Proposition 7

Let $A, A'$ be two commutative rings and $h : A \to A'$ a ring homomorphism. Suppose that $A$ is Noetherian and that $A'$ is a finitely generated $A$-module (with the structure defined by $h$). Let $m$ be an ideal of $A$ and let $m' = mA'$. Then:

(i) For the $m'$-adic topology on $A'$ to be Hausdorff, it is necessary and sufficient that the elements of $1 + h(m)$ be not divisors of $0$ in $A'$.

(ii) If $A$ with the $m$-adic topology is a Zariski ring, then $A'$ with the $m'$-adic topology is a Zariski ring.

(iii) If $h$ is injective (thus identifying $A$ with a subring of $A'$), the $m'$-adic topology on $A'$ induces on $A$ the $m$-adic topology.

Recall that the $m'$-adic filtration on $A'$ coincides with the $m$-adic filtration on the A-module $A'$ (§ 2, no. 1, Example 3). Assertion (i) is thus a special case of Proposition 5 of no. 2 and assertion (iii) a special case of Theorem 2 of no. 2. Finally let us show (ii). Suppose that $A$ is a Zariski ring with the $m$-adic topology and let $E'$ be a finitely generated $A'$-module; it is also a finitely generated $A$-module and the $m$-adic and $m'$-adic filtrations on $E'$ coincide; then $E'$ is Hausdorff with the $m'$-adic topology. Finally the $A$-module $A'$ is Noetherian and hence the ring $A'$ is Noetherian, which completes the proof that $A'$ is a Zariski ring.