\lecture{7: Towards Factorization over Finite Fields}{CS681}{Piyush P Kurur}{Ramprasad Saptharishi}

\section*{Overview}

We shall slowly move into factorization of univariate polynomials
(polynomials with just one variable) over finite fields. We are given
a finite field $K$ and a polynomial over one variable $X$ whose
coefficients are from $K.$ We are to find the factorization of this
polynomial into irreducible factors. 

Before we get into this question, we need to first understand if it
even makes sense. How can we be sure that such a factorization exists?
And even if it did, how do we know if it is unique? 

We shall first answer a lot of questions in the algebra related to it
before going to factorization as such. 

\section{Rings, Ideals, Factorization etc.}

We know that integers can be uniquely factorized into product of prime
powers. However, not all rings are as well-behaved as the integers
are. We first need to ask if the algebraic structure has this property
of unique factorization. Let us look at an example where this fails. 

Look at the set of integers modulo $8$. This is called $\Z_8$ and we
know that this forms a ring. Suppose we look at polynomials over this
ring, polynomials of a single variable $X$ whose coefficients come
from $\Z_8,$ does this ring have the property of unique factorization?
Here is a counter example in $\Z_8[X].$
$$
X^2 - 1 = (X-1)(X+1) = (X-3)(X+3) 
$$

But $\Z_8$ is a bad ring, in the sense that non-zero elements can
multiply to give $0$ ($2\times 4 = 0$ here). As for another example,
look at the set of all number of the form $a+b\sqrt{-5}$ where $a,b\in
\Z.$ This forms a ring and over this ring $6 = 2\cdot 3 =
(1+\sqrt{-5})(1 - \sqrt{-5}).$

Hence it's not always true that factorization is unique. 
However, fortunately for us, we have unique factorization over $K[X]$
whenever $K$ is a field. 

\begin{definition}
A ring is said to be an integral domain if and only if there are no
  non-trivial zero divisors. That is, if $a,b\in R$ such that $ab=R$,
  then either $a=0$ or $b=0.$
\end{definition}

The ring $\Z$ is an integral domain but the ring of integers modulo
$6$ is not (since $2\times 3 = 0$ over the ring). 

Inorder to define factorization, we need a notion of primes over
arbitrary rings. Let us first look at the definition of primes over
integers. Let us first look at the wrong definition. \\

\noindent
A number $p$ is said to be prime if for all $a$ that divides $p$,
either $a=1$ or $a=p.$\\

This translates to the ring definition of a maximal ideal and not a
prime ideal. 

This is the common definition in school but generalization based on
this is erraneous. Though it happens to correct over the set of
integers, it is not true in general. Here is the right definition. 

\begin{definition}
A number $p$ is said to be a prime if and only if for all $a,b$ such
that $p$ divides $ab$, either $p$ divides $a$ or $p$ divides $b.$
\end{definition}

Thus this gives the definition of prime ideals in the setting of
rings. 

\begin{definition}
An ideal $\mathfrak{a} \subseteq R$ is said to be prime if and only if
for all $a,b\in R$ such that $ab\in \mathfrak{a}$, either $a\in
\mathfrak{a}$ or $b\in \mathfrak{a}.$

Any element $p\in R$ that generates a prime ideal is called a prime
element of $R.$
\end{definition}

\begin{definition}
An ideal $\mathfrak{a} \subseteq R$ is said to be maximal if and only
if for every ideal $\mathfrak{a'}\supseteq \mathfrak{a}$, either
$\mathfrak{a'} = 1R = R$ or $\mathfrak{a'} = \mathfrak{a}.$
\end{definition}

This basically means that no proper ideal of $R$ properly contains
$\mathfrak{a}.$ Note that not all prime ideals are maximal. We were
just lucky that this was true on $\Z$ and hence both definitions of
prime numbers were equivalent. This is not true over arbitrary rings. 

\begin{definition}
An ideal $\mathfrak{a} \subseteq R$ is said to be a principle ideal if
the ideal is generated by a single element. That is, $\mathfrak{a} =
aR$ for some $a\in R.$
\end{definition}

\begin{definition}
An integral domain $R$ is said to be a 
\begin{itemize}
\item principle ideal domain (PID) if every ideal in it is principle (every
  ideal is generated by a single element).
\item unique factorization domain (UFD) if every element can be
  uniquely factorized in to product of prime elements of the ring.
\end{itemize}
\end{definition}

