The following was recently on my algebraic geometry homework:



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> Let $k$ be an algebraically closed field, $f\in B=k[x\_1,\ldots,x\_n]$, and $A=B/(f)$. Show that $\Omega\_{A/k}$ is locally free of rank $n-1$ $\iff$ $\nexists\\, p\in k^n$ such that $f(p)=0$ and all $\frac{\partial f}{\partial x\_i}(p)=0$.
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Here, $\Omega\_{A/k}$ is just the module of differentials, not the sheaf of differentials on the corresponding variety (so locally free is meant in the sense of modules). [My solution](https://docs.google.com/viewer?a=v&pid=sites&srcid=ZGVmYXVsdGRvbWFpbnx6ZXZjaG9ub2xlc3xneDoyODYxYjFhNWY3MzJhODYw) (at least seems to) crucially depend on the Nullstellensatz, so my question is, are there any non-algebraically closed fields $k$ for which this result is still true? If so, is there an argument that treats them simultaneously? Or, if not, is there a good intuition for why algebraically closed is necessary? 



Kahler differentials of a hypersurface over a non-algebraically closed field
For $k$ alg. closed you can phrase the statement as $\Omega\_{A/k}$ is loc. free iff Spec$(A)$ is smooth. 'Spec$(A)$ smooth iff $\Omega\_{A/k}$ is loc free' should be true without requiring $k = \bar{k}$. But if $k \ne \bar{k}$ then the condition 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 $k$.


