[:characteristic:]
• Take $\mathbb F_q$ a finite field with $q$ elements. Then:
• (1) for some $p,d$ with $p$ prime we have $q = p^d$
• $p$ is called the **characteristic** of $\mathbb F_q$, denoted $\text{char}(\mathbb F_q)$
• (2) $p$ is the smallest integer with $\underbrace {1 + 1 + \cdots + 1}_{p \text{ times}} \equiv 0$
• (3) $\mathbb F_q$ contains $\mathbb Z/p$ as a subfield
• And thus $\mathbb F_q$ is a linear space of dimension $d$ with coefficients in $\mathbb Z/p$
• (4) Exponentiation by the characteristic distributes over addition
• $(\alpha + \beta)^p \equiv \alpha^p + \beta^p$
•
• Examples:
• $\text{char}(\mathbb Z/k) = k$
• $\mathbb F_2[x]/(x^3 + x + 1) = \{0, 1, x, x+1, x^2, \dots\}$ has characteristic 2
•
• Proofs:
• Of (2): first note that since $\mathbb F_q$ is finite, then repeatedly adding $1$ must eventually hit $0$. Let $m$ be the minimum number of $1$s required. If $m$ were nonprime, then we would have $0 = m = ab = (\underbrace{1 + \cdots + 1}_\text{a \text{ times}})(\underbrace{1 + \cdots + 1}_{b \text{ times}})$. Because we have $0 = ab$ and are working in a field, we may conclude that either $a = 0$ or $b = 0$, contradicting that $m$ is minimal. $\square$
• Of (3): since $p < q$ we may simple take the lowest $p$ elements (indexed $0, 1, 2, \cdots, p-1$) from $\mathbb F_q$ to form a subset; this is a field because, as assumed, $\underbrace{1 + \cdots + 1}_{p \text{ times}} \equiv 0$
• Skipping over the linear space part
• Of (1): this follows from (3); the cardinality of $\mathbb F_q$ is $q$; viewing it as a linear space $(\mathbb Z/p)^d$ we see that the cardinality is also $p^d$; thus $q = p^d$
• Of (4): the binomial theorem gives that $(\alpha + \beta)^p = \sum_{n=0}^p \binom p k \alpha^k \beta^{p-k}$. Coefficients are $\binom p k = \frac {p!}{k!(p-k)!}$; when $k \notin \{0, p\}$ we get a factor of $p$ on the top but *not* the bottom, and since $p \equiv 0$ that means that the coefficient vanishes. Thus all we are left with is $\alpha^p$ and $\beta^p$!
__Referenced by:__