Kahler differentials of a hypersurface over a non-algebraically closed field - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T15:59:30Z http://mathoverflow.net/feeds/question/61995 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/61995/kahler-differentials-of-a-hypersurface-over-a-non-algebraically-closed-field Kahler differentials of a hypersurface over a non-algebraically closed field Zev Chonoles 2011-04-17T04:36:08Z 2011-04-17T19:28:17Z <p>The following was recently on my algebraic geometry homework:</p> <blockquote> <p>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 $\frac{\partial f}{\partial x_i}(p)=0$.</p> </blockquote> <p>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). <a href="https://docs.google.com/viewer?a=v&amp;pid=sites&amp;srcid=ZGVmYXVsdGRvbWFpbnx6ZXZjaG9ub2xlc3xneDoyODYxYjFhNWY3MzJhODYw" rel="nofollow">My solution</a> (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? </p> http://mathoverflow.net/questions/61995/kahler-differentials-of-a-hypersurface-over-a-non-algebraically-closed-field/61997#61997 Answer by solbap for Kahler differentials of a hypersurface over a non-algebraically closed field solbap 2011-04-17T05:28:28Z 2011-04-17T19:28:17Z <p>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.</p> <p>You can try this with $y^2 = (x^2+1)^2$ and the maximal ideal $(y, x^2 + 1)$ in $\mathbb{R}[x,y]$.</p> <p>But if $A(k) = A(\bar{k})$ then the original statement should hold over $k$.</p>