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For $k$ alg. closed you can phrase the statement as $\Omega_{A/k}$ is loc. free iff Spec$(A)$ is smooth. 'Spec$(A)$ smooth implies iff $\Omega_{A/k}$ is loc free' should be true without requiring $k = \bar{k}$. But I think if $k \ne \bar{k}$ then the converse can failcondition on the derivatives is not the same as smoothness. For example if $C$ is a curve defined over $\mathbb{R}$ with smooth $\mathbb{R}$ points but with singular $\mathbb{C}$ points then the condition on $f$ and its derivatives will be satisfied but there will be a maximal ideal of Spec$(A)$ with residue field $\mathbb{C}$ where $\Omega_{A/k}$ will have the wrong rank.

You can try this with $y^2 = (x^2+1)^2$ and the maximal ideal $(y, x^2 + 1)$ in $\mathbb{R}[x,y]$.

But if $A(k) = A(\bar{k})$ then the original statement should hold over $k$.

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For $k$ alg. closed you can phrase the statement as $\Omega_{A/k}$ is loc. free iff Spec$(A)$ is smooth. 'Spec$(A)$ smooth implies $\Omega_{A/k}$ is loc free' should be true without requiring $k = \bar{k}$. But I think the converse can fail. For example if $C$ is a curve defined over $\mathbb{R}$ with smooth $\mathbb{R}$ points but with singular $\mathbb{C}$ points then the condition on $f$ and its derivatives will be satisfied but there will be a maximal ideal of Spec$(A)$ with residue field $\mathbb{C}$ where $\Omega_{A/k}$ will have the wrong rank.

You can try this with $y^2 = (x^2+1)^2$ and the maximal ideal $(y, x^2 + 1)$ in $\mathbb{R}[x,y]$.