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The following was recently on my algebraic geometry homework:

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 $\frac{\partial f}{\partial x_i}(p)=0$.

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 (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?

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If $k$ is not algebraically closed then $f(p)=0$ may be empty, so that the Jacobian condition on the right would be vacuous. The condition on the left, called smoothness, would still have content however. Notice that smoothness is stable under field extensions. So it is equivalent to the Jacobian condition over $\bar k$. – Donu Arapura Apr 17 '11 at 11:38
Regarding intuition, the problem is that $k$-points don't tell you everything about a $k$-variety when $k$ isn't algebraically closed, so there's no reason to expect the condition on the right to behave well for general $k$ as Donu says. Looking at $k$-points is a bad choice of "concretization" of $k$-varieties. A much better one is the functor which sends a $k$-variety to its $\bar{k}$-points equipped with the natural action of $\text{Gal}(\bar{k}/k)$. – Qiaochu Yuan Apr 17 '11 at 20:33
up vote 2 down vote accepted

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 $k$.

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