Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 113 / 1

The points of the spectrum of a commutative ring $A$ are its prime ideals. Its Zariski topology has closed sets $V(I)=\{\mathfrak{p}:I\subseteq \mathfrak{p}\}$ and basis $D(f)=\{\mathfrak{p}:f\notin \mathfrak{p}\}$. The structure sheaf of a scheme can be defined as follows: a section over $U$ assigns an element of $A_{\mathfrak{q}}$ to every $\mathfrak{q}\in U$, and locally that assignment is represented by one fraction $a/s$, with $s$ outside every prime in the neighbourhood. Restrictions forget values. Local representability and equality can be checked on an open cover, so these assignments satisfy the sheaf gluing axiom. On $D(f)$ the sections identify with the localization of a ring $A_f$.

Evaluation at $\mathfrak{p}$ induces the ring homomorphism ${\mathcal{O}}_{X,\mathfrak{p}}\to A_{\mathfrak{p}}$. It is surjective because $a/s\in A_{\mathfrak{p}}$ is represented by a section on $D(s)$. For injectivity, represent a germ by $a/s$ on a neighbourhood of $\mathfrak{p}$. If its value in $A_{\mathfrak{p}}$ vanishes, there is $t\notin \mathfrak{p}$ with $ta=0$. The section is then zero after restricting to $D(st)$ inside that neighbourhood. Its germ is zero. Thus

Put $A=k[t_1,t_2]$. The punctured affine plane is covered by the principal open subschemes $D(t_1)$ and $D(t_2)$. Since $A$ is an integral domain, restriction identifies its global sections withThis intersection is $A$: if $a/t_1^r=b/t_2^s$, then $t_2^{s}a=t_1^{r}b$, and unique factorization forces $t_1^r$ to divide $a$. Therefore the inclusion $j:X\hookrightarrow \mathrm{Spec}A$ induces the identity on global regular functions.

If $X$ were an affine scheme, affine scheme reconstruction from global sections would make $j$ an isomorphism of schemes, because its global-section map is an isomorphism. But its image omits the closed point $(t_1,t_2)$. The punctured affine plane is therefore not affine. This is a useful example of global regular functions failing to distinguish a nonaffine open subscheme from its affine ambient space.

An isomorphism of schemes over $\mathbb{R}$ induces a bijection on rational points. A real point of this projective plane curve would be represented by a nonzero real triple with $t_0^2+t_1^2+t_2^2=0$. Each summand is nonnegative, so every coordinate would vanish. Thus $X(\mathbb{R})=\varnothing$, whereas $[1:0]\in {\mathbb{P}}^1(\mathbb{R})$. The two real schemes are not isomorphic. The obstruction is the real conic without real rational points; nonsingularity alone does not make a real conic a projective line over its ground field.