Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 126 / 1

For a ring homomorphism $A\to B$, the Module of Kähler differentials ${\Omega }_{B/A}$ is the $B$-module generated by symbols $db$, subject toEquivalently, it represents $A$-linear derivations:For $A\to B\to C$, the Transitivity exact sequence for Kähler differentials isIf $B\to C=B/I$ is surjective, the Conormal exact sequence for Kähler differentials is

Let $L/K$ be a finite field extension. By the primitive element theorem, its maximal separable field extension $L_s/K$ is simple, and transitivity reduces the calculation to a simple algebraic extension. If $L=K(\alpha )$ with minimal polynomial $m_{\alpha }$, thenThus a separable simple extension has zero differentials. Conversely, if $L/K$ is not separable, the purely inseparable part has a generator whose minimal polynomial has zero formal derivative in positive characteristic, producing a nonzero differential. Hence

Now let $\mathrm{char}K=p>0$. If $L=K(a)$, $a^p\notin K$, and $a^{p^2}\in K$, then the minimal polynomial is $T^{p^2}-a^{p^2}$ and has zero derivative, soIf $L=K(a,b)$, where $a\notin K$, $b\notin K(a)$, and $a^p,b^p\in K$, then $[K(a):K]=[L:K(a)]=p$. Both defining equations have zero derivative, and

For the three morphisms, compute the Sheaf of relative Kähler differentials on coordinate rings.

If $\mathrm{char}k\ne 2$, this is $k[y]/(y)dy$ and its support is the origin $V(y)=V(x,y)$. If $\mathrm{char}k=2$, it is free of rank one and its support is all of $X$.

Its support is the origin $V(x)$ in every characteristic; the module is $k[x]/(x)$ unless $\mathrm{char}k=2$, when it is $k[x]/(x^2)$.

Its support is the entire component $V(x)\subset X$.

A morphism of schemes $f:X\to Y$ is a flat morphism when every local-ring map ${\mathcal{O}}_{Y,f(x)}\to {\mathcal{O}}_{X,x}$ makes ${\mathcal{O}}_{X,x}$ a flat module.

In (i), the coordinate map is $k[t]\to k[y]$, $t\mapsto y^2$, andIt is therefore a free module of rank two and the morphism is flat, including in characteristic two.

In (ii), $k[x]$ is finite over the cusp ring $R=k[x^2,x^3]$ and has generic rank one. Were it flat, finite flatness over the local ring at the cusp would make it free of rank one. Its fiber there is insteadwhich has dimension two, so this morphism is not flat.

In (iii), the base coordinate $t$ acts as $x$, and the nonzero element $y$ satisfies $ty=xy=0$. Thus the coordinate ring has torsion as a $k[t]$-module. Since $k[t]$ is a principal ideal domain and a module over it is flat exactly when it is torsion-free, this morphism is not flat. Consequently