In the previous posts, we have just talked about one single adic ring. In this post, we will say something about the morphisms between adic rings.
Suppose that two adic rings with their rings of definition respectively, and their ideals of definition respectively. A ring morphism is called an adic morphism if and is an ideal of definition associated to for .
One of the most basic problem concerning this definition is whether the definition depends on the choice of rings of definition. Before all, we have a small lemma which turns out to be very useful later:
In an adic ring , is a subset, and is open in , then for any open neighborhood of in , we have that is also open in for any positive integer .
This is a rather surprising result. Suppose that is a ring of definition for and is an ideal of definition for associated to . Up to changing for its some power , we can assume that . Write the finite set of generators of the ideal . Then since , we can find a finite set of such that . Since is finite, thus bounded, so for some . So we have that , thus showing that is open. Then is also open, and we can deduce that is open for all .
So, in the proof we used in an essential way the fact that is finitely generated.
Then let’s start with some simple properties for adic morphisms.
is an adic morphism, then (1) is bounded(sending bounded sets to bounded sets);(2) if is a ring of definition for and is an open subset of such that , then there exists a ring of definition for such that ;(3)if is a ring of definition for and the associated ideal of definition, and a ring of definition for such that , then is an ideal of definition for associated to .
So the last point shows that the definition of adic morphism does not depend on the choice of rings of definition.
The first point is rather straightforward, since in the definition, , so any subset of is sent to a bounded subset of . For any bounded subset (not necessarily contained in ), for some , so is bounded. Yet is an ideal of definition, thus showing that is bounded. The second point is easy. For the third point, we need a lemma:
Lemma: two rings of definition , and an open ideal of , then is an ideal of definition associated to if and only if is an ideal of definition associated to .
Proof: If is an ideal of definition, then for any open neighborhood of in , is again open in , thus . Moreover, is bounded, thus for some . On the other hand, is open, and is a subring, thus defines a fundamental system of neighborhoods of in , so is an ideal of definition associated to . Conversely, if is an ideal of definition, then and that is open show that defines a fundamental system of neighborhoods of in , thus is an ideal of definition associated to .
Now return to our proof to the third point. We take the original notations in the definition of adic morphism. And we consider , it is also a ring of definition. If we write , then . Since is an ideal of definition associated to , so for some , so , so is open in . And thus is open in , so according to one proposition above, we have that is open in . Moreover, for any open set containing , then . Thus . We can choose large enough such that , thus we have that , showing that defines a fundamental system of neighborhoods of in , and thus is an ideal of definition associated to for , and so is which completes the proof.
The following are some simple consequences of the above results:
three adic rings, and ring morphisms, then (0)if is adic morphism, then it is continuous; (1)if are adic morphism, then so is ;(2) are both continuous, and is adic, then so is ;(3) are open subrings such that , then is adic if and only if is adic.
The zeroth and the first points are easy. For the second point, since ring of definition for is bounded, so is . Since ring of definition for is open, so is , thus there exists a ring of definition for such that . And it is not difficult to find an ideal of definition associated to such that is an ideal of definition for . The third point is not difficult, either.
The last proposition concerns with adic morphisms between Tate rings:
two adic rings and is a continuous ring morphism, if is Tate ring , then is Tate ring, too, and is adic morphism.
The proof is not difficult. Since has a unit element that is topologically nilpotent, then its image in is also a unit, and also topologically nilpotent, thus showing that is Tate ring. What is more, if is a ring of definition containing , then is an open subring, so it is adic ring. Since contains , a unit topologically nilpotent element, so is Tate ring, so there is a ring of definition for such that . Since is open in , this shows that is a ring of definition for . Then it is easy to show that is adic, and so is .
In the next post, I will say something on the completion of adic rings.