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Marco Golla
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Suppose I have a morphism $f:X\rightarrow Y$ such that the relative sheaf of differentials $\Omega_{X/Y}$ is locally free. Does it follow that $f$ is smooth?

The answer is no, but for a silly reason. You could have some non-reducedness $\mathrm{Spec}\left(k\left[e\right]/e^{2}\right)$ over $\mathrm{Spec}\left(k\right)$ has a free sheaf of differentials, but isn't smooth). But what if you add the hypothesis that the rank of $\Omega_{X/Y}$ is $\mathrm{dim}\left(X\right)-\mathrm{dim}\left(Y\right)$$\dim X-\dim Y$?

Edit: As Jonathan points out in his answer, I was careless with my counterexample. It only works if $\mathrm{Char}\left(k\right)=2$$\mathrm{char}\,k=2$.

Suppose I have a morphism $f:X\rightarrow Y$ such that the relative sheaf of differentials $\Omega_{X/Y}$ is locally free. Does it follow that $f$ is smooth?

The answer is no, but for a silly reason. You could have some non-reducedness $\mathrm{Spec}\left(k\left[e\right]/e^{2}\right)$ over $\mathrm{Spec}\left(k\right)$ has a free sheaf of differentials, but isn't smooth). But what if you add the hypothesis that the rank of $\Omega_{X/Y}$ is $\mathrm{dim}\left(X\right)-\mathrm{dim}\left(Y\right)$?

Edit: As Jonathan points out in his answer, I was careless with my counterexample. It only works if $\mathrm{Char}\left(k\right)=2$.

Suppose I have a morphism $f:X\rightarrow Y$ such that the relative sheaf of differentials $\Omega_{X/Y}$ is locally free. Does it follow that $f$ is smooth?

The answer is no, but for a silly reason. You could have some non-reducedness $\mathrm{Spec}\left(k\left[e\right]/e^{2}\right)$ over $\mathrm{Spec}\left(k\right)$ has a free sheaf of differentials, but isn't smooth). But what if you add the hypothesis that the rank of $\Omega_{X/Y}$ is $\dim X-\dim Y$?

Edit: As Jonathan points out in his answer, I was careless with my counterexample. It only works if $\mathrm{char}\,k=2$.

If \Omega_X$\Omega_{X/Y}$ is locally free of rank dim$\mathrm{dim}\left(XX\right)-\mathrm{dim}\left(YY\right)$, is X->Y$X\rightarrow Y$ smooth?

Suppose I have a morphism f:X→Y$f:X\rightarrow Y$ such that the relative sheaf of differentials ΩX/Y$\Omega_{X/Y}$ is locally free. Does it follow that f$f$ is smooth?

The answer is no, but for a silly reason. You could have some non-reducedness (Spec(k[e]/(e^2))$\mathrm{Spec}\left(k\left[e\right]/e^{2}\right)$ over Spec(k)$\mathrm{Spec}\left(k\right)$ has a free sheaf of differentials, but isn't smooth). But what if you add the hypothesis that the rank of ΩX/Y$\Omega_{X/Y}$ is dim(X)-dim(Y)$\mathrm{dim}\left(X\right)-\mathrm{dim}\left(Y\right)$?

Edit: As Jonathan points out in his answer, I was careless with my counterexample. It only works if char(k)=2$\mathrm{Char}\left(k\right)=2$.

If \Omega_X/Y is locally free of rank dim(X)-dim(Y), is X->Y smooth?

Suppose I have a morphism f:X→Y such that the relative sheaf of differentials ΩX/Y is locally free. Does it follow that f is smooth?

The answer is no, but for a silly reason. You could have some non-reducedness (Spec(k[e]/(e^2)) over Spec(k) has a free sheaf of differentials, but isn't smooth). But what if you add the hypothesis that the rank of ΩX/Y is dim(X)-dim(Y)?

Edit: As Jonathan points out in his answer, I was careless with my counterexample. It only works if char(k)=2.

If $\Omega_{X/Y}$ is locally free of rank $\mathrm{dim}\left(X\right)-\mathrm{dim}\left(Y\right)$, is $X\rightarrow Y$ smooth?

Suppose I have a morphism $f:X\rightarrow Y$ such that the relative sheaf of differentials $\Omega_{X/Y}$ is locally free. Does it follow that $f$ is smooth?

The answer is no, but for a silly reason. You could have some non-reducedness $\mathrm{Spec}\left(k\left[e\right]/e^{2}\right)$ over $\mathrm{Spec}\left(k\right)$ has a free sheaf of differentials, but isn't smooth). But what if you add the hypothesis that the rank of $\Omega_{X/Y}$ is $\mathrm{dim}\left(X\right)-\mathrm{dim}\left(Y\right)$?

Edit: As Jonathan points out in his answer, I was careless with my counterexample. It only works if $\mathrm{Char}\left(k\right)=2$.

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Anton Geraschenko
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Anton Geraschenko
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Anton Geraschenko
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