Remark 1

Let $A$ and $A'$ be two rings not reduced to 0. Suppose that the characteristic of $A$ is defined and that there is a homomorphism $u$ of $A$ into $A'$. The image under $u$ of the prime subfield of $A$ is a subfield $P'$ of $A'$, isomorphic to $P$, and hence prime. It follows that the characteristic of $A'$ is defined and is equal to that of $A$. If $A$ and $A'$ are of characteristic 0 (resp. $p \neq 0$), the mapping $u$ is a homomorphism of algebras over $\mathbf{Q}$ (resp. $\mathbf{F}_p$).