Huber spaces-4

In the previous posts, we have just talked about one single f-adic ring. In this post, we will say something about the morphisms between f-adic rings.

Suppose that A,B two f-adic rings with A_0,B_0 their rings of definition respectively, and I_0,J_0 their ideals of definition respectively. A ring morphism F:A\rightarrow B is called an adic morphism if F(A_0)\subset B_0 and F(I_0)B_0 is an ideal of definition associated to B_0 for B.

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 f-adic ring A, U\subset A is a subset, and U\cdot A is open in A, then for any V open neighborhood of 0 in A, we have that U^n\cdot V is also open in A for any positive integer n.

This is a rather surprising result. Suppose that A_0 is a ring of definition for A and I_0 is an ideal of definition for A associated to A_0. Up to changing I_0 for its some power I_0^n, we can assume that I_0\subset V\bigcap (U\cdot A). Write G the finite set of generators of the ideal I_0. Then since G\subset I_0\subset U\cdot A, we can find a finite set G' of A such that G\subset U\cdot G'. Since G' is finite, thus bounded, so G'\cdot I_0^N\subset V for some N. So we have that I_0^{N+1}=G\cdot I_0^N\subset U\cdot G'\cdot I_0^N\subset U\cdot V, thus showing that V\cdot U is open. Then U\cdot U\cdot V is also open, and we can deduce that U^n\cdot V is open for all n.

So, in the proof we used in an essential way the fact that I_0 is finitely generated.

Then let’s start with some simple properties for adic morphisms.

F:A\rightarrow B is an adic morphism, then (1) F is bounded(sending bounded sets to bounded sets);(2) if A_0 is a ring of definition for A and B' is an open subset of B such that F(A_0)\subset B', then there exists a ring of definition B_0 for B such that F(A_0)\subset B_0\subset B';(3)if A_1 is a ring of definition for A and I_1 the associated ideal of definition, and B_1 a ring of definition for B such that F(A_1)\subset B_1, then F(I_1)B_1 is an ideal of definition for B associated to B_1.

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, F(A_0)\subset B_0, so any subset of A_0 is sent to a bounded subset of B. For any bounded subset S\subset A(not necessarily contained in A_0), S\cdot I_0^N\subset A_0 for some N, so F(S\cdot I_0^N)\subset (F(I_0)\cdot B_0)^N\cdot F(S) is bounded. Yet F(I_0)\cdot B_0 is an ideal of definition, thus showing that F(S) is bounded. The second point is easy. For the third point, we need a lemma:

Lemma: two rings of definition A_0\subset A_1\subset A, and I_0\subset A_0 an open ideal of A_0, then I_0 is an ideal of definition associated to A_0 if and only if I_0\cdot A_1=I_0A_1 is an ideal of definition associated to A_1.

Proof: If I_0 is an ideal of definition, then for any open neighborhood U of 0 in A, U\bigcap A_1 is again open in A_1, thus I_0^N\subset U\bigcap A_1\subset U. Moreover, A_1 is bounded, thus (I_0\cdot A_1)^M=I_0^M\cdot A_1\subset U for some M>N. On the other hand, I_0 is open, and A_1 is a subring, thus ((I_0\cdot A_1)^n) defines a fundamental system of neighborhoods of 0 in A, so I_0\cdot A_1 is an ideal of definition associated to A_1. Conversely, if I_0A_1 is an ideal of definition, then I_0\subset I_0A_1  and that I_0 is open show that (I_0^n) defines a fundamental system of neighborhoods of 0 in A, thus I_0 is an ideal of definition associated to A_0.

Now return to our proof to the third point. We take the original notations in the definition of adic morphism. And we consider A_2=A_0\cdot A_1, it is also a ring of definition. If we write B_2=B_0\cdot B_1, then F(A_2)\subset B_2. Since I_1 is an ideal of definition associated to A_1, so I_0^M\subset I_1^N\subset I_0 for some M,N, so F(I_0^M)B_0\subset F(I_1^N)B_0\subset F(I_1)B_0, so F(I_1^N)B_0 is open in B. And thus F(I_1^N)B is open in B, so according to one proposition above, we have that F(I_1^N)B_1 is open in B. Moreover, for any open set U\subset B_1 containing 0, then F(I_1^{LN})B_0\subset F(I_0)^LB_0\subset U. Thus F(I_1^{LN})B_1\subset F(I_1^{LN})B_0B_1\subset F(I_0^L)B_0B_1. We can choose L large enough such that F(I_0^L)B_0B_1\subset U, thus we have that F(I_1^{LN})B_1\subset U, showing that (F(I_1^{LN})B_1) defines a fundamental system of neighborhoods of 0 in B, and thus F(I_1^{LN})B_1 is an ideal of definition associated to B_1 for B, and so is F(I_1)B_1 which completes the proof.

The following are some simple consequences of the above results:

A,B,C three f-adic rings, and F:A\rightarrow B,G:B\rightarrow C ring morphisms, then (0)if F is adic morphism, then it is continuous; (1)if F,G are adic morphism, then so is G\circ F;(2)F,G are both continuous, and G\circ F is adic, then so is G;(3) A'\subset A,B'\subset B are open subrings such that F(A')\subset B', then F is adic if and only if F|_{A'}: A'\rightarrow B' is adic.

The zeroth and the first points are easy. For the second point, since A_0 ring of definition for A is bounded, so is F(A_0). Since C_0 ring of definition for C is open, so is G^{-1}(C_0), thus there exists a ring of definition B_0 for B such that F(A_0)\subset B_0\subset G^{-1}(C_0). And it is not difficult to find an ideal of definition J_0 associated to B_0 such that G(J_0)C_0 is an ideal of definition for C_0. The third point is not difficult, either.

The last proposition concerns with adic morphisms between Tate rings:

A,B two f-adic rings and F:A\rightarrow B is a continuous ring morphism, if A is Tate ring , then B is Tate ring, too, and F is adic morphism.

The proof is not difficult. Since A has a unit element a that is topologically nilpotent, then its image in B is also a unit, and also topologically nilpotent, thus showing that B is Tate ring. What is more, if B_0 is a ring of definition containing a, then A'=F^{-1}(B_0) is an open subring, so it is f-adic ring. Since A' contains a, a unit topologically nilpotent element, so A' is Tate ring, so there is a ring of definition A'_0 for A' such that A'=A_0'[a^{-1}]. Since A' is open in A, this shows that A'_0 is a ring of definition for A. Then it is easy to show that F_{A'}:A'\rightarrow B_0 is adic, and so is F.

In the next post, I will say something on the completion of f-adic rings.

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