Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 20 / 1

A morphism of schemes is a morphism of locally ringed spaces. Thus it consists of a continuous map $f:|X|\to |Y|$ and a homomorphism of sheaves of ringssuch that, for every $x\in X$, the induced map on stalksis a local homomorphism: it carries the maximal ideal of the source into the maximal ideal of the target. Equivalently, the inverse image of the target maximal ideal is the source maximal ideal. The locality condition on every stalk distinguishes a morphism of schemes from a general morphism of ringed spaces.

The affine-target adjunction for schemes gives the natural bijectionA morphism of schemes $f$ gives the homomorphism on global sections induced by $f^{\#}$, using $\Gamma (\mathrm{Spec}A,\mathcal{O})=A$.

Conversely, let $\alpha :A\to \Gamma (X,{\mathcal{O}}_X)$ be a ring homomorphism. Choose an affine open subscheme cover $U_i=\mathrm{Spec}{B}_i$ of $X$. Restriction of global sections gives homomorphisms $A\to B_i$, hence morphisms of schemes $f_i:U_i\to \mathrm{Spec}A$. On any affine open subscheme $W\subseteq U_i\cap U_j$, both restrictions correspond to the same homomorphism $A\to \Gamma (W,{\mathcal{O}}_W)$. They therefore agree on $W$. Such affine opens cover the overlap, so the $f_i$ glue uniquely to a morphism of schemes $f:X\to \mathrm{Spec}A$.

The two constructions are inverse: the first recovers $\alpha$ on each $U_i$, hence on all of $X$, and the second recovers every restriction $f{|}_{U_i}$ of a given $f$. This proof needs neither affineness nor quasi-compactness of $X$.

Let $\mathfrak{m}$ be the maximal ideal of the local ring $A$. Every open subset of $\mathrm{Spec}A$ containing the closed point $\mathfrak{m}$ is the whole spectrum of a commutative ring: it contains a principal open subscheme $D(a)$ with $a\notin \mathfrak{m}$, and that $a$ is a unit, so $D(a)=\mathrm{Spec}A$.

Given $f:\mathrm{Spec}A\to {\mathbb{P}}_{\mathbb{Z}}^n$, choose a standard affine open subscheme $D_{+}(x_i)$ containing $f(\mathfrak{m})$. Its preimage is consequently all of $\mathrm{Spec}A$. The affine-target adjunction for schemes expresses $f$ in this chart by elements $b_j\in A$ for $j\ne i$, the images of $x_j/x_i$. It is represented by homogeneous coordinates with $a_i=1$ and $a_j=b_j$.

Conversely, a tuple $(a_0,\dots ,a_n)$ with some $a_i\in A^{\times }$ defines a morphism of schemes into $D_{+}(x_i)$ by $x_j/x_i\mapsto a_j/a_i$. Choosing another unit entry gives the same morphism of schemes, since the usual projective space transition functions identify the ratios. Multiplying all entries by one unit does not change any ratio. If two such tuples define the same morphism of schemes, choose a unit entry $a_i$ in the first and a unit entry $a_j^{\prime }$ in the second. In the second chart, the function $x_i/x_j$ pulls back to $a_i^{\prime }/a_j^{\prime }$. Because the whole map lies in $D_{+}(x_i)$, this ratio is a unit, so $a_i^{\prime }$ is a unit too. Equality in this chart gives $a_j/a_i=a_j^{\prime }/a_i^{\prime }$ for every $j$, hence $a_j^{\prime }=(a_i^{\prime }/a_i)a_j$.

Thus the correspondence is exactlyThis is the projective coordinates over a local ring description.

For a general ring, the key open-neighbourhood argument fails. Even a tuple generating the unit ideal need not have any unit entry. For example, over $A=k\times k$ the pair $((1,0),(0,1))$ defines a map to ${\mathbb{P}}^1$ whose two points have images $[1:0]$ and $[0:1]$. Neither coordinate is a unit, and no common unit multiple changes that fact. The map lies in no single standard chart. More generally, maps into projective space correspond to invertible sheaf quotients of $A^{n+1}$; the quotient need not be a free rank-one module outside the local ring case.