We already saw an example of a ring (and a domain) that was not a
UFD. Here is an example of a ring that is not a PID. Consider a field
$K$ and look at the ring of polynomials on two variables $X,Y$ over
this field. This is denoted by $K[X,Y].$

In this field, look at the ideal generated by $X$ and $Y.$ That is,
the set of polynomials of the form $Xf(X,Y) + Yg(X,Y)$, those
polynomials that do not have a constant term. This is clearly an ideal
but this isn't principle. 

A similar example is over $\Z[X]$ and the ideal being $(p,X)$ where
$p$ is any prime number.

\begin{fact}
For any field $K$, $K[X]$ is a PID.
\end{fact}

\begin{fact}
Any PID is also a UFD
\end{fact}

The two facts together tell us that we can indeed talk of
factorization of polynomials in $K[X].$ Another useful fact is the
following, and this helps us see that factorization makes sense even
on multivariate polynomials. 

\begin{fact}
If $R$ is a UFD, so is $R[X].$
\end{fact}

The following theorems are very useful. 

\begin{theorem}
A ring $R$ is a field if and only if the only ideals of $R$ are the
$0$ ideal and the whole ring $R$.
\end{theorem}
\begin{proof}
First we shall show that a field has no non-trivial ideals. Suppose
The field had some ideal $I$ that contained some element $x\neq
0$. Since it is a field, the inverse of $x$ exists. Since $I$ is an
ideal and $x\in I$ would mean that $xa \in I$ for all $a\in R$ and in
particular $xx^{-1} = 1 \in I.$ But if $1\in I$, then
for every element $a$ in the field, $1a \in I$ which would then force
$I$ to be the entire field. \\

As for the other direction, suppose the ring $R$ was not a field. We
want to show that there exists some non-trivial ideal in this
ring. Since we assumed that it isn't a field, there must be some
non-zero element $a$ whose inverse does not exist. Look at the ideal
generated by it, $aR.$ This ideal certainly contains $a$ and it cannot
$1$ since if it did, it would mean that $a$ is invertible. And hence
this is an ideal that is non-zero and also not the whole of $R$; a
non-trivial ideal. 
\end{proof}

\begin{theorem}
For any ring $R$
\begin{enumerate}
\item if an ideal $\mathfrak{a}$ is prime, then $R/\mathfrak{a}$ is an
  integral domain.
\item if an ideal $\mathfrak{a}$ is maximal, then $R/\mathfrak{a}$ is
  a field. 
\end{enumerate}
\end{theorem}
\begin{proof}
We have to show that if $\mathfrak{a}$ is prime, then $R/\mathfrak{a}$
is an integral domain. Suppose not, then there exists two non-zero
elements $a,b$ such that $ab=0$ in $R/\mathfrak{a}.$ This means that
$a \bmod{\mathfrak{a}} \neq 0$ and $a\bmod{\mathfrak{a}} \neq 0$ but
$ab\bmod{\mathfrak{a}} = 0$ or in other words $ab \in \mathfrak{a}$
but neither $a$ nor $b$ belongs to $\mathfrak{a}.$ This contradicts
the assumption that $\mathfrak{a}$ and hence $R$ has to be an integral
domain.\\ 

As for the case when $\mathfrak{a}$ is maximal, assume that
$R/\mathfrak{a}$ is not a field. Then there exists some non-zero
element that is not invertible. Look at the ideal generated by this
element. As in the earlier theorem, this is a non-trivial ideal
(neither $0$ nor the entire ring). But in the map from $R$ to
$R/\mathfrak{a}$, ideals of $R/\mathfrak{a}$ corresponds to ideals in
$R$ that contain $\mathfrak{a}.$ Since we just found a non-trivial
ideal in $R/\mathfrak{a}$, this would translate to a non-trivial ideal
in $R$ that properly contains $\mathfrak{a}$ thus contradicting the
maximality of $\mathfrak{a}.$ Thus $R$ has to be a field. 
\end{proof}

