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The positive characteristic of a ring which is an integral domain is a prime number. The characteristic cannot be one because $1\ne 0$. If its least positive value were $n=ab$ with $1<a,b<n$, minimality would make $a1$ and $b1$ nonzero, while $(a1)(b1)=n1=0$ would give zero divisors.

The analogous result holds for a nonzero commutative nonunital ring without zero divisors, using the characteristic of a rng. If $n=ab$ were composite and minimal, choose $r$ with $ar\ne 0$. For every $s$, $(ar)(bs)=n(rs)=0$, so cancellation forces $bs=0$ for every $s$, contradicting minimality of $n$.

The characteristic of a rng is the least positive integer annihilating every element of its additive abelian group, or zero if no such integer exists. In a unital ring this equals the characteristic of a ring: $n1=0$ implies $nr=(n1)r=0$, and annihilating every element in particular annihilates $1$. In a finite rng it is the exponent of a finite group of the additive group.

In a nonzero unital ring with finite odd unit group, the unit $-1$ has order dividing two. By Lagrange's theorem, that order also divides $|U(R)|$, so $-1=1$ and the characteristic of a ring is two. Consequently $\overline{a}\mapsto a{1}_R$ defines a unital ring homomorphism ${\mathbb{F}}_2\to R$. This map still exists for the zero ring, whose characteristic is one under the least-positive-integer convention.

The unique structure morphism $X\to \mathrm{Spec}\mathbb{Z}$ factors through the closed subscheme $\mathrm{Spec}(\mathbb{Z}/p\mathbb{Z})$ exactly when the ideal sheaf $p{\mathcal{O}}_X$ is zero. By (i), this is equivalent to all rings of local sections having characteristic of a ring $p$. Therefore

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.

The element $-1_R$ lies in the unit group and satisfies $(-1_R{)}^2=1_R$. Its order of a group element divides two and, by Lagrange's theorem, divides the odd number $|U(R)|$. Its order is therefore one, so $-1_R=1_R$ and $2\cdot 1_R=0$.

It follows thatis a well-defined unital ring homomorphism. For a nonzero ring, this says its characteristic of a ring is two. If the zero ring is permitted, the same map still exists. This is the odd unit group forces characteristic two argument, with that harmless exception to the characteristic wording.

Using the assumed square-free polynomial property, factorinto distinct monic polynomials that are irreducible polynomials, with $n_j=\mathrm{deg}{f}_j$. Part (b)(iii) identifies $T={\mathbb{F}}_2[X]/(X^p-1)$ with ${\prod }_j{K}_j$, where $|K_j|=2^{n_j}$.

By the first isomorphism theorem for rings, the image $S$ from part (c)(i) is $T/J$, where $J$ is the kernel of a ring homomorphism. By part (a)(ii), $J={\prod }_j{J}_j$ for ideals $J_j\subseteq K_j$. A field has only the zero ideal and the whole field as ideals, so each $K_j/J_j$ is either $K_j$ or the zero ring. Consequently, for a subset $D\subseteq \{1,\dots ,r\}$,This is the quotient of a finite product of fields principle. The subset is nonempty because $|U(S)|=p>2$, whereas the zero ring has only one unit. By part (a)(i),Each factor is a positive integer. Since $p$ is a prime number, exactly one factor is $p$ and all remaining factors are one. Thusso $p$ is a Mersenne prime. If $n=ab$ with $a,b>1$, then $2^a-1$ would be a proper divisor of $2^{ab}-1$; thus $n$ is itself a prime number, although only the displayed form is needed.

The assumed square-free polynomial property can also be checked directly: in characteristic of a ring two, the formal derivative of $X^p-1$ is $X^{p-1}$ because $p$ is odd. These two polynomials have greatest common divisor one, since $X$ does not divide $X^p-1$, so no repeated irreducible polynomial factor occurs.

The residue characteristic of a local ring is the characteristic of a ring of its residue field. A finite extension of ${\mathbb{Q}}_p$ has residue characteristic $p$.