## Finite and Cyclotomic Fields

### Finite fields are perfect

Lemma: Let $F$ be a finite field. Then every element of $F$ is a $p^{th}$ power.

Proof: The map $\phi(x)=x^{p}$ is a field homomorphism from $F$ to itself. Since it is injective and $F$ is finite, it is surjective.

Lemma: Every irreducible polynomial over $F$ is separable.

Proof: If $f(x)$ is irreducible and inseparable, then $f’(x)=0$ so $f(x)=g(x^p)$. But then $f(x)=g(x)^p$, contradicting irreducibility.

### Existence and uniqueness of finite fields

Proposition: Let $p$ be a prime. Then there is a unique (up to isomorphism) finite field with $p^n$ elements for every $n\ge 1$.

Proof: If $F$ is a finite field of characteristic $p$, it is a finite dimensional vector space over $F_p$ so has $p^{n}$ elements where $n=[F:F_p]$. Consider the splitting field $F$ of the polynomial $x^{p^{n}}-x$ over $F_p$. It is separable since its derivative is $-1$. Thus it has $p^{n}$ distinct roots. Notice that, if $\alpha$ and $\beta$ are roots of this polynomial, so are $\alpha\beta$, $\alpha+\beta$, and $\alpha^{-1}$. Thus the $p^{n}$ roots of the polynomial form a field. Thus $F$ is exactly this set of $p^{n}$ roots. Finally, let $F$ be any finite field of characteristic $p$ with $p^{n}$ elements. The nonzero elements of $F$ satisfy $x^{p^{n}-1}-1=0$ since $F^{*}$ is a finite abelian group with $p^{n}-1$ elements. Therefore (including zero) the elements of $F$ are the roots of $x^{p^{n}}-x$ so $F$ is the splitting field of this polynomial. Since splitting fields are unique, all finite fields of order $p^{n}$ are isomorphic.

We commonly write $F_{p^{d}}$ or $\mathbb{F}_{p^{d}}$ for this unique field with $p^{d}$ elements.

### Multiplicative groups of finite fields are cyclic

Suppose $F$ has $p^{n}$ elements. Suppose that $d|(p^{n}-1)$ so $x^{d}-1$ divides $x^{p^{n}-1}-1$. If $F^{\times}$ has an element of order $d$, it generates a cyclic subgroup of order $d$, and the elements of that cyclic subgroup are all roots of $x^{d}-1$. In this case the number of elements of order $d$ is $\phi(d)$. If $\psi(d)$ is the number of elements of order $d$, then we see that $\psi(d)$ is either $\phi(d)$ or zero, and in particular is at most $\phi(d)$.

Now by Lagrange’s Theorem

$p^{n}-1=\sum_{d|(p^{n}-1)} \psi(d).$

On the other hand, by counting the elements of order $d$ for each divisor of $p^{n}-1$ in a cyclic group of order $p^{n}-1$, we have

$p^{n}-1=\sum_{d|(p^{n}-1)} \phi(d).$

We conclude that $\psi(d)=\phi(d)$ for all $d$, so $\phi(p^{n}-1)\ge 1$.

### Counting irreducible polynomials mod $p$

How many irreducible polynomials of degree $d$ are there over $F_p$?

Given such a polynomial, you get $d$ elements of the field $F_{p^d}$, all of degree $d$ over $F$.

Conversely, given an element of degree $d$ over $F$ in $F_{p^{d}}$, you get an irreducible polynomial; but there are $d$ elements that give the same polynomial.

So the number of irreducible polynomials is

$\frac{\|\lbrace x\in F_{p^{d}} : F(x)=F_{p^{d}}\rbrace\|}{d}$

If $d$ is prime, then an element of $F_{p^{d}}$ is either of degree one or $d$. There are $p$ elements of degree 1, and $p^{d}-p$ of degree $d$. So the number of irreducible polynomials of prime degree $d$ is $(p^{d}-p)/d$.

For example, if $p=2$ and $d=5$, there are $6$ irreducible polynomials of degree $5$. They are irreducible factors of $x^{32}-x$ mod $2$. According to wolfram alpha, they are:

$\begin{matrix} x^5+x^2+1 & x^5+x^3+1 \\ x^5+x^3+x^2+x+1 & x^5+x^4+x^2+x+1 \\ x^5+x^4+x^3+x+1 & x^5+x^4+x^3+x^2+1\\ \end{matrix}$

### Cyclotomic Polynomials and roots of unity

We let $\mu_n$ denote the complex roots of the polynomial $x^n-1$. These are called the $n^{th}$ roots of unity. If $\zeta\in\mu_n$, then

$\zeta=e^{2\pi i a/n}$

for some integer $a$.

The set $\mu_n$ is in fact a cyclic group of order $n$. Its generators are called the primitive $n^{th}$ roots of unity. If $\zeta\in \mu_n$ is a primitive root of unity then

$\zeta=e^{2\pi i a/n}$

where $(a,n)=1$.

### The cyclotomic polynomials

The cyclotomic polynomial $\Phi_{n}(x)$ is the polynomial whose roots are the primitive $n^{th}$ roots of unity.

