Past exam of the introduction to schemes course of the University of Oxford / 2017 / 1

Apply the surjectivity of $f(Y)$ to the identity morphism $1_Y$. It gives a morphism of schemes $g:Y\to X$ satisfying $f\circ g=1_Y$. To obtain the other identity, compare $g\circ f$ and $1_X$ in $X(X)$. Their images under $f(X)$ agree, sinceThe injectivity of $f(X)$ therefore gives $g\circ f=1_X$. ThusOnly surjectivity for the test scheme $Y$ and injectivity for the test scheme $X$ were needed. This is the isomorphism-detection argument behind the Yoneda lemma for the functor of points.

No. Let $k={\mathbb{F}}_2$ be the field with two elements, put $A=k[\epsilon ]/({\epsilon }^2)$, and take the closed immersioninduced by the ring homomorphism $q:A\twoheadrightarrow k$ with $q(\epsilon )=0$.

For every field $K$, a ring homomorphism $\alpha :A\to K$ must send $\epsilon$ to zero: $\alpha (\epsilon {)}^2=0$, and a field has no nonzero nilpotent elements. Consequently $\alpha$ factors uniquely through $q$. Passing from ring homomorphisms to morphisms of schemes between affine schemes gives a bijectionThis concerns all morphisms of schemes, without requiring them to be over $k$: if $K$ does not have characteristic two, both sets are empty; otherwise each contains exactly one element.

Nevertheless, $f$ is not an isomorphism of schemes. Its map on global sections is $q$, whose kernel of a ring homomorphism contains the nonzero element $\epsilon$, so $q$ is not a ring isomorphism. The target is the nonreduced scheme defined by the dual numbers, whereas the source is a reduced scheme. This illustrates why field-valued points do not detect nilpotents.

Yes. First, the hypothesis implies that $f(T)$ is injective for every scheme $T$, even when $T$ is not an affine scheme. Indeed, suppose $a,b:T\to X$ satisfy $f\circ a=f\circ b$. Choose an affine open cover $T={\bigcup }_{\lambda }{V}_{\lambda }$. On each $V_{\lambda }$, injectivity of $f(V_{\lambda })$ gives $a{|}_{V_{\lambda }}=b{|}_{V_{\lambda }}$. By gluing morphisms of schemes, $a=b$.

Now choose an affine open cover $Y={\bigcup }_i{U}_i$, and denote its inclusion maps by $j_i:U_i\hookrightarrow Y$. Surjectivity of $f(U_i)$ gives morphisms of schemes $g_i:U_i\to X$ with $f\circ g_i=j_i$. On $U_i\cap U_j$, the two restrictions have the same composite with $f$. By the injectivity just established for arbitrary test schemes,In particular, the overlaps need not be affine schemes. Gluing morphisms of schemes supplies $g:Y\to X$ with $g{|}_{U_i}=g_i$, and $f\circ g=1_Y$ because this holds on every $U_i$.

Finally,Injectivity of $f(X)$, already proved without an affineness assumption, gives $g\circ f=1_X$. ThereforeThis proves the relevant instance of affine test schemes determine scheme morphisms: the restricted functor of points still retains the information needed to construct the inverse.