Let $A$ be a commutative algebra (of essentially finite type, perhaps I have to add other reasonable assumptions) over a field, and I assume that $\Omega^1(A)$ is a projective module. Then the definition of $\Omega^1$ gives the following exact sequence of bimodules $ 0 \to I^2 \to I \to \Omega^1 \to 0, $ where $I$ is the kernel of multiplication homomorphism $A \otimes A \to A$. Module $\Omega^1$ is projective, thus, if we consider the sequence as sequence of (for example) left modules, sequence splits $ s: \Omega^1 \to I. $ Splitting $s$ induces a map of bimodules $ A \otimes \Omega^1 \to I. $ In one paper I found a claim that the last map is surjective, but I can't see why it is surjective. Any suggestions?
Kähler differentials, smooth algebras (one specific question)
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abstract-algebra
commutative-algebra
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0Decomposable elements $b \otimes x$ go to $s(x) (1 \otimes b)$. – 2012-12-20