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The absolute Frobenius morphism of a scheme of characteristic p is the identity on the underlying topological space and raises every local function to its $p$th power. On an affine chart $\mathrm{Spec}A$ it is induced contravariantly by the Frobenius endomorphism $A\to A$, $a\mapsto a^p$. It need not be an isomorphism of schemes, even when it acts invertibly on global sections.

The algebraic Frobenius endomorphism induces this scheme map contravariantly on affine charts.

Over a field of characteristic $p$, additive polynomials are sums of $p$-power monomials, $L(X)={\sum }_i{a}_i{X}^{p^i}$. They act linearly over the prime field. The identity follows from the Frobenius endomorphism; the converse follows by comparing mixed binomial coefficients. Their zero sets are additive groups, and a nonzero linear coefficient makes all roots simple.

A root $\beta$ of the binary Golay generator polynomial has order $23$, and the Frobenius endomorphism shows that $\beta ,{\beta }^2,{\beta }^3,{\beta }^4$ are roots: $2^8\equiv 3(\mathrm{mod}23)$. The BCH bound gives minimum Hamming distance at least five. The circular autocorrelation of the binary Golay generator makes its parity extension doubly even, excluding original weights five and six. The generator itself has Hamming weight seven, proving that the distance is exactly seven.

// Target: algebra.bigb

For an algebraic group defined over ${\mathbb{F}}_q$, its Frobenius morphism acts as entrywise $q$th powers in a standard matrix realization. Its differential is zero, and its fixed-point group is finite. A conjugate $F^{\prime }=\phi F{\phi }^{-1}$ defines an isomorphic rational structure. This group morphism must be distinguished from a field Frobenius endomorphism alone and from the convention called geometric Frobenius in Galois theory, which is inverse to the arithmetic generator on field points.

Every cyclotomic polynomial ${\Phi }_n$ is an irreducible polynomial over $\mathbb{Q}$. If a monic irreducible factor $f\in \mathbb{Z}[X]$ contains a primitive root $\zeta$, then it also contains ${\zeta }^p$ for every prime number $p\nmid n$: otherwise, writing ${\Phi }_n=fg$, reduction modulo $p$ and the Frobenius endomorphism give $\bar{f}\mid \bar{g}(X{)}^p$. Some irreducible factor would then divide both $\bar{f}$ and $\bar{g}$, giving a repeated factor of $X^n-1$, contrary to separability of roots of unity. Iterating over the prime factors of every integer coprime to $n$ shows that $f$ contains all primitive $n$th roots, so $f={\Phi }_n$.

For a prime number $p$, the binomial coefficient modulo $p$ factors into the corresponding digit binomial coefficients in the base-p expansions of the nonnegative integers $n,k$. Pad both expansions to the same length and use $\left(\genfrac{}{}{0ex}{}{a}{b}\right)=0$ when $b>a$. The identity follows by factoring $(1+x{)}^n$ into digit powers and using the Frobenius endomorphism; digit uniqueness identifies each coefficient.

Reduction modulo the prime $p$ preserves integer addition, multiplication, and one:Every congruence class has an integer representative, so the ring homomorphism is onto. Its kernel is exactly $p\mathbb{Z}=(p)$. The quotient is a field because every class not divisible by $p$ has a multiplicative inverse by Bézout's identity.

For the coefficientwise extension to the polynomial ring, addition is preserved term by term. If $f={\sum }_i{a}_i{x}^i$ and $g={\sum }_j{b}_j{x}^j$, the coefficient of $x^k$ in their product is ${\sum }_{i+j=k}{a}_i{b}_j$. Reducing it modulo $p$ gives ${\sum }_{i+j=k}{\bar{a}}_i{\bar{b}}_j$, the coefficient in $\hat{\varphi }(f)\hat{\varphi }(g)$. Thus multiplication and one are preserved as well. Lift each coefficient of a polynomial over ${\mathbb{Z}}_p$ to an integer to prove surjectivity. The kernel isIndeed, its elements are precisely the polynomials whose coefficients are all divisible by $p$.

In the binomial theorem expansion of $(x-1{)}^p$, every intermediate coefficient $\left(\genfrac{}{}{0ex}{}{p}{j}\right)$ is divisible by $p$: the numerator contains the prime factor $p$, while $j!(p-j)!$ does not for $1\le j<p$. HenceFor $p=2$, the last equality uses $1=-1$ in characteristic two. This is an instance of the Frobenius endomorphism.

