Finite Fields

Subjects: Field Theory
Links: Characteristic of a Ring, Splitting Fields and Normal Field Extensions, Separable Field Extensions, Galois Field Extensions

Let us note that finite fields must have nonzero characteristic. In this note $F$ be a finite subfield with characteristic $p$. Note that the prime field of $F$ is isomorphic to $\mathbb{Z}/p\mathbb{Z}$.

Obs: The polynomial $x^{p^n}-x$ over $(\mathbb{Z}/p\mathbb{Z})[x]$ has derivative $p^n{x}^{p^n-1}-1=-1$ since the field has characteristic $p$. Since the derivative has no roots at all, we see that the polynomial has no multiple roots, and hence it is separable.

Let $n>0$ be any positive integer and consider the splitting field of the polynomial $x^{p^n}-x$ over $\mathbb{Z}/p\mathbb{Z}$. Since it is separable, then it has $p^n$ different roots. If $\alpha$,and $\beta$ be any two roots of this polynomial, then ${\alpha }^{p^n}=\alpha$ and ${\beta }^{p^n}=\beta$. This means that $\alpha +\beta$. $\alpha \beta$, and $\alpha /\beta$ (when $\beta \ne 0$) are roots. Hence the set $\mathbb{F}$ consisting of the $p^n$ distinct roots of $x^{p^n}-x$ over $\mathbb{Z}/p\mathbb{Z}$ is the splitting field. Since the number of elements is $p^n$, we have that $[\mathbb{F}:\mathbb{Z}/p\mathbb{Z}]=n$, which shows that there exist fields of degree $n$ over $\mathbb{Z}/p\mathbb{Z}$ for any $n>0$.

Let now $\mathbb{F}$ be a any finite field of characteristic $p$. If $\mathbb{F}$ is of dimension $n$ over its prime subfield $\mathbb{Z}/p\mathbb{Z}$, then $\mathbb{F}$ has $p^n$ elements. Since ${\mathbb{F}}^{\times }$ is a cyclic group of order $p^n-1$, we have that ${\alpha }^{p^n-1}=1$ for every $\alpha \in {\mathbb{F}}^{\times }$ , so that ${\alpha }^{p^n}=\alpha$ for every $\alpha \in \mathbb{F}$. Since we have seen that the splitting field has order $p^n$ this shows that $\mathbb{F}$ is the splitting field for $x^{p^n}-x$. Since splitting fields are unique up to isomorphism, this proves that finite fields of any order $p^n$ exists and are unique up to isomorphism.

Th: Let $F$ be finite field. If $|F|=p^n$, then $F$ is isomorphic to the splitting field of $x^{p^n}-x\in (\mathbb{Z}/p\mathbb{Z})[x]$. We are going to denote this as ${\mathbb{F}}_{p^n}$.

Cor: In ${\mathbb{F}}_{p^n}$, we see that $$x^{p^n-1}-1 = \prod_{\alpha\in \Bbb F_{p^n}^\times} (x-\alpha).$$If we evaluate at $0$, we get the equality $$\prod_{\alpha\in \Bbb F_{p^n}^\times} \alpha = (-1)^{p^n} = -1.$$This is a generalisation of Wilson's Theorem.

Def: Let $F$ be a field of characteristic $p$. The map $\varphi :F\to F$ such that $\varphi (a)=a^p$ is a monomorphism called the Frobenious endomorphism of $F$.

Obs: If $F$ is a finite field, then $F$ is perfect, meaning, every finite extension is separable

Prop: If $f(x)\in {\mathbb{F}}_p$, then $f(x^p)=f(x{)}^p$.

Cor: $${pn \choose pi} \equiv {n \choose i} \pmod p.$$

Prop: Let $\varphi$ be the Frobenius endomorphism $x\mapsto x^p$ on the field $F$. We see that $phi$ is an automorphism. If $|F|=p^n$, then ${\varphi }^n={\text{id}}_F$, and it is the smallest positive integer that does this.

If we look at the matrix representation of the Frobenius endomorphism in ${\mathbb{F}}_{p^n}$ as a ${\mathbb{F}}_p$-linear transformation, we see that has a minimal polynomial of $x^n-1$, and thus has rational form of the form: $$\begin{pmatrix}0 & 0 & \cdots & 0 & 1 \ 1 & 0 &\cdots & 0 & 0 \ \vdots &\vdots & \ddots& \vdots & \vdots \
0 & 0 & \cdots &1&0 \end{pmatrix} $$
We can see that $F/(\mathbb{Z}/p\mathbb{Z})$ is a separable, normal extension, thus it is a Galois extension, and $[F:\mathbb{Z}/p\mathbb{Z}]=|\text{Gal}(F/(\mathbb{Z}/p\mathbb{Z}))|$, and $\varphi$ is an element of $\text{Gal}(F/(\mathbb{Z}/p\mathbb{Z}))$, since $\phi {|}_{\mathbb{Z}/p\mathbb{Z}}={\text{id}}_{\mathbb{Z}/p\mathbb{Z}}$. This means that $\text{Gal}(F/(\mathbb{Z}/p\mathbb{Z}))\cong C_n$, where $C_n$ is the cyclic group or order $n$

Wedderburn's Theorem on Finite Skew-Fields: A finite skew-field is a field.

