Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 20 / 2

For a point $y\in Y$ with residue field $\kappa (y)$, the scheme-theoretic fibre is the base change of a morphism of schemesThe map $\mathrm{Spec}\kappa (y)\to Y$ is the canonical point map. This fibre product of schemes retains the structure sheaf, including any nilpotents, rather than just the set $f^{-1}(y)$. For an affine scheme map induced by $A\to B$, at $\mathfrak{p}\in \mathrm{Spec}A$ it is

Compute each scheme-theoretic fibre by tensoring with the residue field of the chosen base point.

For the first map, put $L=\kappa (\mathfrak{p})$ for a point $\mathfrak{p}$ of $\mathrm{Spec}k[T]$ and let $t$ be the image of $T$ in $L$. ThenIf the characteristic of a field is not two and $t\ne 0$, the two factors $U-t$ and $U+t$ are coprime, so the Chinese remainder theorem gives $L[U]/(U^2-t^2)\simeq L\times L$: the scheme-theoretic fibre is two distinct $L$-points. If $t=0$, its ring is $L[U]/(U^2)$, a dual number ring, so it is a nonreduced double point. In characteristic two, $U^2-t^2=(U-t{)}^2$ at every point, giving a nonreduced double point in every scheme-theoretic fibre. This covers the generic point, where $L=k(T)$, as well as closed points defined by irreducible polynomials.

For the arithmetic map, the generic scheme-theoretic fibre isOver a closed point $(p)$ it is $\mathrm{Spec}{\mathbb{F}}_p[T]/(T^2+1)$. At $p=2$ this is $\mathrm{Spec}{\mathbb{F}}_2[ϵ]/({ϵ}^2)$, since $T^2+1=(T+1{)}^2$. For odd $p$, the finite field multiplicative group is cyclic, and $-1$ is a square exactly when $p\equiv 1(\mathrm{mod}4)$. ThusIn the second case there is one degree-two closed point over ${\mathbb{F}}_p$, which becomes two points after extending the residue field to an algebraic closure. The case $p=2$ remains nonreduced after such extension.

For $\mathrm{Spec}\mathbb{C}\to \mathrm{Spec}\mathbb{Z}$, the unique source point maps to the generic point $(0)$. Since all nonzero integers are invertible in $\mathbb{C}$,Indeed $\mathbb{C}{\otimes }_{\mathbb{Z}}\mathbb{Q}=\mathbb{C}$, whereas $\mathbb{C}{\otimes }_{\mathbb{Z}}{\mathbb{F}}_p=\mathbb{C}/p\mathbb{C}=0$. The distinction between a reduced split scheme-theoretic fibre and a nonreduced double point is essential in the first two examples.

Necessity follows from commutativity of the fibre product of schemes square: if $p(z)=x$ and $q(z)=y$, then $f(x)=g(y)$.

For sufficiency, put $s=f(x)=g(y)$ and $K=\kappa (s)$, $L=\kappa (x)$, $L^{\prime }=\kappa (y)$. The maps on local rings induce field embeddings $K\hookrightarrow L,L^{\prime }$. The canonical point maps into $X$ and $Y$ therefore give a morphism of schemesThe tensor product of commutative algebras $L{\otimes }_K{L}^{\prime }$ is nonzero. To see the hinted fact directly, choose a $K$-basis of $L$ containing $1$; tensoring that basis with $L^{\prime }$ makes $1\otimes 1$ nonzero. A nonzero unital commutative ring has a prime ideal, so its spectrum of a commutative ring has a point $w$. Its projections to the two field spectra are their unique points. The image $z$ of $w$ consequently projects to $x$ and $y$.

The desired point exists precisely when the base images coincide. This is the point-lifting property of a scheme fibre product. It does not claim that such a point is unique: different prime ideals of the tensor product may give different points over the same pair.

Let $f:X\to Y$ be a surjective morphism of schemes and let $Y^{\prime }\to Y$ be any morphism of schemes. Take any point $y^{\prime }\in Y^{\prime }$ and let $y$ be its image in $Y$. Surjectivity supplies $x\in X$ with $f(x)=y$. By the point-lifting property of a scheme fibre product proved in part (c), there is a point of $X{\times }_Y{Y}^{\prime }$ projecting to both $x$ and $y^{\prime }$. Thus every point of $Y^{\prime }$ is hit by the projection.

This proves that surjectivity is preserved by base change. No restriction is imposed on the base change of a morphism of schemes. No flatness or finite-type assumption is required.