\lecture{9: Uniqueness of $\F_{p^m}$}{CS681}{Piyush P Kurur}{Ramprasad Saptharishi}

Last lecture we closed with two questions of existence and uniqueness
of fields of order $p^m.$ This lecture, we shall the question of
uniqueness and also understand some other properties of extensions. 

\section{Some More Properties of Field Extensions}

Firsly, we need to understand what quotienting means. Suppose we have
a field $K$ and we look at the polynomial ring over this field -
$K[X].$ Let us take some irreducible polynomial $f$ and look at
$K[X]/(f(X)).$ What does this mean?

Quotienting means that you consider all occurences of the ideal
$(f(X))$ as zero. So in particular, if you look at $f(X)$ in this
ring, it would be zero. Hence the variable $X$ can now be thought of
as a root of $f.$

Infact, this is exactly what we do to adjoin roots. Suppose we have a
field $K$ and we need to adjoin some element $\alpha.$ We take the
minimum polynomial $f$ of $\alpha$ and we look at $K[X]/(f(X)).$ Here,
essentially, $X$ is $\alpha.$

This is how we built $\C.$ We looked at $\R[X]$ and quotiented it with
the ideal generated by $X^2 + 1.$ And now note that there is no real
distinction between $-i$ or $i.$ Algebraically, $\R(i) \iso \R(-i) =
\C.$ So the general point to note is that if $\alpha$ and $\beta$ are
both roots of the same irreducible polynomial $f$, then $K(\alpha) \iso
K(\beta) \iso K[X]/(f(X)).$

\subsection{Splitting Fields}

Until we knew that an object called $\C$ existed, we had no idea if
there was an element $i$ such that $i^2 = -1.$ So as such, it doesn't
make sense to adjoin some element until you know where it is
from. When you look at $K[X]/(f(X))$, we said that $X$ can be
identified with a root of $f$, but what root? Where is this root? I
know it's not in $K$ (for if it were, $X-\alpha$ is a factor of $f$
and hence can't be irreducible), but where else is it?

This is where splitting fields come in. 

\begin{definition}
A splitting field of a polynomial $f$ over $K$ is the smallest field
extension $E/K$ that contains all the roots of $f.$

Or equivalently, it's the smallest field $E$  where the polynomial $f$
factorizes into linear factors. 
\end{definition}

The first thing to note is that this is not equivalent to adjoining a
root. To illustrate the difference, take the field $\Q$ and let us
look at the polynomial $X^3 - 2.$ This polynomial is irreducible and
hence we can talk of the field $\Q[X]/(X^3 - 2)$ and identify the
element $X$ with a root of the polynomial say $\sqrt[3]{2}.$ 

But note that this is {\em not} the splitting field. The polynomial
has other roots, namely $\sqrt[3]{2}\omega,\sqrt[3]{2}\omega^2$ where
$\omega$ is a primitive cube root of unity. And it is clear that
$\Q(\sqrt[3]{2})$ is a subfield of the reals and obviously cannot
contain the complex number $\sqrt[3]{2}\omega$ and hence cannot be the
splitting field of the polynomial. 

The splitting field of this polynomial is $\Q(\sqrt[3]{2},\omega)$ and
is a degree $6$ extension over $\Q$ whereas the extension by adjoining
roots are degree $3$ extensions over $\Q.$\\

An important theorem is the following:

\begin{theorem}
If $E$ and $E'$ are two splitting fields of a polynomial $f(X)$ over
$K$, then $E\iso E'.$
\end{theorem}

We omit the proof, but nevertheless the theorem is important and will
be used. Interested readers can refer any abstract algebra books for
the proof of this fact. 

\subsection{The Frobenius Map}

The binomial theorem takes a very simple form over $\F_p.$ We know that
$$
(X+Y)^p = \sum_{i=0}^p \binom{p}{i} x^i y^{p-i}
$$

\begin{lemma}
For $i\neq 0,p$, the coefficient $\binom{p}{i}$ is divisible by $p.$
\end{lemma}
\begin{proof}
The proof is quite obvious. The coeffient is 
$$
\binom{p}{i} = \frac{p\cdot(p-1)\cdots (p-i+1)}{1\cdot 2\cdots i}
$$
and the numerator has a factor of $p$ and the denominator does
not. And since this si an integer, the factor of $p$ will remain
uncancelled and hence will be divisible by $p.$
\end{proof}

This then tells us that in the field $\F_p$,
$$
(X+Y)^p = X^p + Y^p
$$

If you look at it as a automorphism (an isomorphism of $\F_p$ into
itself), this shows the map $x\mapsto x^p$ is an automorphism of
$\F_p.$

This map $\phi(x) = x^p$ is called the Frobenius map. It is a very
important automorphism of finite fields and appears almost
everywhere. 

\subsection{The Multiplicative Group of a Finite Field is Cyclic}

Let $K$ be a finite field of $p^m$ elements. We shall show now that
the multiplicative group $K\setminus\inbrace{0}$ is a cyclic group. 

