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 map
$$
K^2\cong\mathcal I_\eta\hookrightarrow\mathcal O_{X,\eta}=K,
$$
contradicting 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 \b[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.

Solution

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

A finite <locally free sheaf> of ideals on a <reduced scheme> has rank at most one at every point. On a nonempty <affine open subscheme> where its rank is $r$, localize its inclusion into the structure sheaf at a <minimal prime ideal>. The resulting <local ring> is a field $K$, giving an injection $K^r\hookrightarrow K$, hence $r\le1$. This works without a Noetherian assumption. The empty scheme is a vacuous exception to claims phrased as nonexistence of a sheaf of a prescribed rank.

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