A finite field is a field with finitely many elements. Its characteristic is the least positive integer $p$ for which $p\cdot 1=0$; such an integer exists because the additive group is finite. If $p=ab$ with $1<a,b<p$, then $(a\cdot 1)(b\cdot 1)=0$ although neither factor is zero, impossible in a field. Hence $p$ is prime, and the multiples of $1$ form its prime field ${\mathbb{F}}_p$. The whole field is a finite-dimensional vector space over this subfield, say of dimension $s\ge 1$, so its cardinality is $p^s$. Thus $q=p^s$ and its characteristic is $p$.

For existence, let $K$ be a splitting field over ${\mathbb{F}}_p$ of $P(T)=T^{p^s}-T$. Since $P^{\prime }(T)=-1$, it has exactly $p^s$ distinct roots. Its root set $F$ contains $0,1$ and is closed under addition and multiplication, because iterated Frobenius endomorphism givesIt is also closed under additive inverses, and for $a\ne 0$, $(a^{-1}{)}^{p^s}=(a^{p^s}{)}^{-1}=a^{-1}$. Thus $F$ is a field with exactly $p^s$ elements. Since it contains every root, it is the entire splitting field. This proves existence.

For uniqueness, let $L$ be any field with $p^s$ elements. Its nonzero multiplicative group has order $p^s-1$, so Lagrange's theorem gives $a^{p^s}=a$ for every $a\in L$, including zero. Hence $L$ is a splitting field of the same polynomial. Here is the embedding argument giving uniqueness, rather than merely invoking it by name. Extend the identity on ${\mathbb{F}}_p$ from one root at a time. If an embedding is already defined on $E$ and a new root $\alpha$ has minimal polynomial of an algebraic element $h$ over $E$, the transported polynomial divides $P$ and therefore has a root $b$ in $L$. Evaluation at $b$ induces an embedding $E[T]/(h)\to L$, extending the existing embedding to $E(\alpha )$. Repeating for the finitely many roots embeds the constructed field $F$ in $L$. They have the same finite cardinality, so this embedding is an isomorphism. Thus the field of order $p^s$ is unique up to isomorphism.

We prove cyclicity of the multiplicative group, also establishing the more general finite multiplicative subgroup of a field is cyclic result. Let $H$ be a finite subgroup of a field's multiplicative group, and let $e$ be the least common multiple of all element orders. For every prime $\ell$ dividing $e$, choose an element whose order has the maximal $\ell$-power ${\ell }^{a_{\ell }}$; raising it to the factor of its order coprime to $\ell$ gives an element of order ${\ell }^{a_{\ell }}$. Since the group is abelian, the product of these elements has order ${\prod }_{\ell }{\ell }^{a_{\ell }}=e$: for commuting elements of coprime orders, an equation $(ab{)}^t=1$ forces $a^t=b^{-t}$ into the intersection of their cyclic subgroups, which is trivial, so both orders divide $t$. Every member of $H$ is a root of $T^e-1$. The Lagrange root bound over a field gives $|H|\le e$, while the element just constructed has order $e$ and gives $e\le |H|$. Thus equality holds and that element generates $H$. In particular

For the field with nine elements, use ${\mathbb{F}}_3[u]/(u^2+1)$. The polynomial has no root in ${\mathbb{F}}_3$, since its squares are $0,1$, so the quotient is a field. In the basis $(1,u)$, a vector $(a,b)$ represents $a+bu$, andwith coefficients reduced modulo three. Put $\beta =1+u$. Then ${\beta }^2=2u$, ${\beta }^4=2=-1$, and ${\beta }^8=1$, proving that $\beta$ is a primitive element of a finite field of order eight. Its complete power/vector table isFor completeness, the full addition and multiplication tables in these labels are displayed below. The subscripts are element labels; they are not the scalar residues of ${\mathbb{F}}_3$.

