Theorem 1

— Let $A$ be a ring ; suppose that there exists a subfield of $A$. Then $A$ has a unique subfield $P$ which is a prime field. Moreover, $P$ is contained in the centre of $A$ and in every subfield of $A$.

Let $K$ be a subfield of $A$, $C$ the centre of $A$, and put $K' = K \cap C$; then $K'$ is a subfield of $A$. Let $f$ be the unique homomorphism of $\mathbf{Z}$ into $A$ and $p$ its kernel. Every subring of $A$, in particular $K'$, contains $f(\mathbf{Z})$; hence the ideal $p$ is prime (I, p. 116-117). If $p = (0)$, the homomorphism $f$ of $\mathbf{Z}$ into $K'$ is injective; hence it extends (I, p. 116) to an isomorphism $\bar{f}$ of $\mathbf{Q}$ onto a subfield $P$ of $K'$. If $p \neq (0)$, there is a strictly positive integer $p$ such that $p = (p)$ (I, p. 111); if we had $p = ab$ with $a > 1$ and $b > 1$, this would mean $a \notin p, b \notin p$ and $ab \in p$ in contradiction with the fact that $p$ is prime. The number $p$ is therefore prime and by passage to quotients $f$ defines an isomorphism of $F_p = \mathbf{Z}/p$ onto a subfield $P$ of $K'$. In both cases $P$ is a subfield of $A$ contained in the centre $C$ of $A$, and it is a prime field. Let $L$ be a subfield of $A$; then $P \cap L$ is a subfield of $P$, and since $P$ is prime, we have $P \cap L = P$, whence $P \subset L$. If $P'$ is a subfield of $A$ and is a prime field, then by what has been said, $P \subset P'$, whence $P = P'$ because $P'$ is a prime field.