in a lot of places
is that, for any prime $p$, the first quadratic non-residue is less
than $O(\log^2 p).$ 

So, in essence, the witness for the Solovay-Strassen test can be found
without going too far. One just need to go till up to polylog$(n)$ to
get a witness. This thus derandomizes the Solovay-Strassen test. \\

As remarked earlier, the RH or the ERH is a really strong
conjecture. One doesn't need to go as far as the ERH to get a
primality test. Agrawal-Kayal-Saxena show that primality can be solved
in deterministic polynomial time without using any high-end
mathematical machinery. The give a very elegant and simple algorithm. 

\section{The AKS Primality Test: The Idea}

All primality tests have the same form: 
\begin{enumerate}
\item Find a property such that it is satisfied only by primes. 
\item Try to verify that property
\end{enumerate}

In the Solovay-Strassen test, it was the properties of the Legendre
symbol that we checked. The following proposition is the property used
in the AKS test. 

\begin{proposition} The equation ``$(X+a)^n = X^n + a\pmod{n}$'' is
  true for all $a$ if and only if $n$ is prime.
\end{proposition}
\begin{proof}
When $n$ is a prime, then the above equation is just the binomial
theorem for prime numbers. Therefore the equation is indeed true for
all $a$ when $n$ is a prime. \\

Suppose $n$ is not a prime, then we must show that the two polynomials
$(X+a)^n$ and $X^n+a$ are not the same. Suppose $n = pq$ where $p$ is
a prime and $q\neq 1$, then look at the following binomial term:
$$
\binom{n}{p} = \frac{n\cdot (n-1)\cdot (n-2)\cdots
  (n-(p-1))}{p\cdot(p-1)\cdot(p-2)\cdots 1}
$$
Let $p^m$ be the largest power of $m$ that divided $n$. The
only term in the numerator that is divisible by $p$ is the first term
since it is a product of $p$ consecutive integers and therefore only
one of the can be divisible by $p$. Therefore the largest power or $p$
that divides the numerator is $p^m$ since it divides $n$ and none of
the other terms are divisible by $p.$ The denominator has a factor of
$p$ and therefore will cancel off one factor of $p$ from the
numerator. Therefore the largest power of $p$ that divides
$\binom{n}{p}$ is $p^{m-1}.$ And since $p^m\nmid \binom{n}{p}$,
clearly $n\nmid\binom{n}{p}$ and therefore this terms survives. 

Since $p\neq n$, this isn't the last term in the binomial
series. Therefore, the two polynomials are different. 
\end{proof}

There is one major problem here - how do we check if the two
polynomials are the same? We can compute them by repeated
squaring. But note that the polynomial $(X+a)^n$ has $n$
terms. Checking if every term other than the first and the last is
zero would take atleast $O(n)$ time adn therefore will be exponential
in the input size which is $\log n$. We are looking for an algorithm
that runs in $(\log n)^{O(1)}$ time.

The natural fix to this is computing the equation modulo a polynomial
of small degree. Instead of computing $(X+a)^n\bmod{n}$, we compute
$(X+a)^n \bmod{(X^r - 1),n}$ where $r\leq \log^cn.$ 

If $n$ was a prime, then $(X+a)^n = X^n + a \bmod{n}$ and therefore
$(X+a)^n  = X^n + a \bmod{(X^r - 1,n)}.$ The tricky part is the
converse. It is very well possible that some composite number could
satisfy this equation since we are going modulo a very small degree
polynomial. What AKS does here is try this for different $a$'s; they
check if $(X+a)^n = X^n + a \bmod{(X^r - 1,n)}$ for a fixed $r$ and
$a\in \inparen{1,2,\cdots, l}$ where $l\leq \log^{c'}n.$ They then
argue that if all these tests go through, $n$ has to be a prime or a
power of a prime. 

Checking if $n$ is a power of a prime is easy and that would conclude
the algorithm. 

\subsection{Checking if $n = p^k$ for $k\geq 2$}

Now $p\geq 2$ and therefore $k$ can be at most $\log n.$ Suppose we
fix a $k$ and want to find if $n = p^k$ for some $p$, how do we do
that? Just a binary search. 

Try $(n/2)^k$. If this is equal to $n$, you already got it. If it is
less than $n$, then you know that if an $a$ exists, it must be greater
than $n/2.$ Recurse on that half. 

Thus, the number of steps is $\log^2n,$ which is polynomial in the
input size. \\

We need to understand some properties of {\em cyclotomic extensions}
over finite fields. We shall do those next time and that should fix
the choice of $l$, $r$ and other constants that come in the process. 