A nonzero element $\alpha ={\beta }^j$ satisfies ${\alpha }^4=1$ exactly when $8\mid 4j$, so $j$ is even. Thus, with $e=1$,

Work over the finite field ${\mathbb{F}}_2$. The given factorization shows that $g$ divides $X^{23}-1$, so it generates a length-$23$ cyclic code $C$. Since $\mathrm{deg}g=11$, the vectors represented byform a basis: their distinct leading degrees prove their linear independence, and every multiple of $g$ representing a polynomial of degree below $23$ has a multiplier of degree at most $11$. Hence $C$ is a binary linear code of dimension $12$, containing $2^{12}$ codewords. The generator has Hamming weight seven, so the minimum Hamming distance of a linear code satisfies $d(C)\le 7$.

The BCH bound used here is the following consecutive-root theorem. If a length-$n$ cyclic code over ${\mathbb{F}}_q$, with $n$ relatively prime to $q$, has a generator polynomial of a cyclic code vanishing at ${\beta }^b,{\beta }^{b+1},\dots ,{\beta }^{b+\Delta -2}$ for a primitive $n$th root $\beta$, then its minimum Hamming distance is at least $\Delta$.

Choose a root $\beta$ of $g$ in an algebraic closure of ${\mathbb{F}}_2$. The factorization implies ${\beta }^{23}=1$, while $g(1)=1$ in ${\mathbb{F}}_2$, so $\beta \ne 1$. As $23$ is prime, $\beta$ has order $23$. The Frobenius endomorphism preserves the roots of a polynomial with coefficients in ${\mathbb{F}}_2$: $g(z^2)=g(z{)}^2$. Thus every ${\beta }^{2^j}$ is a root. In particular, $2^8=256\equiv 3(\mathrm{mod}23)$ shows thatare roots of $g$. These four consecutive roots give $d(C)\ge 5$ by the BCH bound. A further argument is needed to exclude weights five and six.

Put $S(X)=1+X+\cdots +X^{22}$. Cancelling $X+1$ in the supplied factorization gives $g(X)g^{\mathrm{rev}}(X)=S(X)$. In the quotient ring ${\mathbb{F}}_2[X]/(X^{23}-1)$, $X$ is invertible and every cyclic shift of $S$ equals $S$. Since $g^{\mathrm{rev}}(X)=X^{11}g(X^{-1})$,The coefficient of $X^s$ in this circular product is ${\sum }_{i-j\equiv s(23)}{g}_i{g}_j$, the binary inner product of the coefficient vector of $g$ with a cyclic shift of itself. Every coefficient of $S$ is one. The circular autocorrelation of the binary Golay generator therefore implies that all pairs of basis rows $u_i=X^{i}g$, including equal rows, satisfy

Each $u_i$ has Hamming weight seven. Append its parity bit, which is one, and write ${\hat{u}}_i=(u_i,1)$. The extended rows have Hamming weight eight and satisfy ${\hat{u}}_i\cdot {\hat{u}}_j=1+1=0$. Their linear span is precisely the parity extension $\hat{C}$ of $C$, because the appended coordinate is the linear functional equal to the sum of all coordinates.

These extended generators produce a doubly even code. To prove this rather than assume it, useAll the extended generators have weights divisible by four. A sum of previous generators is orthogonal to the next generator, so the intersection size in this identity is even. Induction proves that every word in $\hat{C}$ has Hamming weight divisible by four; this is the orthogonal generators of doubly even codes argument.

An original word of weight five acquires a parity bit and has extended weight six, while a word of weight six acquires a zero parity bit and still has extended weight six. Both contradict divisibility by four. Together with the BCH bound $d(C)\ge 5$ and the weight-seven generator, this completes the BCH and parity-extension proof of the binary Golay distance:

Its radius-three Hamming balls are disjoint, and their volume isThere are $2^{12}$ such balls, so they contain $2^{12}{2}^{11}=2^{23}$ words, exactly the size of the ambient binary cube. They therefore cover the cube, with every word in exactly one ball. Hence

Assume $K\subseteq L^p$. Every $a\in A$ has a unique $p$th root $r(a)$ in $L$: existence is the field inclusion and uniqueness follows from injectivity of the Frobenius endomorphism in a field. Since $r(a{)}^p=a$, this root satisfies the monic polynomial $T^p-a$ over $B$. Because $B$ is an integrally closed domain, this gives $r(a)\in B$. The identitiesand uniqueness show thatis a ring homomorphism.

