Finite Field Arithmetic

Definition 1. Let $\mathbb F$ be a field with at least two elements. Define the characteristic of the field by

$\text{char}(\mathbb F) := \min \{n \in \mathbb Z^+ : n \cdot 1 = 0\},$

where $1$ denotes the multiplicative identity in $\mathbb F$ . If the right-hand side is empty, we set $\text{char}(\mathbb F) := 0$ by convention.

Problem 1. Show that either $\text{char}(\mathbb F)$ is prime or $\mathbb F \supseteq \mathbb Q$ as an isomorphic sub-field.

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Solution. Suppose the right-hand side is non-empty. Then there exists some smallest $p \in \mathbb Z^+$ such that $p \cdot 1 = 0$ . If $p$ is composite, there exists smaller positive integers $r, s$ such that $p = rs$ , so that $(r \cdot 1) \cdot (s \cdot 1) = (rs) \cdot 1 = 0$ . Since $\mathbb F$ is a field, at least one of $r \cdot 1,s \cdot 1$ is $0$ , so that $\mathrm{char}(\mathbb F) \leq \min\{r,s\} < p$ , a contradiction. Therefore, $p$ is prime.

Now suppose the right-hand side is empty. Then define the monomorphism $\phi : \mathbb Q \to \mathbb F$ by $\phi(p/q) = (p \cdot 1)/(q \cdot 1)$ so that $\mathbb F \supseteq \mathbb Q$ as a subfield.

Problem 2. Deduce that if $\text{char}(\mathbb F) = p$ for some prime $p$ , then $\mathbb F \supseteq \mathbb F_p$ as an isomorphic sub-field.

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Solution. Define the monomorphism $\phi : \mathbb F_p \to \mathbb F$ by $\phi([n]_p) = n \cdot 1$ .

Henceforth, let $p$ denote a prime. and suppose $\mathrm{char}(\mathbb F) = p$ .

Problem 3. Show that for any $u,v \in \mathbb F$ , $(u+v)^p = u^p + v^p$ .

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Solution. Using the binomial theorem,

$\displaystyle (u+v)^p = u^p + v^p + \sum_{k=1}^{p-1} {p \choose k} u^{p-k} v^k.$

It suffices to check that $\displaystyle p \mid {p \choose k}$ for $1 \leq k \leq p-1$ . By computing the binomial coefficient,

$\displaystyle {p \choose k} = p \cdot \frac{(p-1)!}{k! (p-k)!}.$

Since $p$ is prime, none of the numbers in the denominator divides $p$ . Therefore, $\displaystyle p \mid {p \choose k}$ , as required.

Problem 4. Show that $|\mathbb F| = p^n$ for some positive integer $n$ .

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Solution. By Problem 2, $\mathbb F \supseteq \mathbb F_p$ as a subfield. Therefore, there exists $n \in \mathbb Z^+$ such that $[\mathbb F : \mathbb F_p] = n$ . Let $\{v_1,\dots, v_n\}$ be a basis for $\mathbb F$ over $\mathbb F_p$ . Then

$\displaystyle \mathbb F = \left\{ \sum_{i=1}^n c_i v_i : (c_1,\dots, c_n)^{\mathrm T} \in \mathbb F_p^n \right\}.$

In particular, $|\mathbb F| = |\mathbb F_p^n| = |\mathbb F_p|^n = p^n$ .

Problem 5. Show that $\mathbb F^* := \mathbb F \backslash \{0\}$ is cyclic under multiplication.

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Solution. Suppose $|\mathbb F| = p^n$ for some prime $p$ and positive integer $n$ . Then $|\mathbb F^*| = p^n - 1$ . By the fundamental theorem of arithmetic, there exists primes $p_i$ and positive integers $\alpha_i$ such that

$\displaystyle p^n - 1 = \prod_{i=1}^m p_i^{\alpha_i}.$

By the Sylow theorems, for each $i$ , there exists a multiplicative subgroup $G_i$ of $\mathbb F^*$ such that $|G_i| = p_i^{\alpha_i}$ . By the fundamental theorem of finitely generated Abelian groups, there exists $u_i \in \mathbb F^*$ such that $G_i = \langle u_i \rangle$ . In particular, $u_i^{p_i^{\alpha_i}} = 1$ , so that

$\displaystyle u_i^{p^n-1} = \left( u_i^{p_i^{\alpha_i}} \right)^{\prod_{j\neq i} p_j^{\alpha_j}} = 1^{\prod_{j\neq i} p_j^{\alpha_j}} = 1.$

Define $u := \prod_{i=1}^m u_i$ . Then

$\begin{aligned} u^{p^n-1} &= \prod_{i=1}^m u_i^{p^n-1} = \prod_{i=1}^m 1 = 1. \end{aligned}$

Furthermore, for $0 < k < p^n - 1$ , $u^k \neq 1$ . Therefore, $|\langle u \rangle| = p^n - 1$ , so that $\mathbb F^* = \langle u \rangle$ .

Problem 6. For any positive integer $n$ , construct a unique subfield of $\bar{\mathbb F}_p$ with $p^n$ elements, denoted $\mathbb F_{p^n}$ .

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Solution. Define $f(x) = x^{p^n}-x$ . We claim that $\mathbb F_{p^n} := \mathcal Z(f)$ forms a field with $p^n$ elements. Since $f$ does not contain any double root, $|\mathcal Z(f)| = p^n$ . It suffices to check that $\mathcal Z(f)$ forms a field. To that end, given $u,v \in \mathcal Z(f)$ , we leave it as an exercise in induction to check that for any $w \in \mathbb F_p$ , $w^{p^n} = w$ . Particularising to $u+v, uv$ respectively,

$\begin{aligned} f(u+v) &= (u+v)^{p^n} - (u+v) = (u+v) - (u+v) = 0, \\ f(uv) &= (uv)^{p^n} - (uv) = (uv) - (uv) = 0. \end{aligned}$

Therefore, $\mathcal Z(f)$ is closed under addition and multiplication, and clearly Abelian in both operations, so that $\mathcal Z(f)$ forms a field, as required.

—Joel Kindiak, 3 May 26, 1852H

KindiakMath

Finite Field Arithmetic

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Finite Field Arithmetic

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