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.