Past exam of the algebraic Geometry course of the University of Oxford / 2011 / 1

An affine variety in ${\mathbb{A}}_k^n$ is the common zero set of a collection of polynomials in $k[x_1,\dots ,x_n]$, with its Zariski topology. We use the convention of this question that such a variety need not be irreducible. Its vanishing ideal and coordinate ring areThe quotient identifies polynomials that give the same function on $X$, so its elements are exactly polynomial functions on $X$.

For affine embeddings $X\subseteq {\mathbb{A}}^n$ and $Y\subseteq {\mathbb{A}}^m$, a morphism of affine varieties is a map $F=(f_1,\dots ,f_m)$ with $f_i\in A(X)$ and image in $Y$. Pullback of polynomial functions defines the contravariant homomorphismIt is well defined because any polynomial vanishing on $Y$ pulls back to zero on $X$, and it preserves addition, multiplication, one, and the constants in $k$.

The radical of an ideal isThe ideal is radical if $I=\sqrt{I}$. Its zero set is $V(I)=\{a\in k^n:f(a)=0\text{ for every }f\in I\}$, an affine variety by definition.

Every $f\in I$ vanishes at every point of $V(I)$, so $I\subseteq I(V(I))$. If $g^N\in I(V(I))$, then $g(a{)}^N=0$ in the field $k$ for every $a\in V(I)$, so $g(a)=0$ for each such point. Hence $g\in I(V(I))$ and

Part (b) immediately gives $\sqrt{I}\subseteq I(V(I))$. For the reverse inclusion, let $h\in I(V(I))$. The case $h=0$ is immediate, so assume $h\ne 0$. Noetherianity gives finite generators $I=(f_1,\dots ,f_r)$.

Use the Rabinowitsch trick: in the polynomial ring with one auxiliary variable, formApply the assumed maximal-ideal description to polynomial rings, including this ring in $n+1$ variables. If $J$ were proper, it would lie in the maximal ideal of some point $(a,b)$. Then $a\in V(I)$ and $1-bh(a)=0$, impossible because $h$ vanishes on $V(I)$. Therefore $J$ is the whole ring, and there is an identitySubstitute $t=h^{-1}$ in the localization $R_h$. The last term vanishes; clearing the finitely many powers of $h$ from the remaining denominators yields $h^N\in I$ for some $N$. Thus $h\in \sqrt{I}$ andThis proves the strong Hilbert Nullstellensatz from the polynomial-ring weak form by converting nonvanishing into an extra equation.

Here subvariety means a closed affine subvariety. Suppose first that $F^{\#}$ is onto, and write $X\subseteq {\mathbb{A}}^n$, $Y\subseteq {\mathbb{A}}^m$. Put $J=\ker F^{\#}$ and $Z=V_Y(J)$. Choose $g_i\in A(Y)$ with $F^{\#}{g}_i=x_i{|}_X$ and let $G:Z\to {\mathbb{A}}^n$ have components $g_i{|}_Z$.

If $h\in I(X)$, then $F^{\#}(h(g_1,\dots ,g_n))=h{|}_X=0$, so $h(g_1,\dots ,g_n)\in J$. It vanishes on $Z$, proving $G(Z)\subseteq X$. The chosen lifts give $G\circ F={\mathrm{id}}_X$, and $F(X)\subseteq Z$ by the definition of $J$.

Let the components of $F$ be polynomial functions $f_1,\dots ,f_m$ on $X$. For each target coordinate $y_j$, the element $y_j-f_j(g_1,\dots ,g_n)$ has zero pullback, hence belongs to $J$. It vanishes on $Z$, giving $F\circ G={\mathrm{id}}_Z$. Both maps have polynomial components, so $F$ is an isomorphism onto the closed subvariety $Z$.

Conversely, if $F:X\to Z\subseteq Y$ is an isomorphism onto a closed subvariety, restriction $A(Y)\to A(Z)$ is surjective, and the pullback $A(Z)\to A(X)$ of that isomorphism is an isomorphism. Their composition is $F^{\#}$, provingThe word closed matters: an isomorphism onto a merely locally closed subset need not induce a surjective coordinate-ring map. For example ${\mathbb{G}}_m\hookrightarrow {\mathbb{A}}^1$ induces $k[t]\hookrightarrow k[t,t^{-1}]$.

The parametrization lands on the curve because $(t^3{)}^4=(t^4{)}^3$. Its pullback has imageEvery monomial in this subalgebra has exponent $3a+4b$ for nonnegative integers $a,b$. No such exponent is one, so no linear combination of these monomials is $t$. Thus $F^{\#}$ is not surjective and $F$ is not an isomorphism.

It is nevertheless bijective on points. Away from the origin, $x\ne 0$ and $t=y/x$ gives $t^3=x$ and $t^4=y$ using $x^4=y^3$; the origin comes only from $t=0$. The rational inverse cannot extend regularly at the origin: if $t=P(t^3,t^4)/Q(t^3,t^4)$ with $Q(0)\ne 0$, then $tQ(t^3,t^4)$ has a nonzero degree-one term, whereas $P(t^3,t^4)$ has none.This distinguishes a set-theoretic bijection from an isomorphism of varieties in every characteristic.

This is a bijective monomial cusp parametrization with nonregular inverse.