Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 1 / 25I (Algebraic Geometry)

The quotient $A$ is a field generated as a $k$-algebra by the images ${\alpha }_1,\dots ,{\alpha }_n$ of the variables. We prove the relevant form of Zariski lemma.

Choose a maximal algebraically independent subfamily, say ${\alpha }_1,\dots ,{\alpha }_r$, and putEvery remaining ${\alpha }_j$ is algebraic over $K$. After clearing the finitely many denominators in their monic equations, there is a nonzerosuch that all ${\alpha }_j$ with $j>r$ are integral overBecause $A$ is a field containing $R$ and is generated by these integral elements, $A$ is integral over $R$. If a field is integral over a subring, that subring is itself a field: for $0\ne c\in R$, an integral equation for $c^{-1}$ can be multiplied by a power of $c$ to express $c^{-1}$ in $R$.

If $r>0$, uncountability of $k$ lets us choose $a\in k$ such that ${\alpha }_1-a$ does not divide $h$. Then ${\alpha }_1-a$ is not invertible in the localization $R$, contradicting that $R$ is a field. Hence $r=0$, so every generator ${\alpha }_i$, and therefore every element of the finite algebraic extension they generate, is algebraic over $k$. Thus

Use the Rabinowitsch trick. In the polynomial ring $k[x_1,\dots ,x_n,t]$, considerIf $I$ were proper, it would lie in a maximal ideal. Part (a), together with algebraic closedness of $k$, is the Weak Hilbert Nullstellensatz, so that maximal ideal would be evaluation at a point $(a_1,\dots ,a_n,b)\in k^{n+1}$. We would then haveBut the hypothesis gives $f(a)=0$, a contradiction. Therefore $I$ is the unit ideal.

There are polynomials $g_j$ and $h$ such thatSubstitute $t=f^{-1}$ in the localization $k[x_1,\dots ,x_n{]}_f$. This gives $1\in J_f$; clearing a power of $f$ yieldsThis is the Strong Hilbert Nullstellensatz inclusion $I(Z(J))\subseteq \sqrt{J}$.

Suppose first that $\overline{f(X)}=Y$. If $g\in \ker f^{\ast }$, then $g\circ f=0$, so the zero set of the regular function $g$ contains $f(X)$ and hence its closure $Y$. Thus $g=0$ in the coordinate ring $k[Y]$, and $f^{\ast }$ is injective.

Conversely, if $\overline{f(X)}$ were a proper closed subset of $Y$, its vanishing ideal in $k[Y]$ would contain a nonzero regular function $g$. Then $g$ vanishes on $f(X)$, so $f^{\ast }(g)=0$, contradicting injectivity. ThereforeThis is the coordinate-ring criterion for a dominant morphism of affine varieties.

For an irreducible affine variety, define dimension byIrreducibility makes both coordinate rings integral domains. The injective map $f^{\ast }:k[Y]\hookrightarrow k[X]$ extends to an embedding of function fieldsAn algebraically independent family in $k(Y)$ remains algebraically independent in $k(X)$, so the dimension of an irreducible affine variety by transcendence degree gives