\subsection{Some Insights}

This is not completely a part of the course but it would be useful to
know this to understand factorization. In any ring, we can talk of a
tower of prime ideals. What this means is a series of the form $0
\subseteq I_1 \subseteq I_2 \subseteq \cdots \subseteq I_n \subseteq
R$ such that each ideal $I_j$ is a prime ideal. The number $n$ is
called the Krull Dimension of the ring $R.$

The Krull Dimension is actually a local property but for it is well
defined for rings like $K[X_1,X_2,\cdots,X_n]$ (where $K$ is a field)
and $\Z[X].$

If we were to look at $K[X,Y]$, we have the tower $0\subseteq (X)
\subseteq (X,Y) \subseteq K[X,Y].$ The krull dimension of this ring is $2.$
Similarly the ring of polynomials on $n$ variables over a field $K$
will have a krull dimension of $n.$

And the ring $\Z[X]$ has the tower $0 \leq (p) \leq (p,X) \leq \Z[X]$
and hence has krull dimension $2.$ We shall see soon that
factorization of polynomials in $\Z[X]$ is so similar to factorization
of polynomials in $K[X,Y].$ \\

We need to understand the concept of finite fields, extensions, etc
before we get into factorization. We shall first spend some time on
this. 

\section{Finite Fields}

We shall be studying properties of fields that have finite number of
elements in them. A few things to keep in mind, we shall prove them
soon, is that any finite field has its cardinality to be a power of
prime. There cannot exist a finite field whose cardinality is
divisible by two distinct primes. And infact, for any prime $p$ and
$\alpha$, there is exactly one field of size $p^\alpha.$ (and note
that this isn't true on the infinite setting. $\R$ and $\C$ both
have infinite number of elements but are clearly different)

\begin{definition}
A field $E$ is called an extension of a field $K$ if $E$ is a field
that contains $K.$ This (also) is denoted by $E/K.$ 
\end{definition}

There is a notion of a degree of a field extension but one needs to be
familiar with vector spaces to completely understand this. We shall
dwell a little on it. 

\subsection{Vector Spaces}

\begin{definition}
A vector space $V$ over a field $K$, with an additive structure and
multiplication by elements of $K$ (scalars), satisfies the following
conditions:
\begin{itemize}
\item $(V,+)$ is an additive abelian (commutative) group (additive closure, inverse, identity)
\item For any vector $v\in V$ and scalar $\alpha \in K$, the element
  $\alpha v$ is also a vector.
\item For any vectors $u,v\in V$ and scalar $\alpha \in K$, we have
  $\alpha(u+v) = \alpha u + \alpha v.$ 
\item For any vector $u$ and scalars $\alpha,\beta \in K$, we have $(\alpha
  + \beta)u = \alpha u + \beta u$ and $\alpha(\beta u) = (\alpha
  \beta)u.$
\end{itemize}
\end{definition}

Let us look at a few examples to get ourself familiar with this
notion. $\C$ forms a vector space over $\R.$ Clearly the above
properties are satisfied. 

Another example is the plane $\R^2$, set of point $(x,y)$ where both
coordinates are from the reals. Scalar multiplication is defined as
$\alpha(x,y) = (\alpha x, \alpha y).$

Another example is the ring of polynomials $K[X]$ over $K$ where $K$
is a field. Scalar multiplication is just multiplying every
coefficient by the scalar. \\

Next we need a notion of linear independance. 

\begin{definition}
A set $\inbrace{v_1, v_2, \cdots, v_k}$ is said to be linearly
independant if the only way 
$$
c_1v_1 + c_2v_2 + \cdots + c_kv_k = 0
$$
can happen for scalars $c_i$ is when all the $c_i$'s are zero
themselves. That is, no non-trival linearl combination of these
vectors is zero. 
\end{definition}

For example, let us look at each of our examples stated and find a
linearly independant set. In $\C$ over $\R$, look at the set
$\inbrace{3,2+i}.$ Suppose $c_1(3) + c_2(2+i) = 0$, then $(3c_1 +
2c_2) + c_2i = 0$ and this is possible only when both $c_1$ and $c_2$
are zero. Hence the set is linearly independant. 