Lemma: For all $n\ge 1$, the degree of $\Phi_{n}(x)$ is $\phi(n)$, and

$x^{n}-1=\prod_{d|n} \Phi_{d}(x)$

This is because there are $\phi(n)$ primitive roots of unity in $\mu_{n}$ and every $n^{th}$ root of unity is primitive of order $d$ for some $d\vert n$.

### Cyclotomic polynomials have integer coefficients

Lemma: $\Phi_{n}(x)$ is monic and belongs to $\Z[x]$.

Proof: The factorization of $x^{n}-1$ in terms of $\Phi_{d}(x)$ gives a recursive algorithm for computing the $\Phi_{d}$. Clearly $\Phi_{1}(x)=x-1$ belongs to $\Z[x]$. Suppose that $\Phi_{d}(x)$ is monic and belongs to $\Z[x]$ for all $d<n$. Then $x^{n}-1=f(x)\Phi_{n}(x)$ where $f(x)$ is monic with integer coefficients. Then $(x^{n}-1)/f(x)$ is monic with integer coefficients by polynomial division (or by Gauss’s lemma if you want to be fancier).

### The cyclotomic polynomials are irreducible

Theorem: The polynomials $\Phi_{n}(x)$ are irreducible.

Proof: We use results about the reduction mod $p$ of $\Phi_{n}(x)$; in some sense this is a number theoretic result.

Suppose that $\Phi_{n}(x)=f(x)g(x)$ where $f(x)$ is irreducible. Let $\zeta$ be a root of $f(x)$. Choose a prime $p$ not dividing $n$. Then $\zeta^p$ is again a primitive $n^{th}$ root of unity, and therefore a root of either $f(x)$ or $g(x)$. Suppose it’s a root of $g(x)$. Then since $g(\zeta^p)=0$ it follows that $\zeta$ is a root of $g(x^p)$. Now $f(x)$ is irreducible, so it is the minimal polynomial for $\zeta$, and therefore $f(x)$ divides $g(x^p)$:

$g(x^p)=f(x)h(x)$

Reduce this equation modulo $p$, and we have

$\overline{g}(x^p)=\overline{g}(x)^p=\overline{f}(x)\overline{h}(x).$

Then $\overline{f}(x)$ divides $\overline{g}(x)^p$ which means that $\overline{f}(x)$ and $\overline{g}(x)$ have a common factor mod $p$.

Now remember that

$\Phi_{n}(x)=f(x)g(x).$

This tells us that, mod $p$, $\Phi_{n}(x)$ has a multiple root (from the common factor of $f(x)$ and $g(x)$ mod $p$). But that would mean that $x^{n}-1$ has a multiple root mod $p$, which can’t be true. It’s derivative is $nx^{n-1}$ which is not zero mod $p$ since $p$ does not divide $n$. It follows that $\zeta^p$ must be a root of $f(x)$.

Retracing the argument, we’ve shown that, if $\zeta$ is a root of the factor $f(x)$, so is $\zeta^p$ for any $p$ not dividing $n$. If $\alpha$ is any primitive $n^{th}$ root of $1$, then $\alpha=\zeta^a$ for some $a$ relatively prime to $n$. But then $a=p_1\cdots p_k$ where the $p_i$ are primes not dividing $n$ (not necessarily distinct). It follows that $\alpha=((\zeta^{p_{1}})^{p_{2}})^{\ldots}$ is also a root of $f(x)$. In other words, all of the primitive $n^{th}$ roots of one are roots of $f(x)$. That means that $f(x)=\Phi_{n}(x)$ and $g(x)=1$, so $\Phi_{n}(x)$ is irreducible.

Corollary: The field $\Q(\mu_{n})$ has degree $\phi(n)$ over $\Q$.

### The cosine of twenty degrees

In our work on constructibility we claimed that the 60 degree angle could not be trisected because a twenty degree angle is not constructible. Now

$2\cos 20^{\circ}=2\cos{2\pi/18}=e^{\pi i/9}+e^{-\pi i/9}.$

Now $e^{\pi i/9}$ is a primitive $18^{th}$ root of one and there are $\phi(18)=6$ such; they are roots of $\Phi_{18}(x)$. Now

$x^{18}-1=(x^9-1)(x^9+1)$

but also

$x^{18}-1=\Phi_{1}(x)\Phi_{2}(x)\Phi_{3}(x)\Phi_{6}(x)\Phi_{9}(x)\Phi_{18}(x)$

Now:

$\begin{matrix} Phi_{1}(x)=x-1 \\ \Phi_{2}(x)=x+1 \\ \Phi_{3}(x)=x^2+x+1 \\ \Phi_{6}(x)=x^2-x+1 \\ \end{matrix}$

so some algebra tells us that

$\Phi_{18}(x)=\frac{x^9+1}{x^3+1}=x^6-x^3+1.$

This in turn means that, if $\zeta=e^{\pi i/9}$, then $\zeta^{3}+\zeta^{-3}=1$ (divide $\Phi_{18}$ by $x^3$). But

$(\zeta+\zeta^{-1})^3=\zeta^{3}+\zeta^{-3}+3(\zeta+\zeta^{-1})=1+3(\zeta+\zeta^{-1}).$

In other words $2\cos\frac{2\pi i}{18}=\zeta+\zeta^{-1}$ satisfies the cubic polynomial $x^3-3x-1=0$.