Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 113 / 2

For an affine open subscheme $U=\mathrm{Spec}R$, quasi-coherence gives $\mathcal{I}{|}_U=\tilde{J}$ for an ideal $J\subseteq R$. Define $Z_U=\mathrm{Spec}(R/J)$ with its usual closed immersion into $U$. On a principal open subscheme $D(a)$, its restriction is $\mathrm{Spec}(R_a/J_a)$, because localization commutes with taking this quotient. These local constructions therefore agree on overlaps and glue.

The resulting underlying closed set is $Z=\{x:{\mathcal{I}}_x\ne {\mathcal{O}}_{X,x}\}$, locally $V(J)$. With inclusion $i:Z\hookrightarrow X$, its structure sheaf isAt a point corresponding to $\mathfrak{p}\supseteq J$, its stalk is $R_{\mathfrak{p}}/J_{\mathfrak{p}}$, the local ring of $\mathrm{Spec}(R/J)$. Thus the glued ringed space is a scheme and $i$ is a closed immersion, with ideal sheaf of a closed subscheme exactly $\mathcal{I}$. This is the construction of a closed subscheme from a quasi-coherent ideal. If $\mathcal{I}={\mathcal{O}}_X$, the construction gives the empty scheme; if $\mathcal{I}=0$, it gives $X$.

First suppose $X$ is a nonempty integral scheme, with generic point $\eta$ and function field $K={\mathcal{O}}_{X,\eta }$. Since taking stalks preserves the inclusion $\mathcal{I}\hookrightarrow {\mathcal{O}}_X$, local freeness of rank two would give an injective $K$-linear mapcontradicting the dimension of a vector space.

For a nonempty Noetherian scheme that is reduced, there are finitely many irreducible components. Choose one, and remove the union of the others. The resulting nonempty open subscheme is irreducible and reduced, hence integral. Restricting the proposed locally free sheaf to it gives the contradiction above. Therefore no locally free ideal of rank two exists on a nonempty reduced Noetherian scheme.

The rank bound for locally free ideals on reduced schemes in fact removes the Noetherian assumption: on an affine open where the rank is fixed, localization at a minimal prime ideal gives a field, so a free ideal has rank at most one. Nonemptiness is necessary for the literal statement: on the empty scheme the zero sheaf is vacuously locally free of every stipulated rank. The question is interpreted with this usual nonempty hypothesis.