Consider lemma 10.130.10 from this Stacks Project site.

Lemma. Let $\alpha:R\to S,\pi:S\to T$ be two ring maps and assume that $\pi$ is surjective. If there is a $R$-linear $\iota:T\to S$ such that $\pi\circ\iota = 1_T$, then there is a following short split exact sequence of $T$-modules $$0\to I/I^2\to\Omega_{S/R}\otimes_S T\to\Omega_{T/R}\to 0,$$ where $I=\ker(\pi).$

Assume that we have ring $K$ and two $K$-algebras $A,B$ and $K$-algebra homomorphism $\phi:A\to B.$ (Everything here is commutative, associated and unitial). In addition define $$\pi:B\otimes_K A\to B,\quad\iota:B\to B\otimes_K A$$ respectively by $$\pi(b\otimes a)=\phi(a)b,\quad \iota(b)=b\otimes 1.$$ We will now apply lemma to two cases:

First: Let $\alpha:K\to B\otimes_K A$ be natural map form $K$-algebra structure of $B\otimes_K A.$ Then $$0\to I/I^2\to\Omega_{(B\otimes_K A)/K}\otimes_{(B\otimes_K A)} B\to\Omega_{B/K}\to 0$$ is short split exact seqence of $B$-modules.

Second: Let $\alpha:A\to B\otimes_K A$ be given by formula $\alpha(a)=\phi(a)\otimes 1.$ Then $$0\to I/I^2\to\Omega_{(B\otimes_K A)/A}\otimes_{(B\otimes_K A)} B\to\Omega_{B/A}\to 0$$ is short split exact seqence of $B$-modules.

Everything looks ok, but then we put $B=K$ then the first exact sequence degenerates to $$0\to I/I^2\to\Omega_{A/K}\otimes_A K\to 0$$ and the second one to $$0\to I/I^2\to 0.$$

However $\Omega_{A/K}\otimes_A K$ does not have to vanish in general.

Question. Were is the flaw in the above reasoning?


1 Answer 1


The second exact sequence is incorrect because in case $B = K$ the map $\alpha : A \to A$ is not the identity $\operatorname{id}_{A}$ but rather the composite $A \to K \to A$ where the first map is $\phi$. Thus $\Omega_{A/A} \simeq \Omega_{A/K}$.

  • $\begingroup$ You are right. Due to the misleading notation $\Omega_{S/R}$ I recklessly canel out some objects using algebraic properties of tensor. $\Omega_{S/R}$ should be rather denoted by $\Omega_\alpha.$ For example the second sequence is correct if you interprate $\Omega_{(B\otimes_K A)/A}$ correctely. Don't you have problems with that? Or maybe you think in terms of $\Omega_\alpha$? $\endgroup$ Nov 21, 2017 at 18:55
  • $\begingroup$ Yes, it is an abuse of notation, but it is standard. Your question confused me for a while :) $\endgroup$ Nov 21, 2017 at 19:04
  • $\begingroup$ You are wellcome. $\endgroup$ Nov 21, 2017 at 19:07

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