= Solution
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.
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