The proof of this theorem is fairly involved, we actually use the class equation, and cyclotomic polynomials.

Prop: Any finite field is isomorphic to ${\mathbb{F}}_{p^n}$ for some prime $p$ and some integer $n\ge 1$. The field ${\mathbb{F}}_{p^n}$ is the splitting field over ${\mathbb{F}}_p$ of the polynomial $x^{p^n}-x$, with cyclic Galois group of order $n$ generated by the Frobenius automorphism ${\sigma }_p$. The subfields of ${\mathbb{F}}_{p^n}$ are all Galois over ${\mathbb{F}}_p$ and are in one to one correspondence with the divisors $d$ of $n$. That are the fields ${\mathbb{F}}_{p^d}$, the fixed fields of ${\sigma }_p^d$.

Cor: Let $q=p^n$ be a power of a prime and let ${\mathbb{F}}_q={\mathbb{F}}_{p^n}$ be the finite field with $q$ elements. Let ${\sigma }_q={\sigma }_p^n$ be the $n$th power of the Frobenius automorphism ${\sigma }_p$, called the $q$-Frobenius automorphism. The following statements are true

• ${\sigma }_q$ fixes ${\mathbb{F}}_q$.
• Every finite extension of ${\mathbb{F}}_q$ of degree $m$ is the splitting field of $x^{q^m}-x$ over ${\mathbb{F}}_q$.
• Every finite extension of ${\mathbb{F}}_q$ of degree $m$ is cyclic with ${\sigma }_q$ as generator.
• The subfields of the unique extension ${\mathbb{F}}_q$ of degree $m$ are in bijective correspondence with the divisors of $m$.

Cor: The irreducible polynomial $\text{x { <a class="tag" onclick="toggleTagSearch(this)" data-content="\#4">\#4</a>} +1 in Bbb Z[x]}$ is reducible modulo every prime $p$.

Prop: If $a,b\in \mathbb{Z}$ such that $a$ and $b$ are not perfect squares, then $a$, $b$ or $ab$ is a square in ${\mathbb{F}}_p,$for every prime $p$. This happens because of the law of quadratic reciprocity.

Cor: Let $K=\mathbb{Q}(\theta )=\mathbb{Q}(\sqrt{\alpha },\sqrt{\beta })$ with $\alpha ,\beta \in \mathbb{Z}$, is a biquadratic extension and that $\theta =a+b\sqrt{\alpha }+c\sqrt{\beta }+d\sqrt{\alpha \beta }$, where $a,b,c,d\in \mathbb{Z}$ are integers. Then $\text{Irr}(\theta ,\mathbb{Q})$ has degree $4$, but is reducible modulo every prime $p$. This is impart because either $\alpha$, $\beta$ or $\alpha \beta$ is a square in ${\mathbb{F}}_p$.

Cor: If $a,b\in \mathbb{Z}$ such that $a$ and $b$ are not perfect squares, then polynomial $$(x^2-a)(x^2-b)(x^2-ab)$$has no roots in $\mathbb{Q}$, but it has at least one root in ${\mathbb{F}}_p$, for every prime $p$.
Prop: The finite field ${\mathbb{F}}_{p^n}$ is simple. In particular, there exists an irreducible polynomial of degree $n$ over ${\mathbb{F}}_p$ for every $n\ge 1$.

Prop: Let $\psi (n)$ be the number of irreducible polynomials of degree $n$ in ${\mathbb{F}}_p[x]$. Then $$p
{ #n}
= \sum_{d\mid n}d\psi(d), $$and by Möbius inversion formula, we have that $$ n \psi(n) = \sum_{d \mid n }\mu(d) p^{n/d}, \quad \text{or}\quad \psi(n) =\frac1n \sum_{d \mid n }\mu(d) p^{n/d} $$
We have seen that ${\mathbb{F}}_{p^n}/{\mathbb{F}}_{p^m}$ is a field extension iff $m\mid n$. In particular, given two finite filds ${\mathbb{F}}_{p^n}$ and ${\mathbb{F}}_{p^m}$ there is a third finite field containing an isomorphic copy of them, namely, ${\mathbb{F}}_{p^{nm}}$. This gives us a partial ordering on these fields and allows us to think in their direct limit, since $({\mathbb{N}}^{+},\mid )$ is a directed set. Since these give all the finite extensions of ${\mathbb{F}}_p$, we see that the limit of ${\mathbb{F}}_{p^n}$ for all $n$ is an algebraic closure of ${\mathbb{F}}_p$, unique up to isomorphism: $$\overline{\Bbb F_p} = \varinjlim_{n\in\Bbb N⁺} \Bbb F_{p^n}.$$This provides a simple description of the algebraic closure of ${\mathbb{F}}_p$.

Obs: No finite field can be algebraically closed.

Prop: The splitting field of the polynomial $x^p-x-a$ over ${\mathbb{F}}_p$ where $a\ne 0$, $a\in {\mathbb{F}}_p$,