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Consider the following pushout diagram $\require{AMScd}$ \begin{CD} A @>f>> B\\ @V g V V @VV V\\ C @>> > D \end{CD} in the category of $\textbf{Rings}$ where $f,g$ both are flat monomorphisms. The diagram is also a pullback diagram.

I'm trying to show that it's a pullback diagram but could not.

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  • $\begingroup$ I believe that the statement is true. I did not find any argument to prove it. If it is not true, please give a counter-example. $\endgroup$
    – Nil123
    Commented Dec 18, 2020 at 14:52
  • $\begingroup$ What are you trying to prove? That it's also a pullback diagram? $\endgroup$
    – Harry Wilson
    Commented Dec 18, 2020 at 14:59
  • $\begingroup$ Yes, I'm trying to prove that it's a pullback diagram @HarryWilson $\endgroup$
    – Nil123
    Commented Dec 18, 2020 at 15:01
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    $\begingroup$ The statement is wrong: Take $A = \mathbb Z, B = C = \mathbb Q$. Then $D = \mathbb Q$, and the maps from $B$ and $C$ are isomorphisms, so that the pullback is $\mathbb Q$. $\endgroup$
    – Bertram Arnold
    Commented Dec 18, 2020 at 15:37
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    $\begingroup$ I claim that the multiplication map $\mathbb Q\otimes_{\mathbb Z}\mathbb Q\to\mathbb Q$ and the "left inclusion" $\mathbb Q\to\mathbb Q\otimes_{\mathbb Z}\mathbb Q$ are inverse bijections. The composition $\mathbb Q\to\mathbb Q$ is obviously the identity, so that the other composition is surjective since $\frac{m_1}{n_1}\otimes \frac{m_2}{n_2} = n_2\frac{m_1m_2}{n_1n_2}\otimes \frac{1}{n_2} = \frac{m_1m_2}{n_1n_2}\otimes 1$. But a retraction where the inclusion map is surjective is already an isomorphism. $\endgroup$
    – Bertram Arnold
    Commented Dec 18, 2020 at 18:07

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As I've written in the comments, the pushout square $$ \begin{CD} \mathbb Z @>>> \mathbb Q\\ @VVV @VVV\\ \mathbb Q @>>> \mathbb Q \end{CD} $$ is not a pullback square, providing a counterexample. However, there are additional hypotheses which work, namely that one of the maps is faithfully flat.

Suppose first that one of the maps (say $f$) has a section $r:B\to A$, i.e. $rf =\operatorname{id}_A$. Then we have a splitting of $A$-modules $B = A \oplus\ker r$, and the pushout square becomes $$ \begin{CD} A @>i_1>> A\oplus \ker r\\ @VgVV @Vg\oplus hVV\\ C @>i_1>> C\oplus (\ker r\otimes_A C) \end{CD} $$ which obviously is a pullback square.

Since the $A$-module $B$ is flat, the functor $-\otimes_A B$ preserves finite limits. In particular, writing $A' = B\times_D C$ which receives a map $A\to A'$, the induced map $A'\otimes_A B\to (B\otimes_A B)\times_{D\otimes_A B} (C\otimes_A B)$ is an isomorphism. The composition of this map with the left inclusion $B\to A'\otimes_A B$ is then exactly the map from the top left corner to the pullback in the diagram obtained by taking the pushout over $A$ with $B$ everywhere; in particular, the top horizontal map $B\to B\otimes_A B$ has a section, namely the multiplication. By the result of the previous paragraph, the map from $B$ into the pullback, and hence also into $A'\otimes_A B$, is an isomorphism. Thus the map $A\to A'$ becomes an isomorphism after tensoring with $B$; if $B$ is faithfully flat, it must have been an isomorphism in the first place.

This argument is inspired by (and probably an application of) the theory of faithfully flat descent, compare the Stacks project.

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