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For an injective ring homomorphism between nonzero integral domains $A\to B$, the scheme-theoretic fibre over the generic point of $\mathrm{Spec}A$ is $\mathrm{Spec}(S^{-1}B)$ with $S=A\setminus \{0\}$. Localization at these nonzero elements is a nonzero integral domain, so the generic fibre is a nonempty integral scheme. A surjective morphism between these affine schemes necessarily gives an injective ring map: a prime above $(0)$ contains its kernel.

The rings $A,B$ are nonzero integral domains. Let $\eta$ be the generic point of $Y$, corresponding to $(0)\subseteq A$. Surjectivity provides a prime ideal $\mathfrak{q}\subseteq B$ with ${\alpha }^{-1}(\mathfrak{q})=0$. Since $\ker \alpha \subseteq {\alpha }^{-1}(\mathfrak{q})$, the ring homomorphism $\alpha$ is injective.

Put $K=\mathrm{Frac}A$ and $S=A\setminus \{0\}$. The scheme-theoretic fibre at $\eta$ isEvery element of $\alpha (S)$ is nonzero, so the localization $S^{-1}B$ is a nonzero integral domain. Its zero ideal is prime, ensuring that its spectrum is nonempty; an affine spectrum of an integral domain is an integral scheme. This proves the integrality of the generic fibre of an affine dominant morphism. In fact the proof only needs injectivity of $\alpha$, rather than surjectivity at every point.

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.

On an integral scheme with generic point $\eta$ and function field $K$, a rational section of a line bundle $\mathcal{L}$ is an element of the one-dimensional $K$-space ${\mathcal{L}}_{\eta }$. A nonzero rational section writes as $a_i{e}_i$ in local trivializations, with $a_i\in K^{\ast }$. Ratios of these coefficients are regular units, and therefore define a Cartier divisor.