Lemma 85.15.10. 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. $\text{Spf}(\varphi ) : \text{Spf}(B) \to \text{Spf}(A)$ is representable by algebraic spaces,

2. $\text{Spf}(\varphi ) : \text{Spf}(B) \to \text{Spf}(A)$ is representable (by schemes),

3. $\varphi$ is taut, see Definition 85.4.11.

Proof. Parts (1) and (2) are equivalent by Lemma 85.15.7.

Assume the equivalent conditions (1) and (2) hold. If $I \subset A$ is a weak ideal of definition, then $\mathop{\mathrm{Spec}}(A/I) \to \text{Spf}(A)$ is representable and a thickening (this is clear from the construction of the formal spectrum but it also follows from Lemma 85.5.6). Then $\mathop{\mathrm{Spec}}(A/I) \times _{\text{Spf}(A)} \text{Spf}(B) \to \text{Spf}(B)$ is representable and a thickening as a base change. Hence by Lemma 85.5.6 there is a weak ideal of definition $J(I) \subset B$ such that $\mathop{\mathrm{Spec}}(A/I) \times _{\text{Spf}(A)} \text{Spf}(B) = \mathop{\mathrm{Spec}}(B/J(I))$ as subfunctors of $\text{Spf}(B)$. We obtain a cartesian diagram

$\xymatrix{ \mathop{\mathrm{Spec}}(B/J(I)) \ar[d] \ar[r] & \mathop{\mathrm{Spec}}(A/I) \ar[d] \\ \text{Spf}(B) \ar[r] & \text{Spf}(A) }$

By Lemma 85.12.4 we see that $B/J(I) = B \widehat{\otimes }_ A A/I$. It follows that $J(I)$ is the closure of the ideal $\varphi (I)B$, see Lemma 85.4.13. Since $\text{Spf}(A) = \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Spec}}(A/I)$ with $I$ as above, we find that $\text{Spf}(B) = \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Spec}}(B/J(I))$. Thus the ideals $J(I)$ form a fundamental system of weak ideals of definition (see Lemma 85.5.6). Hence (3) holds.

Assume (3) holds. We are essentially just going to reverse the arguments given in the previous paragraph. Let $I \subset A$ be a weak ideal of definition. By Lemma 85.12.4 we get a cartesian diagram

$\xymatrix{ \text{Spf}(B \widehat{\otimes }_ A A/I) \ar[d] \ar[r] & \mathop{\mathrm{Spec}}(A/I) \ar[d] \\ \text{Spf}(B) \ar[r] & \text{Spf}(A) }$

If $J(I)$ is the closure of $IB$, then $J(I)$ is open in $B$ by tautness of $\varphi$. Hence if $J$ is open in $B$ and $J \subset J(B)$, then $B/J \otimes _ A A/I = B/(IB + J) = B/J(I)$ because $J(I) = \bigcap _{J \subset B\text{ open}} (IB + J)$ by Lemma 85.4.2. Hence the limit defining the completed tensor product collapses to give $B \widehat{\otimes }_ A A/I = B/J(I)$. Thus $\text{Spf}(B \widehat{\otimes }_ A A/I) = \mathop{\mathrm{Spec}}(B/J(I))$. This proves that $\text{Spf}(B) \times _{\text{Spf}(A)} \mathop{\mathrm{Spec}}(A/I)$ is representable for every weak ideal of definition $I \subset A$. Since every morphism $T \to \text{Spf}(A)$ with $T$ quasi-compact factors through $\mathop{\mathrm{Spec}}(A/I)$ for some weak ideal of definition $I$ (Lemma 85.5.4) we conclude that $\text{Spf}(\varphi )$ is representable, i.e., (2) holds. This finishes the proof. $\square$

Comment #1949 by Brian Conrad on

In the final paragraph, the mention of Lemma 11.4 seems not relevant (the proof gives a direct construction of a fiber product, rendering moot any need to invoke Lemma 11.4, it seems). When Lemma 4.8 is invoked there, it should also be mentioned that tautness of $\varphi$ is being used (to ensure that $J(I)$ is open in $B$).

Comment #2003 by on

OK, the key point in the last paragraph is that $J(I)$ is open as you say. But we cannot use the discussion in the previous paragraph for the construction of the fibre product, since that discussion was done under the assumption (1). Thus we do have to use Lemma 85.12.4 (11.4) to get the representability. I have edited this and I think it is a lot clearer now. Thanks!

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