On $D(a)$ the rule sends a fraction to the fraction of its unique roots. Since $r(a{)}^p=a$, inverting $a$ is equivalent to inverting $r(a)$, so the rule localizes. Also, for a prime $\mathfrak{q}\subset B$, $r(a)\in \mathfrak{q}$ exactly when $a\in \mathfrak{q}$; hence the induced point map agrees with that of $f$. The localized rules consequently glue to a morphism $h:X\to Y$. Its pullback followed by the Frobenius endomorphism sends $a$ to $r(a{)}^p=a$, which provesIf another morphism satisfies this equation, it has the same point map because the Absolute Frobenius morphism fixes points, and its pullback of $a$ must be a $p$th root of $a$ in the normal domain $B$. That root is unique. Thus $h$ is unique.

The Absolute Frobenius morphism is the identity on the underlying topological space and acts on the structure sheaf by $a\mapsto a^p$. On an affine scheme it is induced by the ring homomorphism $A\to A$, $a\mapsto a^p$. This is additive in characteristic $p$, and the inverse image of a prime ideal $\mathfrak{p}$ is again $\mathfrak{p}$, since $a^p\in \mathfrak{p}$ exactly when $a\in \mathfrak{p}$. These affine maps agree under localization and hence glue.

A perfect ring of characteristic p is one whose Frobenius endomorphism is bijective. Since the Absolute Frobenius morphism fixes points, its map on the stalk at $x$ is precisely the Frobenius endomorphism of ${\mathcal{O}}_{X,x}$. A morphism of sheaves is an isomorphism exactly when it is an isomorphism on all stalks. ThereforeIn the reverse direction the inverse sheaf map, together with the identity on points, gives the inverse morphism of locally ringed spaces. This is perfection in the Frobenius sense, not the unrelated homological use of the word perfect.

For the function field assertions below, interpret the normal schemes as integral, as is implicit in writing a single field $k(X)$ or $k(Y)$. General normal schemes need not be irreducible; without this interpretation those fields are not defined globally as single fields. Set $K=k(Y)\subseteq L=k(X)$. For an affine open $V=\mathrm{Spec}A$ in $Y$, finiteness gives $f^{-1}V=\mathrm{Spec}B$, with $A\subseteq B$ finite and both rings normal domains. In fact $B$ is the integral closure of $A$ in $L$: any element of $L$ integral over $A$ is also integral over $B$ and therefore belongs to $B$. This local description yields the two forms of Frobenius factorization through a finite normal cover.

If $A$ is nonempty, take $a\in A$. The stated implication repeatedly gives $a+jt\in A$ for $j=0,1,\dots ,p-1$. These residues are all distinct: equality of two would give $(j-\ell )t=0$ in the prime field ${\mathbb{Z}}_p$, and the nonzero $t$ can be cancelled. They therefore exhaust the field. ThusEquivalently, a nonzero element generates the additive cyclic group of prime order.

Now assume $0<k<p$. If $A+s=A+t$ with $s\ne t$, translating by $-s$ gives $A=A+(t-s)$, contradicting the preceding result. Hence all $p$ translates are distinct. Translation defines equivalence classes on the collection of all $k$-element subsets: two subsets are equivalent if one is a translate of the other, and each class contains exactly $p$ subsets. Since there are $\left(\genfrac{}{}{0ex}{}{p}{k}\right)$ such subsets, their partition into classes givesThis is the free translation action on nontrivial subsets of a prime cyclic group.

The binomial theorem now gives, in the prime field,Starting with $0^p=0$ and applying this identity successively to the representatives $0,1,\dots ,p-1$ proves $a^p=a$ for every residue $a$. For nonzero $a$, cancellation giveswhich is Fermat's little theorem, or $a^p\equiv a(\mathrm{mod}p)$ for every integer $a$ and $a^{p-1}\equiv 1$ when $p\nmid a$.

The same vanishing of intermediate binomial coefficients proves $(R+S{)}^p=R^p+S^p$ for any commuting elements in a ring whose ring characteristic is $p$, in particular for polynomials over ${\mathbb{Z}}_p$. Applying this repeatedly to the terms of $Q$, and applying Fermat's little theorem to each coefficient, gives the Frobenius endomorphism identityThis is an equality of formal polynomials. It does not assert $x^p=x$ as a polynomial; that latter equality holds only after evaluating at elements of ${\mathbb{Z}}_p$.

