Definition 87.6.1. Let $A$ and $B$ be pre-adic topological rings. A ring homomorphism $\varphi : A \to B$ is adic1 if there exists an ideal of definition $I \subset A$ such that the topology on $B$ is the $I$-adic topology.
87.6 Adic ring maps
Let us make the following definition.
If $\varphi : A \to B$ is an adic homomorphism of pre-adic rings, then $\varphi $ is continuous and the topology on $B$ is the $I$-adic topology for every ideal of definition $I$ of $A$.
Lemma 87.6.2. Let $A \to B$ and $B \to C$ be continuous homomorphisms of pre-adic rings. If $A \to B$ and $B \to C$ are adic, then $A \to C$ is adic.
Proof. Omitted. $\square$
Lemma 87.6.3. Let $A \to B$ and $B \to C$ be continuous homomorphisms of pre-adic rings. If $A \to C$ is adic, then $B \to C$ is adic.
Proof. Choose an ideal of definition $I$ of $A$. As $A \to C$ is adic, we see that $IC$ is an ideal of definition of $C$. As $B \to C$ is continuous, we can find an ideal of definition $J \subset B$ mapping into $IC$. As $A \to B$ is continuous the inverse image $I' \subset I$ of $J$ in $I$ is an ideal of definition of $A$ too. Hence $I'C \subset JC \subset IC$ is sandwiched between two ideals of definition, hence is an ideal of definition itself. $\square$
Lemma 87.6.4. Let $\varphi : A \to B$ be a continuous homomorphism between pre-adic topological rings. If $\varphi $ is adic, then $\varphi $ is taut.
Proof. Immediate from the definitions. $\square$
The next lemma says two things
the property of being adic ascents along taut maps of complete linearly topologized rings, and
the properties “$\varphi $ is taut” and “$\varphi $ is adic” are equivalent for continuous maps $\varphi : A \to B$ between adic rings if $A$ has a finitely generated ideal of definition.
Because of (2) we can say that “tautness” generalizes “adicness” to continuous ring maps between arbitrary linearly topologized rings. See also Section 87.23.
Lemma 87.6.5. Let $\varphi : A \to B$ be a continuous map of linearly topologized rings. If $\varphi $ is taut, $A$ is pre-adic and has a finitely generated ideal of definition, and $B$ is complete, then $B$ is adic and has a finitely generated ideal of definition and the ring map $\varphi $ is adic.
Proof. Choose a finitely generated ideal of definition $I$ of $A$. Let $J_ n$ be the closure of $\varphi (I^ n)B$ in $B$. Since $B$ is complete we have $B = \mathop{\mathrm{lim}}\nolimits B/J_ n$. Let $B' = \mathop{\mathrm{lim}}\nolimits B/I^ nB$ be the $I$-adic completion of $B$. By Algebra, Lemma 10.96.3, the $I$-adic topology on $B'$ is complete and $B'/I^ nB' = B/I^ nB$. Thus the ring map $B' \to B$ is continuous and has dense image as $B' \to B/I^ nB \to B/J_ n$ is surjective for all $n$. Finally, the map $B' \to B$ is taut because $(I^ nB')B = I^ nB$ and $A \to B$ is taut. By Lemma 87.5.10 we see that $B' \to B$ is open and surjective. Thus the topology on $B$ is the $I$-adic topology and the proof is complete. $\square$
[1] This may be nonstandard terminology.
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)