Proposition 1

Let $A$ and $B$ be local rings such that $A$ is a subring of $B$. The following conditions are equivalent: (a) $m(A) \subset m(B)$; (b) $B$ dominates $A$; (c) the ideal $Bm(A)$ generated by $m(A)$ in $B$ does not contain 1.

If $m(A) \subset m(B)$, $m(B) \cap A$ is an ideal of $A$ which does not contain 1 and which contains the maximal ideal $m(A)$; it is therefore equal to it and (a) implies (b). If $B$ dominates $A$, the ideal $Bm(A)$ is contained in $m(B)$ and hence does not contain 1; thus (b) implies (c). If (c) holds, $Bm(A)$ is contained in the unique maximal ideal $m(B)$ of $B$, whence (a).

Note that, if $K$ is a ring, the relation "B dominates A" is an order relation on the set of local subrings of $K$.

Let $A$ and $B$ be two local rings such that $B$ dominates $A$. The canonical injection $A \to B$ defines by taking quotients an isomorphism of the field $\kappa(A)$ onto a subfield of $\kappa(B)$; this isomorphism allows us to identify $\kappa(A)$ with a subfield of $\kappa(B)$.

Examples (1) Let $A$ be a Noetherian local ring and $\hat{A}$ its completion; the local ring $\hat{A}$ then dominates $A$ (Chapter 111, § 3, no. 5, Proposition 9). (2) Let $B$ be an integral domain, $A$ a subring of $B$, $p'$ a prime ideal of $B$ and $p = A \cap p'$. Then $pA_p \subset p'B_{p'}$, so that $B_{p'}$ dominates $A_p$.