Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 113 / 1

An open immersion $j:U\to X$ is a morphism of schemes that identifies $U$, including its structure sheaf, with an open subscheme of $X$.

Let $X={\mathbb{A}}_k^2$ and $U=X\setminus \{(0,0)\}$. The inclusion is an open immersion and $X$ is an affine scheme, but $U$ is not affine. Indeed, regular functions extend across the missing codimension-two point, soWere $U$ affine, the inclusion would therefore correspond to an isomorphism of coordinate rings and would be an isomorphism $U\cong X$, a contradiction.

For an example with both schemes affine, the principal open subschemeis a nontrivial open immersion. The affine line is connected because $k[t]$ has no nontrivial idempotent elements.

The even-degree subring is the second Veronese subringwhere $u,v,w$ have degree one in the regraded ring. The canonical invariance of the Proj construction under passage to a Veronese subring givesOn homogeneous points this is the degree-two Veronese embeddingIt is an isomorphism onto the closed subschemeThus the quotient homomorphism $k[u,v,w]\twoheadrightarrow R^{(2)}$ supplies the requested closed immersion of $\mathrm{Proj}{R}^{(2)}$ into the projective plane.

For a point $p\in {\mathbb{A}}_k^1$, let $a$ be the image of $t$ in its residue field $\kappa (p)$. The scheme-theoretic fiber isIf $p\ne (t)$, then $a\ne 0$. The quadratic polynomial $xy-a{z}^2$ is irreducible. Indeed, after setting $z=0$, any hypothetical linear factors must restrict, up to nonzero scalars, to $x$ and $y$; comparing the $xz$ and $yz$ coefficients then forces both $z$ coefficients to vanish, contradicting the nonzero $z^2$ coefficient. Hence its homogeneous coordinate ring is an integral domain, so $Y_p$ is an integral scheme.

At the origin $p=(t)$, the fiber is $V_{+}(xy)$, the union of the two distinct projective lines $x=0$ and $y=0$, and is therefore not irreducible. It is nevertheless a reduced scheme because the ideal $(xy)=(x)\cap (y)$ equals its radical. Every other fiber is integral and hence reduced. Thus the fiber is integral exactly away from the origin, and it is reduced at every point of ${\mathbb{A}}_k^1$.