For $1\le j\le p-1$, the identity for binomial coefficientsshows that $p$ divides $j\left(\genfrac{}{}{0ex}{}{p}{j}\right)$. Since $p$ is a prime number and $j$ is not divisible by $p$, cancellation modulo $p$ givesThe binomial theorem consequently gives $(1+x{)}^p\equiv 1+x^p$ as a coefficientwise polynomial modular congruence. Apply the same identity with $x$ replaced by $x^{p^r}$ and use mathematical induction to obtainThis is an iteration of the Frobenius endomorphism in characteristic $p$. Comparing the intermediate coefficients gives the prime-power binomial divisibility result

Put $q=p^n$. Every extension of $k={\mathbb{F}}_p$ has characteristic $p$, so the iterated Frobenius endomorphism satisfiesThe root set is $K=\{a\in F:a^q=a\}$. It contains $0$ and $1$. If $a,b\in K$, thenso it is closed under subtraction and multiplication. If $a\in K$ is nonzero, thenThus $K$ is closed under multiplicative inverses as well, and is therefore a field. This is the polynomial characterization of a finite field; in fact it is the finite field with $q$ elements, since $X^q-X$ has degree $q$ and no repeated roots.

On every open set $U$, define $F_U^{\#}(a)=a^p$. In characteristic of a ring $p$, the binomial theorem gives $(a+b{)}^p=a^p+b^p$, so these are ring homomorphisms; they commute with restrictions and hence define a morphism of sheaves of rings.

On an affine chart $\mathrm{Spec}A$, the inverse image of a prime ideal $\mathfrak{p}$ under the Frobenius endomorphism isby primality. The induced continuous map is therefore the identity. These local morphisms agree on overlaps, giving the Absolute Frobenius morphism $F_X:X\to X$. Its action on the underlying space and on every local section was prescribed, so the morphism is unique.

Let $L$ be a splitting field of $f(X)=X^m-1$ over $k$. Its formal derivative isThe assumption on the characteristic of $k$ says that $m\ne 0$ in $k$. Any root $\alpha$ of $f$ is nonzero, and hence $f^{\prime }(\alpha )=m{\alpha }^{m-1}\ne 0$. Thus $f$ and $f^{\prime }$ have no common root, so $f$ is a separable polynomial. Since it has degree $m$ and splits over $L$, it has exactly $m$ distinct roots. This is the separability of roots of unity.

Fix a primitive root of unity ${\zeta }_m=e^{2\pi i/m}$. Define the $m$th cyclotomic polynomial byIt is monic and has degree given by the Euler totient function $\phi (m)$. Partitioning all $m$th roots of unity according to their exact multiplicative order gives the cyclotomic factorizationWe now prove by mathematical induction that ${\Phi }_m\in \mathbb{Z}[X]$. The base case is ${\Phi }_1=X-1$. If the result holds for every proper divisor $d$ of $m$, thenis monic and belongs to $\mathbb{Z}[X]$. Polynomial division of $X^m-1$ by the monic integer polynomial $G$ produces a quotient and remainder in $\mathbb{Z}[X]$. The factorization over $\mathbb{C}$ says that the remainder is zero and the quotient is ${\Phi }_m$, proving the claim.

It remains to prove irreducibility of cyclotomic polynomials. Let $f\in \mathbb{Z}[X]$ be a monic irreducible polynomial dividing ${\Phi }_m$, let $\epsilon$ be a root of $f$, and write ${\Phi }_m=fg$. We claim that ${\epsilon }^p$ is a root of $f$ for every prime number $p\nmid m$. Otherwise ${\epsilon }^p$ is a root of $g$, so $\epsilon$ is a root of $g(X^p)$. Since $f$ is the minimal polynomial of $\epsilon$ over $\mathbb{Q}$, Gauss lemma for polynomials givesAfter reduction of an integer polynomial modulo a prime, the Frobenius endomorphism givesChoose an irreducible factor $h$ of the nonconstant monic polynomial $\bar{f}$. Then $h\mid \bar{g}$, so $h^2\mid \bar{f}\bar{g}={\bar{\Phi }}_m$. The cyclotomic factorization would then make $h^2$ divide $X^m-1$ over ${\mathbb{F}}_p$. This is impossible because its derivative $m{X}^{m-1}$ is coprime to it when $p\nmid m$.