And again, look at the set $\inbrace{(1,0),(0,1)}$ in $\R^2.$ This
again is linearly independant since the only way $c_1(1,0) + c_2(0,1)
= (c_1,c_2) = (0,0)$ is when both $c_1$ and $c_2$ are zero.

In the third example, look at the set $\inbrace{1,X,X^2}.$ $c_1 + c_2X
+ c_3X^2$ can be the zero polynomial if and only if all the $c_i$'s
are zero. \\

This is the notion of linear independance. With a little bit of
thought, any vector that can be represented as a linear sum from such a
set is infact uniquely represented so. 

For example, let us assume that $\inbrace{v_1, v_2,\cdots, v_k}$ was a
linearly independant set. Let $v = c_1v_1 + c_2v_2 + \cdots + c_kv_k.$
Suppose this could be represented as a linear sum in a different way,
we shall obtain a contradiction.
\bea{
v & = & c_1v_1 + c_2v_2 + \cdots + c_kv_k\\
& = & c_1'v_1 + c_2'v_2 + \cdots + c_k'v_k\\
\implies 0 &= &(c_1 - c_1')v_1 + \cdots + (c_k - c_k')v_k
}
And if the two representations were indeed different, there is atleast
one $i$ such that $c_i \neq c_i' \implies (c_i - c_i')\neq 0$ but this
would give a non-trivial linaer combination of the $v_i$'s to become
zero. This contradicts our assumption that they were linearly
independant. Hence such linear representations are unique. \\

An example is that every point $(x,y)$ can be represented uniquely as
a linear sum of $(1,0)$ and $(0,1)$ (it is just $x(1,0) +
y(0,1)$). The students are encouraged to also check it for $\C$ with
our linearly independant set being $\inbrace{3,2+i}.$

Let us look at our example of $K[X]$. We saw that $\inbrace{1,X,X^2}$
was a linearly independant subset but the term $X^5$ can never be
written as a linear sum of $1,X,X^2.$ Thus, the set
$\inbrace{1,X,X^2}$ doesn't cover or {\em span} $X^5.$ Since $X^5$ is
not spanned by the set $\inbrace{1,X,X^2}$, we can add it to the set
and it would still be linearly independant. 

We can keep adding elements to our linearly independant set in this
way by picking up some vector that is not spanned by it and adding
it. This process can go on indefinitely as well. For the moment let us
look at the case where this process stops after finite number of
steps. Now we have a set that is linearly independant and it also
spans the entire space. 

An example would be to look at $\C$ over $\R.$ Start with $3$. The
linear span of this is just elements of the form $3c$ where $c$ is a
real number. Hence it does not span elements like $2+i.$ Hence we can
add $2+i$ to this set and still have a linearly independant set. Now
this set $\inbrace{3,2+i}$ is linearly independant and also spans the
entire space. Any complex number $a+ib$ is equal to $b(2+i) + \frac{a
  - 2b}{3}3.$

Such a set that spans the space and also is linearly independant is
called a {\em basis} of the vector space $V$ over $K$. And every
vector in the vector space can be expressed as a linear combination of
the basis elements, and uniquely so. 

The number of basis elements is called the dimension of the vector
space. But wait, how do we know that every basis will have the same
number of elements? Is it possible that I can find three complex
numbers that are linearly independant over $\R$ and span $\C$? The
answer is no. It is not so hard to see that all basis must have the
same number of elements. Thus the dimension of the vector space is
independant of the choice of basis is hence well-defined.\\

The vector space $K[X]$ over $K$ has infinite dimension and its basis
could be chosen as $\inbrace{1,X,X^2,\cdots}.$ And a polynomial, say
$80 + 2X + 0X^3 + 3X^4$ can be represented by the tuple
$(80,2,0,3,0,0,0,\cdots)$ and every polynomial has a corresponding
tuple. 

Suppose we choose $\inbrace{1,i}$ as a basis for $\C$ over $\R$, then
every element $a+bi$ can be expressed as $a(1) + b(i).$ Now we can
represent the number $a+bi$ as the tuple $(a,b).$

So essentially, a vector space $V$ over $K$ is one where each element
in it can be represented as a tuple, whose entries come from $K.$ The
arity of the tuple is the dimension of the vector space.

