# 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.

$F:A\rightarrow B$ 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.