Huber spaces-3

in this post we will say something on Tate rings.

A ring A is a Tate ring if it is an f-adic ring and there is a unit element a\in A^* such that a\in A^{oo}.

Note that for any ring of definition A_0 for a Tate ring A, A_0\cdot A=A_0 is open, thus A^{oo}\subset \sqrt{A_0}, yet according to the definition, A^{oo}\bigcap A^*\neq\emptyset, so A_0 contains a unit element.

Suppose that a_0\in A_0\bigcap A^*\bigcap A^{oo}, then the morphism f: A\rightarrow A, a\mapsto a a_0 is an isomorphism. So, it sends the open set A_0 to an open set a_0A_0. Thus a_0A_0 is open in A_0, too. Suppose one ideal of definition associated to A_0 for A is I_0, then I_0^n\subset a_0A_0 for some n. Yet on the other hand, a_0^k\rightarrow 0, so a_0^k\in I_0 for some k, thus a_0^kA_0\subset I_0A_0=I_0, this shows that the I_0-adic topology on A_0 and the a_0A_0-adic topology on A_0 are the same. So, the ideals (a_0^mA_0) defines a fundamental system of neighborhoods of 0 for A. What is more, for any open neighborhood U of 0 in A, and any element a\in A, there is another neighborhood V of 0 in A such that aV\subset U. Yet there is some positive integer k such that a_0k\in V, thus a_0^k a\in U, this shows that a_0^n a\rightarrow 0. So, a_0^N a\in A_0 for some N. This means that A=A_0[a_0^{-1}]. It is not difficult to show that

A ring A is Tate ring if and only if there is a ring of definition A_0 with an ideal of definition a_0A_0 such that A=A_0[a_0].

So, in this sense, Tate rings are very special f-adic rings: the ideal of definition is a principal ideal, and in some sense, we can define its all integer powers, I_0^n(n\in \mathbb{Z}) such that they form a fundamental system of neighborhoods of 0 and A=\bigcup_{n\in\mathbb{Z}} I_0^n.

In the next post, we will define morphisms between f-adic rings.


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