Characteristic of a ring ($\mathrm{char}R$)

The characteristic of a ring $R$ is the least positive integer $n$ satisfying $n\cdot 1_R=0$, or zero if no such integer exists. A nonzero ring of positive characteristic has prime characteristic when it is an integral domain.

In a finite unital ring $R$, some positive multiple of $1$ is zero, because two of the finitely many successive multiples coincide. If $n=\mathrm{char}R$, the permutation $r\mapsto r+1$ has every cycle of a permutation of length exactly $n$: returning to $r$ after $k$ steps means $k1=0$, whose least positive solution is $n$. These disjoint cycles partition the finite set $R$, so $n$ divides its cardinality. The zero ring has cardinality and characteristic both one.

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 $n$.

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.