# 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.