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I have a question regarding a proof from Bosch's Algebraic Geometry book, namely section 9.5, Proposition 3, part (ii):

Let $f:X\rightarrow Y=\text{Spec}(B)$ be a morphism of schemes. $f$ is quasi-affine if and only if $X$ is quasi-affine.

My question is about the proof of the if - part, i.e. assuming $f$ is quasi-affine. The argument is as follows. Write $A = \mathcal{O}_X(X)$ and look at the canonical decomposition $$f: X\rightarrow \text{Spec}(A)\rightarrow Y = \text{Spec}(B).$$ For $g\in B$ consider the open set $D(g) = \text{Spec}(B_g)$ and its preiamge in $\text{Spec}(A)$, namely $\text{Spec}(A\otimes_B B_g)$. Denote by $X_g$ the preiamge $f^{-1}(D(g))$ in $X$ and write $f_g = f|_{X_g}$. Then $f_g$ decomposes into $X_g\rightarrow \text{Spec}(A\otimes_B B_g)\rightarrow \text{Spec}(B_g)$. Then the author takes a finite affine cover $(U_i)_{i\in I}$ of $X$. Since $f$ is separated, all intersections $U_i\cap U_j$ are affine and he consideres the exact sequence $$ A = \mathcal{O}_X(X)\rightarrow \prod_{i\in I}\mathcal{O}_X(U_i)\rightrightarrows \prod_{i,j\in I} \mathcal{O}_X(U_i\cap U_j).$$

Tensoring by $B_g$ over $B$ preserves exactness of the sequence. So far so good, no questions up until now: He then claims: ''This shows that the morphism $A\otimes_B B_g\rightarrow \mathcal{O}_X(X_g)$ is an isomorphism''

Why is that? I have seen this argument a few times now in different proofs and was never able to make any sense of it. Any comment is highly appreciated! :)

Bosch's definition of a quasi-affine scheme and quasi-affine morphism (from the comments):

A scheme $X$ is quasi-affine if it is quasi-compact and there is an open immersion $X\to\operatorname{Spec} R$ into some affine scheme.

A morphism $f:X\to Y$ is quasi-affine if there exists a quasi-affine covering $\{V_i\}_{i\in I}$ of $Y$ so that $f^{-1}(V_i)$ is quasi-affine for all $i\in I$.

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Teddyboer
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  • Can you add Bosch's definition of a quasi-affine morphism to your post? It would seem to me that the definition you would want here is that $X\to Y$ factors as the composition of a quasi-compact open immersion and an affine morphism, which pretty much trivializes the problem and would not require so much in the way of proof. But the argument presented here suggests this is not the definition. – KReiser May 19 '20 at 06:42
  • Yes, of course. According to Bosch, $X$ is quasi-affine if it is quasi-compact and there is an open immersion $X\rightarrow \text{Spec}(R)$ into some affine scheme. Equivalently, if the induced morphism $X\rightarrow \text{Spec}(\mathcal{O}X(X))$ derived from the identity on $\mathcal{O}_X(X)$ is an open immersion. And $f$ is quasi-affine if there exists a quasi-affine covering $(V_i){i\in I}$ of $Y$ such that the preimages $f^{-1}(V_i)$ are quasi-affine. – Teddyboer May 19 '20 at 08:20
  • I think the compositio of $f$ is as follows: look at $f^{\sharp}(Y) = \text{id}_{\mathcal{O}_X(X)}\circ f^{\sharp}(Y)$. Then this composition on the level of rings incduces a composition $X\rightarrow \text{Spec}(\mathcal{O}_X(X))\rightarrow Y$. And we want to show that the first arrow is indeed an open immersion. – Teddyboer May 19 '20 at 08:34
  • Actually, it turns out this is an application of the qcqs lemma which you asked about and I answered here: $f$ separated implies $f$ is quasi-separated, and $f$ is quasi-compact by assumption. So by applying the lemma we get that $\mathcal{O}_X(X)_g\cong \mathcal{O}_X(X_g)$, but the former is exactly $A\otimes_B B_g$. I am inclined to mark this as a duplicate of that question - what do you think? – KReiser May 20 '20 at 06:27
  • Yes I see the relation to the qcqs lemma, but my question was about Bosch's argument. From the exactness of the sequence above (after tensoring with $B_g$) he concludes that $A\otimes_B B_g\rightarrow \mathcal{O}_X(X_g) $ is an isomorphism. He doesn't quote anything so I assume it's obvious sonehow. But I don't see it. – Teddyboer May 21 '20 at 08:56
  • It's the same answer as that question: both $A\otimes_B B_g$ and $\mathcal{O}_X(X_g)$ equalize that diagram, but equalizers are unique up to unique isomorphsim, so they must be isomorphic. – KReiser May 21 '20 at 09:07

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