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The additive sequence $0,1,2\cdot 1,\dots$ must repeat because $K$ is finite, so the characteristic is positive. If its least positive value were composite, $r=ab$ with $1<a,b<r$, then $(a{1}_K)(b{1}_K)=r{1}_K=0$ while neither factor would be zero. A field has no zero divisors, giving a contradiction. Hence $\boxed{\mathrm{char}K=p\text{ is prime}}$.

Write $q=p^r$, where $p$ is the characteristic. In a splitting field of $X^{q^m}-X$ over $F$, let $K$ be its set of roots. The derivative of this polynomial is $-1$, so all $q^m$ roots are distinct. The Frobenius power is additive and multiplicative: $(x+y{)}^{q^m}=x^{q^m}+y^{q^m}$ and $(xy{)}^{q^m}=x^{q^m}{y}^{q^m}$. Thus roots are closed under addition, negatives and multiplication; for a nonzero root, $(x^{-1}{)}^{q^m}=x^{-1}$. They include $0,1$ and form a field.

Every $a\in F$ satisfies $a^q=a$, by Lagrange's theorem on $F^{\times }$ and separately at zero. Iterating gives $a^{q^m}=a$, so $F\subseteq K$. The constructed field therefore contains $F$ and has exactly $\boxed{q^m\text{ elements}}$.

The four points form a projective frame: no three are collinear. By the allowed uniqueness theorem and cross-ratio invariance, useIntersecting the indicated projective lines givesA point of $Q_0{Q}_2$ has coordinates $[a:b:a+b]$. Intersecting with $P_0{P}_1$, whose equation is $z=0$, gives $Q_3=[1:-1:0]$. On that line use its first two homogeneous coordinates. The cross-ratio from (a)(ii) isThe characteristic of a field being zero ensures that $Q_3\ne Q_1$. The construction is the diagonal-point property of a complete quadrangle, producing harmonic conjugate points.

The displayed bilinear form is the polar form of a quadratic form. In odd characteristic of a field, $Q(x)=B_Q(x,x)/2$ recovers the quadratic form; in characteristic of a field two, different forms can have the same alternating bilinear form. Over ${\mathbb{F}}_2$, two such forms differ by a linear map to the field, giving the affine parametrization of binary quadratic refinements.

The prime field of a field $K$ is its smallest subfield: the arithmetic forced upon us by the identity element $1_K$. Every subfield must contain $1_K$, its repeated sums, their additive inverses, and the multiplicative inverses of nonzero results. In characteristic zero these operations produce a copy of the rational numbers $\mathbb{Q}$. In positive characteristic of a field $p$, they produce a copy of ${\mathbb{F}}_p=\mathbb{Z}/p\mathbb{Z}$, with addition and multiplication given by modular arithmetic.

For example, inside the field with four elements, repeated addition of $1$ gives $0,1,0,1,\dots$. Thus its prime field is $\{0,1\}$, and the other two elements cannot be reached by repeatedly adding $1$. This is why a finite field need not consist entirely of ordinary integer residues: a larger finite field extends the arithmetic of its prime field with additional elements.

The positive characteristic of a field must be a prime number. If its least vanishing multiple of $1_K$ were $rs$ with $1<r,s<rs$, then $(r{1}_K)(s{1}_K)=0$ would exhibit zero divisors. The absence of zero divisors rules this out. The prime field supplies the scalars when viewing $K$ as a vector space; its basis then explains the possible sizes of a finite field.

Let $R$ be a complete discrete valuation ring of residue characteristic of a field $p$, with maximal ideal $\mathfrak{m}$, and let $F$ be a one-dimensional formal group law over $R$. Its multiplication series is $[n{]}_F(T)=nT+O(T^2)$. If $p\nmid n$, the invertible morphism criterion for formal group laws constructs its inverse with coefficients in $R$. Both formal power series converge on $\mathfrak{m}$, since their coefficients are integral and the powers of an element of $\mathfrak{m}$ tend to zero. Their compositions are the identity there. Thus multiplication by $n$ is bijective, including when $p=2$ and $n$ is odd. Every finite-order element of $F(\mathfrak{m})$ consequently has order of a group element equal to a prime power with prime $p$.

These roots are taken in an algebraic closure, not merely in ${\mathbb{F}}_q$. If the characteristic of a field is $p$ and $n=p^{a}m$ with $p\nmid m$, then $X^n-1=(X^m-1{)}^{p^a}$; hence there are $m$ distinct roots. They form a cyclic group by the finite multiplicative subgroup of a field is cyclic theorem. In particular they have order $n$ when $\gcd (n,q)=1$.

A linear operator is semisimple if it becomes diagonalizable over an algebraic closure; equivalently, its minimal polynomial has no repeated root over that closure. If an invertible operator $A$ has finite order $r$ prime to the characteristic of a field, its minimal polynomial divides $X^r-1$. The derivative $r{X}^{r-1}$ is coprime to this polynomial, so $A$ is semisimple. Its invariant irreducible summands over a finite field become multiplication by scalars in suitable finite field extensions.

A filling $T$ of shape $\lambda$ and content $\alpha$ defines a module homomorphism ${\Theta }_T:M^{\lambda }\to M^{\alpha }$: given a numbered tableau, sum over distinct assignments obtained by permuting $T$ within its rows, and put each numbered entry into output row $j$ when its assigned label is $j$. Restricting to $S^{\lambda }$ gives ${\hat{\Theta }}_T$. For Semistandard Young tableaux $T$, these restrictions are linearly independent over every field. They form a basis of ${\mathrm{Hom}}_{F{S}_n}(S^{\lambda },M^{\alpha })$ if the characteristic of a field is not two, or if $\lambda$ is a two-regular partition of an integer. For two-singular shapes in characteristic two, they need not span. Column antisymmetrization and triangular leading tabloids establish independence; straightening the images of a cyclic polytabloid establishes spanning under the stated hypotheses.

On $y^2=x^3+ax+b$ in characteristic of a field other than two, take the elementary symmetric polynomials $s_i$ of $x(P),x(Q),x(P+Q)$. If the line through $P,Q$ is $y=\lambda x+\mu$, its third intersection is $-(P+Q)$, so coefficient comparison gives $s_1={\lambda }^2$, $s_2=a-2\lambda \mu$, $s_3={\mu }^2-b$. Hence $(s_2-a{)}^2=4s_1(s_3+b)$. This is an identity of rational functions; tangent cases follow by specialization, and coordinates at infinity are handled through the projective addition map.

The scheme $C^{(d)}$ parametrizes effective degree-$d$ Cartier divisors on a smooth algebraic curve. Near a Cartier divisor $\sum m_i{P}_i$, its completed local ring is a power-series ring in the elementary symmetric coordinates of $m_i$ local parameters at each $P_i$. It is therefore smooth of dimension $d$, even when the characteristic of a field divides $d!$. For projective $C$ it is projective.

The tabloid bilinear form makes the tabloid basis orthonormal and restricts to an invariant bilinear form on each Specht module. In positive characteristic of a field, its radical controls the corresponding simple quotient.