Proposition 1

— Let $A$ be a ring not reduced to 0. a) For $A$ to be of characteristic 0 it is necessary and sufficient that the mapping $x \mapsto n \cdot x$ of $A$ into itself should be bijective, for every integer $n \neq$

b) Let $p$ be a prime number. For $A$ to be of characteristic $p$ it is necessary and sufficient that $p \cdot x = 0$ for all $x \in A$.

Let $f$ be the unique homomorphism of $\mathbf{Z}$ into $A$; we have $n \cdot x = f(n)x$ for every integer $n$ and every $x$ in $A$. For $A$ to be of characteristic 0, it is necessary and sufficient that $f$ should extend to a homomorphism of $\mathbf{Q}$ into $A$, that is, $f(n)$ should be invertible in $A$ for every $n \neq 0$ (I, p. 113); this proves $a$. Similarly for $A$ to be of characteristic $p$ it is necessary and sufficient that $f$ should annihilate $p\mathbf{Z}$, that is, $f(p) = 0$, or also that $p \cdot x = 0$ for all $x \in A$; this proves $b$.

Let us take for $A$ a not necessarily commutative field. The centre of $A$ is a (commutative) field; therefore the characteristic and the prime subfield of $A$ are defined.