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

A quasi-projective variety is a locally closed subset of projective space: equivalently an open subset of some projective variety, with the induced algebraic variety structure. Affine varieties are examples, via the standard affine chart in projective space.

It is irreducible if it is nonempty and cannot be written as the union of two proper closed subsets. Equivalently, every two nonempty open subsets meet, or every nonempty open subset is dense. Closed and open are interpreted in the induced Zariski topology of $X$.

For a multiplicatively closed set $S$ containing one, the localization of a ring $S^{-1}R$ consists of fractions $r/s$ with $r\in R,s\in S$, under the equivalenceThe factor $u$ is needed when the ring has zero divisors. Addition and multiplication areThe canonical map sends $r$ to $r/1$ and makes every element of $S$ a unit. It is universal with this property: a ring homomorphism $\varphi :R\to T$ taking $S$ into units extends uniquely by $r/s\mapsto \varphi (r)\varphi (s{)}^{-1}$. Usually zero is excluded from $S$; if it is included, this construction is the zero ring.

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.

Polynomial functions are regular, giving an injective map $A(X)\hookrightarrow {\mathcal{O}}_X(X)$; injectivity follows from quotienting by the full vanishing ideal. To prove surjectivity, let $f$ be globally regular and consider the denominator idealIt is an ideal because sums and polynomial multiples preserve this property.

Fix $p\in X$. Locally $f=a/b$ on a neighbourhood $V$ of $p$, with $b$ nonzero there. Principal affine opens form a basis, so choose $h\in A(X)$ with $h(p)\ne 0$ and $D(h)\subseteq V$. Thenas functions on all of $X$: on $D(h)$ it follows from the local fraction, and off $D(h)$ both sides vanish because $h=0$. Hence $hb\in J$ and $(hb)(p)\ne 0$.

Thus $J$ has no common zero on $X$. By the weak Nullstellensatz for $A(X)$, every proper ideal lies in a maximal ideal belonging to a point of $X$; therefore $J$ cannot be proper. We get $1\in J$, so $f$ itself is polynomial. This provesThe argument works for reducible affine varieties as well: it uses functions and denominator ideals rather than assuming a single fraction field.

This is the denominator ideal proof for global regular functions.

For any affine variety and point $p$, with ${\mathfrak{m}}_p=\{f\in A(X):f(p)=0\}$, the standard localization description isAlso ${\mathcal{O}}_X(D(g))\cong A(X{)}_g$ for a principal open subset. One way to see the latter is to realize $D(g)$ as the affine variety $\{(x,z):g(x)z=1\}$; its coordinate ring is $A(X)[z]/(gz-1)=A(X{)}_g$ and part (d) applies.

For this cusp use $x=t^3,y=t^2$, identifying $A(X)$ with $k[t^2,t^3]$ inside $k(t)$. The parametrization gives exactly the relation $x^2=y^3$; its kernel is generated by this binomial. For example reduce a polynomial modulo $x^2-y^3$ to $a(y)+xb(y)$; substitution has even powers from $a(t^2)$ and odd powers at least three from $t^{3}b(t^2)$, so it can vanish only when both polynomials are zero.

Write $g=x-y=t^2(t-1)$. The open set corresponds to $t\ne 0,1$, andTo simplify this ring without assuming the normalization coordinate is present, observe that in $A(X)$Thus $y$ becomes a unit after inverting $g$, with inverse $(y^2-2x+y)/g^2$. Consequently $t=x/y$ belongs to the localization. Since $g=t^2(t-1)$ is a unit, both $t$ and $t-1$ become units. Conversely every original generator lies in the corresponding two-point localization of $k[t]$. HenceFinally $p=(1,1)$ corresponds to $t=1$. It lies outside $U$, so compute its local ring from $X$, at $\mathfrak{m}=(x-1,y-1)$. The element $y$ is a unit there, hence $t=x/y$ belongs to $A(X{)}_{\mathfrak{m}}$. Every polynomial in $t$ nonzero at one is then a unit. Conversely a denominator from $A(X)\setminus \mathfrak{m}$ remains nonzero at $t=1$. These inclusions proveThis smooth point has the ordinary one-variable local ring, even though the global curve has a cusp at its origin.

The localization calculation describes regular functions on a cusp with two points removed.