Infact, we shall show a much stronger theorem.

\begin{theorem}
Let $K$ be any field (not necessarily finite) and let $A$ be any
finite subgroup of $K^\star = K \setminus\inbrace{0}.$ Then the subgroup
$A$ is cyclic. (That is, there exists an element $a\in A$ such that
every other element in $A$ is a power of $a$)
\end{theorem}
\begin{proof}
Let $|A| = M.$ We need to show that there exists an element in $A$ of
order $M.$ Suppose not, let $a$ be the element of highest order in
$A.$ Let the order of $a$ be $m < M.$

Now pick any $x\in A$. We claim that the order of $x$ must divide $M.$
To prove this, let the order of $x$ be $n.$ Let the gcd of $m$ and $n$
be $d.$

By Euclid's lemma, there exists integers $p,q$ such that $pn + qm =
d.$ Thus, let us look at the element $g = x^pa^q.$ Then $g^m =
x^{pm}a^{qm} = a^{qm} = d$ since $qm = d \bmod{n}.$ And similarly $g^n
= x^d.$ Thus, the order of $g$ is infact equal to $\frac{mn}{d} =
lcm(m,n).$ And since we assumed that $a$ was an element of maximum
order, the lcm of $m$ and $n$ has to be $m$ and therefore $n$ has to
divide $m.$\\

Having established this, we see that the order of every element must
divide $m$ and therefore $x^m = 1$ for every $x\in A.$ And therefore,
the polynomial $X^m - 1$ has every element of $A$ as a root. But if we
were to assume that $m<M$, then a polynomial of degree $m$ would have
$M>m$ roots! And this clearly cannot happen in a field. And therefore,
$m = M$ and hence the group $A$ is cyclic. 
\end{proof}


\section{Uniqueness of $\F_{p^m}$}

Now we come to the proof that every field of $p^m$ elements are
isomorphic. So essentially, there is only $1$ field of size $p^m$ and
hence would make sense to refer to {\em the} field of size $p^m$ as
$\F_{p^m}.$To avoid cluttering of subscripts and superscripts, let $q
= p^m.$ We need to show that any two fields of size $q$ are
isomorphic.

Now let $F$ be any field of order $q.$ Then every element $a\in F$
satisfies the property that $a^q = a.$ Hence in particular, every
element of the field $F$ is a root of $X^q - X.$ And therefore,
considering this polynomial over $\F_p$, $F$ is a splitting field of
$X^q - X$ over $\F_p.$ And by a theorem stated earlier, all splitting
fields of a polynomial are isomorphic. And therefore, any two fields
of order $q$ are isomorphic. 

\section{More about $X^q - X$}

Here is a very important property of this polynomial $X^q - X.$

\begin{theorem}
Let $Irr(K,d)$ be the set of all irreducible polynomials of degree $d$
over $K.$ Then, the following equation holds in $\F_p[X]:$
$$
X^{p^m} - X = \prod_{\substack{f\in \text{Irr}(\F_p,d)\\d\mid m}} f(X)
$$
\end{theorem}
\begin{proof}
We shall show that the LHS is equal to the RHS by comparing the roots
on both sides. For the roots to first exist, we shall go to some large
field. 

Firstly, note the polynomial $X^{q} - X$ is satisfied by every
element of $\F_q.$ And by the same degree argument, the roots of
the polynomial is precisely all the elements of $\F_q.$ We shall
first show that every element $\alpha \in \F_q$ is also a root of
the RHS. 

Since $\alpha \in \F_q$, pick up the minimum polynomial $f(X)$ of
$\alpha$ over $\F_p.$ We have seen earlier that this polynomial must
be irreducible. Hence $\F_p[X]/(f(X))$ corresponds to the field
$\F_p(\alpha).$ Let the degree of $f$ be $d.$

We know that 
\bea{
\insquar{\F_q:\F_p} & = &
\insquar{\F_{q}:\F_p(\alpha)}\insquar{\F_p(\alpha):\F_p}\\
\implies m & = & \insquar{\F_q:\F_p(\alpha)}\cdot d\\
\implies d\mid m & & }

And hence, since the degree of this irreducible polynomial divides
$m$, it appears in the product of the RHS as well. \\

The other way is easy too. Pick up any factor $f(X)$ on the RHS. Let
its degree be $d.$ Let $\alpha$ be one of the roots. Then we know that
$\F_p(\alpha)$ must be the field $\F_{p^d}.$ But since $d$ divides
$m$, this is a subfield of $\F_q.$ And therefore every element of
$\F_{p^d}$, in particular $\alpha$, must be in $\F_q$ as well.
\end{proof}

\subsection{Extracting Factors}

The formula outlined in the previous section is extremely useful in
factoring. It helps us pull out factors of the same degree. We shall
see a quick sketch here and discuss this in detail next class. 

Let us say we have some polynomial $f$ over $\F_p.$ How do we extract
all linear factors over $\F_p$? The idea is pretty simple. We know that
$X^p - X$ splits as linear factors over $\F_p.$ And more over, the
formal derivative of it is $pX^{p-1} - 1 = -1 \neq 0$ and therefore it
has no repeated roots as well. 

Then, we have 
$$
g(X) = gcd(f,X^p - X)
$$

Then if $\alpha \in \F_p$ was any root of $f$, then clearly since
$\alpha$ also satisfies $X^p - X$, $\alpha$ will be a root of $g$ as
well. And more importantly, since $X^p - X$ is has no repeated roots,
$g$ will retain the same property as well. Hence every root of $f$
over $\F_p$ appears exactly once in $g.$

Having removed all degree $1$ factors, we can all extract degree $2$
factors by taking the gcd with $X^{p^2} - X.$\\

We shall discuss this idea of factoring in detail soon.

