Huber spaces-2

In the previous post we have introduced the adic rings, in this post we will introduce a type of ring which includes adic rings as open subrings.

A topological ring A is called f-adic ring if it contains an open subring A_0 for which the induced subspace topology is an I_0-adic topology where I_0 is a finitely-generated ideal of A_0.

The A_0 in the definition is called a ring of definition for A and the ideal I_0 is called an ideal of definition for A. We have three equivalent characterization for the ring of definition. That is

If A is an f-adic ring, then the following three statements are equivalent: (1)A_0\subset A is a ring of definition; (2)A_0 is open and an adic ring; (3)A_0 is open and bounded;

Proof: Before we show (1) implies (2), we have a little remark: (I^n) defines a fundamental system of neighborhoods of 0 of A(note here, not of A_0). This is rather trivial once we notice that each element  of A is bounded and I^n are all open in A. The implication of (1) to (2) is trivial. That (2) implies (3) is also trivial. So it remains to show the implication of (3) to (1). Suppose a ring of definition for A is B_0 and an ideal of definition for A in B_0 is J, since A_0 is open, we have that J^n\subset A_0 for some n. So, if G_J is the finite set of generators of J, then G_J^n\subset A_0, too. So, we define I_0 to be the ideal genertated by G_J^n(I_0=G_J^nA_0), which is finitely generated. Moreover, I_0\subset J^n, and I_0=G_J^nA_0\supset G_J^nJ^n\supset J^{2n}(here we use a rather simple yet important fact, that is, G_JJ=J^2, which garantes that G_JJ is open, otherwise it will be very difficult to show this point), thus proving that the I_0-adic topology and the induced subspace topology on A_0 are the same. What is more, I_0 is finitely generated, thus A_0 is a ring of definition.

The followings are some simple properties of f-adic rings:

A is an f-adic ring, (1)if A_0,A_1 are two rings of definitions for A, then A_0\bigcap A_1, A_0\cdot A_1 are again rings of definition;(2) Subset B of A is bounded and contains 1, and subset C of A is an open subring, what is more, B\subset C, then there exists a ring of definition A_0 for A such that B\subset A_0\subset C;(3) A^o is the union of rings of definition for A.

The proofs are not very difficult. For the first point, it is sufficient to show that A_0\bigcap A_1,A_0\cdot A_1 are open and bounded(to show that A_0\cdot A_1 is bounded, note that for any U an open neighborhood of 0 in A, there exists V another open neighborhood such that A_0\cdot V\subset U. Note that, A_0\cdot V is an open set! Thus there exists a third open neighborhood W of 0 in A such that A_1\cdot W\subset A_0\cdot U\subset V, thus we have that A_0\cdot A_1\cdot W\subset A_0\cdot A_0\cdot U=A_0\cdot U\subset V, thus showing the boundedness of A_0\cdot A_1). For the second point, we can take a ring of definition, say A_0 for A, then noting that since C is an open subring, A_0\bigcap C is a ring of definition for C, making C into an f-adic ring. It is easy to verify that B\cdot (A_0\bigcap C) is also a ring of definition for C, thus also for A(since C is open in A); For the third point, since any a\in A^o is power-bounded, thus there exists a ring of definition A_0 containing (a^n). So A^o is contained in the union of rings of definition. Conversely, since any ring of definition is bounded, so is any element in it, thus showing that the union of rings of definition is contained in A^o, and this finishes the proof for the third point.

The following result says something on A^o and A^{oo}. Combining the first and the third points, we get that A^o is actually an open subring of A.

A is f-adic ring, then A^o is integrally closed; for any subset S\subset A, S\cdot A is open if and only if A^{oo}\subset \sqrt{S\cdot A}.

For the first point, we observe that if a\in A satisfies a^n+b_{n-1}a^{n-1}+...+b_1a+b_0=0 with b_0,b_1,...,b_{n-1}\in A^o, we choose A_0 a ring of definition containing b_0,...,b_{n-1}, then A_0[a]=A_0\cdot\{1,a,a^2,...,a^{n-1}\}. Yet the second set \{1,a,a^2,...,a^{n-1}\} is a finite set, thus bounded, and so A_0[a] is bounded. It is clearly open, thus it is a ring of definition, and this shows that a is power-bounded, thus a\in A^o, so A^o is integrally closed. For the second point, if S\cdot A is open, then for any a\in A^{oo}, a^n\rightarrow 0, so a^N\in S\cdot A, thus a\in \sqrt{S\cdot A}, and so A^{oo}\subset \sqrt{S\cdot A}. If A^{oo}\subset \sqrt{S\cdot A}, then note that since (I_0^n) defines a fundamental system of neighborhoods of 0, so each element in I_0 is topologically nilpotent, thus I_0\subset A^{oo}. That is to say, I_0\subset \sqrt{S\cdot T}. Yet I_0 is finitely generated, so there exists an integer n such that I_0^n\subset S\cdot T, thus showing that S\cdot T is open.

In the next post, I will say something about Tate-rings, which are a special kind of f-adic rings.


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