Lemma 87.29.2. Let $S$ be a scheme. Let $\varphi : A \to B$ be a continuous map of weakly admissible topological rings over $S$. The following are equivalent

1. $\text{Spf}(\varphi ) : Y = \text{Spf}(B) \to \text{Spf}(A) = X$ is of finite type,

2. $\varphi$ is taut and $B$ is topologically of finite type over $A$.

Proof. We can use Lemma 87.19.8 to relate tautness of $\varphi$ to representability of $\text{Spf}(\varphi )$. We will use this without further mention below. It follows that $X = \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Spec}}(A/I)$ and $Y = \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Spec}}(B/J(I))$ where $I \subset A$ runs over the weak ideals of definition of $A$ and $J(I)$ is the closure of $IB$ in $B$.

Assume (2). Choose a ring map $A[x_1, \ldots , x_ r] \to B$ whose image is dense. Then $A[x_1, \ldots , x_ r] \to B \to B/J(I)$ has dense image too which means that it is surjective. Therefore $B/J(I)$ is of finite type over $A/I$. Let $T \to X$ be a morphism with $T$ a quasi-compact scheme. Then $T \to X$ factors through $\mathop{\mathrm{Spec}}(A/I)$ for some $I$ (Lemma 87.9.4). Then $T \times _ X Y = T \times _{\mathop{\mathrm{Spec}}(A/I)} \mathop{\mathrm{Spec}}(B/J(I))$, see proof of Lemma 87.19.8. Hence $T \times _ Y X \to T$ is of finite type as the base change of the morphism $\mathop{\mathrm{Spec}}(B/J(I)) \to \mathop{\mathrm{Spec}}(A/I)$ which is of finite type. Thus (1) is true.

Assume (1). Pick any $I \subset A$ as above. Since $\mathop{\mathrm{Spec}}(A/I) \times _ X Y = \mathop{\mathrm{Spec}}(B/J(I))$ we see that $A/I \to B/J(I)$ is of finite type. Choose $b_1, \ldots , b_ r \in B$ mapping to generators of $B/J(I)$ over $A/I$. We claim that the image of the ring map $A[x_1, \ldots , x_ r] \to B$ sending $x_ i$ to $b_ i$ is dense. To prove this, let $I' \subset I$ be a second weak ideal of definition. Then we have

$B/(J(I') + IB) = B/J(I)$

because $J(I)$ is the closure of $IB$ and because $J(I')$ is open. Hence we may apply Algebra, Lemma 10.126.9 to see that $(A/I')[x_1, \ldots , x_ r] \to B/J(I')$ is surjective. Thus (2) is true, concluding the proof. $\square$

Comment #1968 by Brian Conrad on

At the end of the sentence preceding this lemma, perhaps replace the period with a colon to make clear that this lemma is proving the assertion in that sentence.

The notation $X$ and $Y$ in the proof is never defined, so in the statement of (1) it is better to write "${\rm{Spf}}(\varphi): Y = {\rm{Spf}}(B) \rightarrow {\rm{Spf}}(A) = X$". (There is also the aesthetic issue that notation such as Spf, Spec, _{red}, etc. should never be in italics mode, even within the statement of the Theorem -- much as cos, sin, etc. are never put in italics -- but it seems too late to make this "correction".)

Comment #1969 by Brian Conrad on

Whoops, I forgot to enter a few more corrections:

1. In the 2nd paragraph of the proof, on line 4 replace $B/I$ with $A/I$.

2. On line -2 of the proof, put parentheses around $A/I'$ for clarity, and insert the missing period at the end of the sentence.

Comment #2021 by on

About the colon at the end of a sentence before a lemma. I've never understood why you have to do this, so I am going to leave it as it is, unless a bunch of other people chime in to agree with you.

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