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Let $A$ and $B$ be finitely generated $\mathbf{Z}$-algebra. Suppose that there exists two coprime integers $m$ and $n$ and an isomorphism of $\mathbf{Z}$-algebra $\phi:A\otimes_{\mathbf{Z}}\mathbf{Z}[1/n]\simeq B\otimes_{\mathbf{Z}}\mathbf{Z}[1/m] $. Then we can glue $A$ and $B$ along $\phi$ so that we obtain a scheme $Y$ over $Spec(\mathbf{Z})$.

Q1: Under what general conditions do we have $Y$ affine ?

Q2: In the case where $Y=Spec(C)$ is affine then how do we construct the $\mathbf{Z}$-algebra $C$ from the triple $(A,B,\phi)$?

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    $\begingroup$ You probably mean "an iso. of algebras $A\otimes_{\bf Z}{\bf Z}[1/(nm)]\simeq B\otimes_{\bf Z}{\bf Z}[1/(nm)]$". If that is the case, then the morphism $Y\to {\rm Spec}\ {\bf Z}$ is affine (see ex. 5.17, p. 128 in Hartshorne). You can then use Ex. 4.1, p. 222 in Hartshorne and Serre's criterion for affineness to conclude that $Y$ is affine. $\endgroup$
    – Damian Rössler
    Commented Sep 2, 2013 at 17:31
  • $\begingroup$ A toy example I had in mind was $Z[1/2][1/3]\simeq Z[1/3][1/2]$ where $n=3$ and $m=2$. If we glue them along the isomorphism then we get $Z$. Anyhow thanks a lot for the reference, I'll look at it $\endgroup$
    – Hugo Chapdelaine
    Commented Sep 2, 2013 at 17:47
  • $\begingroup$ So I think that proof you were suggesting is correct. So it seems to me that there should be an (obvious) direct construction of $C$ but I don't see it... $\endgroup$
    – Hugo Chapdelaine
    Commented Sep 2, 2013 at 17:55
  • $\begingroup$ By Zariski glueing on affine schemes,$$C=\{(a,b)\in A\times B\mid \phi(a\otimes1)=b\otimes1\}.$$ $\endgroup$
    – Laurent Moret-Bailly
    Commented Sep 3, 2013 at 9:53
  • $\begingroup$ Beautiful! I never internalized fibered products in the category of rings but now with this example I see their importance! Please Laurent, copy and paste your equality so that I can checkmark your answer. $\endgroup$
    – Hugo Chapdelaine
    Commented Sep 3, 2013 at 12:42

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By Zariski glueing on affine schemes, $$C=\{(a,b)\in A×B\mid\phi(a\otimes1)=b\otimes1\}.$$ Of course the beautiful thing in this story is the answer to Q1: if $f:Y\to X$ is an affine morphism and $X$ is an affine scheme, then $Y$ is affine. Once you know this, the only possible answer to Q2 is the above.

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