Thus, in the finite setting, if $V$ is a finite dimensional (say
$d$-dimensional) vector space over a finite field $K$, then the number
of elements of $V$ is $|K|^d.$ This is clear since it is just the
number of $d$-tuples whose entries come from $K.$

\subsection{Field Extensions}

Let $E/K$ be a field extension. This just means that both $E$ and $K$
are fields and that $E$ contains $K.$ 

Now observe that for all $\alpha,\beta \in K$ and $u,v\in E$,
$\alpha(u+v) = \alpha u + \alpha v$ and $(\alpha + \beta) u = \alpha u
+ \beta u$ etc. Thus all the conditions to call this a vector space
hold. Thus, we can think of $E$ as a vector space over $K.$ 

An example of this we have seen already. $\C$ is a field that contains
$\R.$ And $\C$ actually is a vector space over $\R.$ Another example
would be to look at 
$$
\Q[\sqrt{2}] = \setdef{a+b\sqrt{2}}{a,b\in \Q}
$$
It is easy to check that this is a field and this clearly contains
$\Q.$ And this also naturally forms a vector space over $\Q.$ 

\begin{definition}
The dimension of $E$ as a vector space over $K$ is called the degree
of the extension $E/K$. This is denoted by $[E:K].$
\end{definition}

$\C$ over $\R$ is a 2-dimensional extension. $\R$ over $\Q$ is an
infinite dimensional extension. $\Q[\sqrt{2}]$ over $\Q$ is a $2$
dimensional extension. 

\subsubsection*{Adjoining Elements: An informal discussion}

The field $\C$ is just taking $\R$ and adding the element $i$ to
it. Once we add $i$ to $\R$, we just take all possible linear
combinations, products, inverses to make it a field. We let the set
$\R \cup {i}$ grow into the smallest field containing $\R$ and $i.$
This is formally referred to as $\R(i)$, the field got by adjoining
$i$ to $\R.$

It is easy to check that $\Q(\sqrt{2})$ is infact $\Q[\sqrt{2}] =
\setdef{a+b\sqrt{2}}{a,b\in Q}.$ And similarly one can also check that
$\Q(\sqrt[3]{2}) = \setdef{a + b\sqrt[3]{2} + c\sqrt[3]{2^2}}{a,b,c\in
  \Q}.$ From this is it easily seen that $\Q(\sqrt[3]{2})$ is a degree
$3$ extension over $\Q.$

Given such an adjointed field extension, is it easy to find out the
degree? The answer is yes. All we need to do is choose an easy basis
for the vector space. For example, let us look again at
$\Q(\sqrt[3]{2}).$ Let  $\alpha = \sqrt[3]{2}.$ We want the degree of
the extension $\Q(\alpha)/\Q.$ Now consider the set
$\inbrace{1,\alpha, \alpha^2, \alpha^3,\cdots...}.$ When does this
fail to be a linearly independant subset? We know that $\alpha^3 - 2 =
0$ and hence it loses its linear independance after $\alpha^3.$ This
is because $\alpha$ was a root of $X^3 - 2$, a degree $3$ polynomial
over the $\Q.$

Instead if we were to look any $\alpha$, any equation of linear
dependance would look like $a_0 + a_1\alpha + a_2\alpha^2 + \cdots a_k\alpha^k = 0$
and this would just mean that $\alpha$ is a root of the polynomial
$a_0 + a_1X + a_2X^2 + \cdots a_kX^k = 0.$ Thus, the degree of such an
extension $\Q(\alpha)/\Q$ is just the degree of the smallest degree
polynomial of which $\alpha$ is a root.

$\C = \R(i)$ and $i$ has $X^2 + 1$ as its minimum polynomial and thus
$[\C:\R] = 2.$ If we were to look at $\Q(\pi)$, $\pi$ is not a root of
any polynomial with coefficients in $\Q$ (this is also referred as
'$\pi$ is transcendental'). Thus the set
$\inbrace{1,\pi,\pi^2,\cdots}$ would be an infinite linearly
independant subset. And hence the extension $\Q(\pi)$ over $\Q$ is of
infinite degree. \\


We shall look at more properties of finite fields and extensions next
time. 

