Kahler differentials of a hypersurface over a non-algebraically closed field - MathOverflow

Kahler differentials of a hypersurface over a non-algebraically closed field

2011-04-17T04:36:08Z

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?

Answer by solbap for Kahler differentials of a hypersurface over a non-algebraically closed field

2011-04-17T05:28:28Z

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 $k$.