[:characteristic:] • Take Fq\mathbb F_q a finite field with qq elements. Then: • (1) for some p,dp,d with pp prime we have q=pdq = p^dpp is called the characteristic of Fq\mathbb F_q, denoted char(Fq)\text{char}(\mathbb F_q) • (2) pp is the smallest integer with 1+1++1p times0\underbrace {1 + 1 + \cdots + 1}_{p \text{ times}} \equiv 0 • (3) Fq\mathbb F_q contains Z/p\mathbb Z/p as a subfield • And thus Fq\mathbb F_q is a linear space of dimension dd with coefficients in Z/p\mathbb Z/p • (4) Exponentiation by the characteristic distributes over addition • (α+β)pαp+βp(\alpha + \beta)^p \equiv \alpha^p + \beta^p • • Examples: • char(Z/k)=k\text{char}(\mathbb Z/k) = kF2[x]/(x3+x+1)={0,1,x,x+1,x2,}\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 Fq\mathbb F_q is finite, then repeatedly adding 11 must eventually hit 00. Let mm be the minimum number of 11s required. If mm were nonprime, then we would have 0=m=ab=(1++1a  times)(1++1b times)0 = m = ab = (\underbrace{1 + \cdots + 1}_\text{a \text{ times}})(\underbrace{1 + \cdots + 1}_{b \text{ times}}). Because we have 0=ab0 = ab and are working in a field, we may conclude that either a=0a = 0 or b=0b = 0, contradicting that mm is minimal. \square • Of (3): since p<qp < q we may simple take the lowest pp elements (indexed 0,1,2,,p10, 1, 2, \cdots, p-1) from Fq\mathbb F_q to form a subset; this is a field because, as assumed, 1++1p times0\underbrace{1 + \cdots + 1}_{p \text{ times}} \equiv 0 • Skipping over the linear space part • Of (1): this follows from (3); the cardinality of Fq\mathbb F_q is qq; viewing it as a linear space (Z/p)d(\mathbb Z/p)^d we see that the cardinality is also pdp^d; thus q=pdq = p^d • Of (4): the binomial theorem gives that (α+β)p=n=0p(pk)αkβpk(\alpha + \beta)^p = \sum_{n=0}^p \binom p k \alpha^k \beta^{p-k}. Coefficients are (pk)=p!k!(pk)!\binom p k = \frac {p!}{k!(p-k)!}; when k{0,p}k \notin \{0, p\} we get a factor of pp on the top but not the bottom, and since p0p \equiv 0 that means that the coefficient vanishes. Thus all we are left with is αp\alpha^p and βp\beta^p! Referenced by: