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The following is a subquestion of a unanswered question posted two days ago by some user in math.stackexchange (answer that question if you prefer).

For a (not necessarily cocartesian) commutative square of commutative rings, $$\begin{array}{ccc} A&\rightarrow&B\\\ \downarrow&&\downarrow\\\ A'&\rightarrow&B' \end{array}$$ in EGA 0$_{IV}$20.5.4 there is a map written as $d_{B/A} \otimes 1: B' \to \Omega_{B/A} \otimes_BB'$, where $d_{B/A}: B \to \Omega_{B/A}$ is the canonical derivation. Since $d_{B/A}$ is not a $B$-module homomorphism, how is this map defined?

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  • $\begingroup$ Well, $B'=B\otimes_B B'$ so $d_{B/A}\otimes 1$ makes sense. $\endgroup$ Feb 23, 2017 at 17:19
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    $\begingroup$ @Matthieu Romagny Both $B$ and $\Omega_{B/A}$ are $B$-modules, bur the map $d_{B/A}: B \to \Omega_{B/A}$ is not of $B$-linear (only $A$-linear). $\endgroup$
    – user409777
    Feb 23, 2017 at 18:06
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    $\begingroup$ Let me mention - it might be a clue - that they say $d_{B/A}\otimes1$ (whatever it is) is $A'$-linear $\endgroup$ Feb 23, 2017 at 19:02
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    $\begingroup$ But $A'$ is any ring "between" $A$ and $B'$. It can be $A'=B=B'$ and then they say that (if as I think in this case $d_{B/A}=d_{B/A}\otimes 1$) that $d_{B/A}$ is $B$-linear! The definition of $d_{B/A}\otimes 1$ only makes sense to me if $B'=B\otimes_AA'$. $\endgroup$
    – user409777
    Feb 23, 2017 at 19:27
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    $\begingroup$ I think this is a mistake in EGA, but one that seems to be harmless. In the diagram 20.5.4.2, presumably the ring $B'$ in the lower left should be $B$, and the bottom horizontal arrow should be $v$. All of this theory is reviewed in Illusie's "Complexe Cotangent et Deformations I", so you could double-check there. $\endgroup$ Feb 24, 2017 at 12:33

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