Past exam of the algebraic Geometry course of the University of Oxford / 2011 / 3 / c

A regular function on $U$ is a function $f:U\to k$ locally representable, in an affine chart of the ambient projective space, as a quotient $a/b$ of polynomial functions with denominator nonzero on that neighbourhood. Equivalently it is locally a quotient of homogeneous forms of the same degree. The equal-degree condition makes the quotient independent of the chosen projective representative.

The ring ${\mathcal{O}}_{X,p}$ is the ring of germs of such functions near $p$. Two representatives $(U,f)$ and $(V,g)$ define the same germ when they agree on some smaller neighbourhood of $p$. Addition and multiplication use a common neighbourhood. It is a local ring: its unique maximal ideal consists of germs vanishing at $p$, and any germ nonzero at $p$ has a regular reciprocal on a smaller neighbourhood.

A morphism $F:X\to Y$ is a continuous map in the Zariski topologies such that for every open $V\subseteq Y$ and every regular function $h$ on $V$, the pullback $h\circ F$ is regular on $F^{-1}(V)$.

Equivalently, near every point, choose affine charts of source and target; the target coordinates of $F$ must then be regular functions of the source coordinates. These local conditions agree on overlaps. In particular a morphism is defined at every point of its source, not merely on a dense open subset.

A rational map from irreducible $X$ to $Y$ is an equivalence class of morphisms $F_U:U\to Y$ defined on nonempty open subsets $U\subseteq X$. Two representatives are equivalent if they agree on a nonempty open subset of their intersection.

Irreducibility makes all these nonempty opens dense and ensures their intersections are nonempty, so the equivalence expresses agreement at general points. A rational map need not extend across all of $X$, and it need not be dominant. An injective homomorphism of function fields additionally requires a dominant morphism of affine varieties, or the corresponding condition on a rational map.