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 $K=\operatorname{Frac}A$ and $S=A\setminus\{0\}$. The <scheme-theoretic fibre> at $\eta$ is
$$
\boxed{X_\eta=X\times_Y\operatorname{Spec}K
=\operatorname{Spec}(B\otimes_AK)=\operatorname{Spec}(S^{-1}B).}
$$
Every 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.

Solution

Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 113 / 1 / iii / Solution

For an injective <ring homomorphism> between nonzero <integral domains> $A\to B$, the <scheme-theoretic fibre> over the <generic point> of $\operatorname{Spec}A$ is $\operatorname{Spec}(S^{-1}B)$ 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.

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 Incoming links: Integrality of the generic fibre of an affine dominant morphism