Thus $f({\epsilon }^p)=0$ for every prime $p\nmid m$. Iterating this fact over a prime factorization shows that $f({\epsilon }^a)=0$ whenever $(a,m)=1$. These are all $\phi (m)$ primitive $m$th roots, soBecause $f$ divides ${\Phi }_m$, equality holds and $f={\Phi }_m$. Hence every cyclotomic polynomial is irreducible over $\mathbb{Q}$.

Finally, direct multiplication givesAlsoApplying the cyclotomic factorization to the two factors on the right givesand thereforeSince $X^2-X+1={\Phi }_6(X)$, cancellation yieldsThe thirtieth cyclotomic polynomial is irreducible over $\mathbb{Q}$ by the theorem just proved, so $F$ is irreducible.

Write $q=p^r$, where $p=\mathrm{char}K$. Forthe derivative isso $h$ is a separable polynomial and all its roots in its splitting field $L$ are distinct.

Let $R$ be the set of these roots. The Frobenius endomorphism and the identitiesshow that $R$ is closed under addition and multiplication. It is also closed under additive inverses and, for $x\ne 0$, inversion. Thus $R$ is a finite field. Since $h$ splits into $q^n$ distinct linear factors, $|R|=q^n$. The field generated by all roots is therefore $R$ itself, so $L=R$ andHence the polynomial characterization of a finite field gives

Now let $f\in K[X]$ be irreducible of degree $n$, and let $\alpha$ be a root in an algebraic closure. The minimal polynomial of an algebraic element givesso $K(\alpha )$ has $q^n$ elements. Its nonzero elements form a finite group under multiplication, of order $q^n-1$; hence ${\alpha }^{q^n}=\alpha$. Therefore $\alpha$ is a root of $X^{q^n}-X$, and its minimal polynomial $f$ divides every polynomial over $K$ that vanishes at $\alpha$. Thus the irreducible factors of a finite-field Frobenius polynomial satisfy

For any prime $p$, takeThe Eisenstein criterion, applied at the prime element $t$ of ${\mathbb{F}}_p[t]$, shows that $f$ is irreducible over $K$ and therefore has degree $p$. If $\alpha$ is a root in any extension field, the Frobenius endomorphism givesIt follows that this purely inseparable polynomial over a rational function field has at most one distinct root in every extension, and that root has multiplicity $p$, as required.

For $a\in k$ and each $n$, choose a lift $x_n\in {\mathcal{O}}_K$ of the unique $p^n$th root $a^{p^{-n}}$, which exists because $k$ is a perfect field. DefineChanging $x_n$ by an element of the maximal ideal changes its $p^n$th power by an element whose valuation tends to infinity, so the limit exists and is independent of all choices. In characteristic $p$, the Frobenius endomorphism satisfies $(x+y{)}^{p^n}=x^{p^n}+y^{p^n}$, making $[\cdot ]$ a ring homomorphism lifting the identity on $k$. If $s:k\to {\mathcal{O}}_K$ is any other such section, thenfor every $n$, and the same limiting construction forces $s(a)=[a]$. This proves uniqueness of the Teichmuller lift.

Choose a uniformizer $t$. Repeatedly subtracting the lift of the residue and dividing by $t$ gives every $x\in {\mathcal{O}}_K$ a unique convergent Teichmuller expansionBecause the lift is a ring map, this identifies ${\mathcal{O}}_K$ with $k[[t]]$ and its fraction field with the Laurent series field $k((t))$. This is the equal-characteristic complete discretely valued field classification.

If $K$ is locally compact, its compact valuation ring has only finitely many disjoint residue-class balls. Thus $k$ is finite, as also follows from the local compactness criterion for a complete non-Archimedean field.

An irreducible degree-$n$ polynomial over ${\mathbb{F}}_q$ is separable and its roots lie in ${\mathbb{F}}_{q^n}$, with any one root generating that field. Hence its splitting field is ${\mathbb{F}}_{q^n}$ andgenerated by the Frobenius automorphism $x\mapsto x^q$.