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If $k$ is a Noetherian ring, then do the Kahler forms of a smooth affine $k$-algebra of dimension $d$ vanish above $d$?

I mean is: $\Omega^{d+1}_{A|k}\cong 0$?

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    $\begingroup$ Yes. This just follows from the fact that $\Omega_{A/k}$ is locally free of rank $d$, and so locally your module is trivial, so is actually trivial. $\endgroup$
    – Alex Youcis
    Jul 16, 2014 at 3:49
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    $\begingroup$ @AlexYoucis: This is a complete answer. Not a comment. $\endgroup$
    – Martin Brandenburg
    Jul 16, 2014 at 7:56

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This follows precisely from the fact that $\Omega_{A/k}^1$ is locally free of rank $d$ (by smoothness). So, locally $\wedge^{d+1}\Omega_{A^k}^1$ is locally zero, and so globally zero.

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