The Stacks project

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. $\square$

Comments (3)

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 and in the proof is never defined, so in the statement of (1) it is better to write "". (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 with .

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

Comment #2021 by on

Thanks, your comments are fixed here.

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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