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Let $R$ be a ring (commutative with unit), and let $M$ be an $R$-module. Let $S = R \oplus M$ be made into a ring with the obvious product, where $M^2 = 0$. In Eisenbud's book (Commutative Algebra with a View Toward Algebraic Geometry), Exercise 16.2 asks to compute $\Omega_{S/R}$, the universal $S$-module of $R$-linear differentials.

The hint in the back of the book says just one short statement with no further explanation: $\Omega_{S/R} = M$. I don't think this is right, although it is easy to check that the projection $S \to M$ is a derivation. If $\Omega_{S/R} = M$, then for any $R$-linear derivation $d : S \to N$, and $m_1, m_2 \in M$, we would have $m_1 dm_2 = 0$. I don't think this relation can be derived.

If I use $m^2 = 0$, I get $m\,dm = 0$ (assuming characteristic $\ne 2$), and if I use $m_1 m_2 = 0$ then I get $m_1 d m_2 + m_2 d m_1 = 0$. No other relations seem derivable in general.

So is this an error in the book?

Ted
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  • On the other hand, given any $R$-derivation $\partial : S\to N,$ there exists a unique map of $R$-modules $f_\partial : M\to N$ such that $f_\partial\circ d = \partial,$ where $d : S\to M$ is the projection, defined by $f_\partial(m) = \partial(0,m)$. However, it doesn't seem that $f_\partial$ is an $S$-module homomorphism in general, unless I'm mistaken about what the $S$-module structure on $M$ is. – Stahl Oct 12 '20 at 09:09
  • One final potentially relevant remark: let $S$ be any $R$-algebra, and let $M$ be any $S$-module. Then $R$-linear derivations $\partial : S\to M$ are equivalent to $R$-algebra sections $\sigma : S\to S[M]$ of the projection $\pi_M : S[M]\to S.$ – Stahl Oct 12 '20 at 09:15
  • Dear @Ted, using the description $\Omega_{S/R}\cong I/I^2$, where $I\subset S\otimes_RS$ is the kernel of the multiplication map $S\otimes_RS\to S$, it can be seen that $M$ acts trivially on $\Omega_{S/R}$. More directly, you could consider the map $M\to I$, $m\mapsto m\otimes 1-1\otimes m$. (Btw, I refrain from posting this as an answer since I still hope that somebody will give a derivation-intrinsic explanation.) – Ben Oct 13 '20 at 10:00

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