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

On the principal open subscheme $D(a)\subseteq \mathrm{Spec}A$, the induced morphism of sheaves is the ring homomorphismIf $\alpha$ is injective, this map is injective by exactness of localization: it is the localization of the injective $A$-module map $A\to B$. Explicitly, if the displayed fraction vanishes, some power of $\alpha (a)$ annihilates $\alpha (b)$; injectivity gives $a^{m}b=0$, so the original fraction also vanishes. The principal opens form an open basis, so the kernel sheaf is zero on a basis and hence zero everywhere. Conversely, an injective morphism of sheaves induces an injection on global sections. On the whole affine scheme these sections are $A\to B$, exactly $\alpha$. Thus

Take the dual numbers $A=k[\epsilon ]/({\epsilon }^2)$ over any field $k$, let $B=k$, and let $\alpha$ send $\epsilon$ to zero. Its kernel is the nonzero ideal $(\epsilon )$, but both spectra of rings have exactly one point: every prime ideal contains the nilpotent element $\epsilon$, and the only prime of $A$ is $(\epsilon )$. Consequently the induced map is a homeomorphism for the Zariski topology, although it is not an isomorphism of schemes:More generally, quotienting by a nilpotent ideal gives a nilpotent thickening with unchanged underlying topology. The structure sheaf retains information that the topology alone cannot recover.

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.

Use the ring homomorphism $k[t]\to k[x]$ given by $t\mapsto x^2$. The source scheme $X=\mathrm{Spec}k[x]$ is an integral scheme. At the rational point $y=(t)$, its residue field is $k$, and its scheme-theoretic fibre isThe class of $x$ is nonzero but has square zero, so this is not a reduced scheme. This example works over every field. An integral total space therefore need not have reduced fibres.