Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 101 / 4

For an algebraically closed field $k$, the Weak Hilbert Nullstellensatz says that every maximal ideal of $k[x_1,\dots ,x_n]$ isfor a unique $a\in k^n$. Equivalently, every proper ideal has a common zero. The Strong Hilbert Nullstellensatz says that for every ideal $I$,

To deduce the strong form, let $f$ vanish on $V(I)$ and introduce a variable $y$. The equations in $I$ together with $1-yf$ have no common zero: a common zero would satisfy both $f=0$ and $yf=1$. The weak theorem therefore givesSubstitute $y=f^{-1}$ in the localization $k[x_1,\dots ,x_n,f^{-1}]$. The last term vanishes, and clearing a power of $f$ yields $f^N\in I$. Thus $f\in \sqrt{I}$. The reverse inclusion is immediate, completing the Rabinowitsch trick proof.

TakeThenEvery prime ideal in the image ofis disjoint from the multiplicative set $\{1,t_1,t_1^2,\dots \}$. In particular, the prime ideal $(t_1)$ is not in the image, so the contraction map is not surjective.

Let $d=\mathrm{deg}f$ and let $f_d$ be the nonzero highest homogeneous part. Choose one coordinate, after a permutation, such thatis not the zero polynomial. A polynomial of degree at most $d$ in each variable cannot vanish on the entire grid $\{0,1,\dots ,d{\}}^{n-1}$, by induction on the number of variables. Hence there are $a_i\in \{0,1,\dots ,d\}$ such that

SetThis is given, up to the initial coordinate permutation, by an integer matrix $M$ with determinant $\pm 1$ andIn the inverse coordinates $t_i=y_i+a_i{y}_n$, the coefficient of $y_n^d$ in $f$ is the nonzero real number $f_d(a_1,\dots ,a_{n-1},1)$. Dividing by it makes the defining equation monic in $y_n$. Thus $A$ is integral over $\mathbb{R}[y_1,\dots ,y_{n-1}]$ by linear Noether normalization for a hypersurface. The Lying-over theorem now makessurjective.

Pass to the integral extensionThe ring $B/I$ is an integral domain, and the ideal $J/I$ has zero contraction to the base. If $J/I$ contained a nonzero element $b$, choose an integral equation for $b$ of least degree:Its constant term is nonzero, since otherwise the domain property would let us cancel $b$ and obtain an equation of lower degree. Butis a nonzero element of $(J/I)\cap (A/(I\cap A))$, a contradiction. Therefore $J/I=0$ and $I=J$. This proves ideal contraction rigidity under an integral extension.

It is enough to prove that every prime ideal $\mathfrak{q}$ of $S$ is the intersection of the maximal ideals containing it. Replace $R\subseteq S$ byThe new extension is integral, both rings are domains, and the base remains a Jacobson ring.

Let $0\ne s\in S$. Choose an integral equation of least degreeAs above, $a_0\ne 0$. Since the zero ideal of $R$ is the intersection of its maximal ideals, choose a maximal ideal $\mathfrak{m}$ with $a_0\notin \mathfrak{m}$. By the Lying-over theorem, some maximal ideal $\mathfrak{n}$ of $S$ contracts to $\mathfrak{m}$. If $s\in \mathfrak{n}$, the integral equation would imply $a_0\in \mathfrak{n}\cap R=\mathfrak{m}$, a contradiction. Thus every nonzero $s$ is omitted by some maximal ideal, so their intersection is zero. Therefore $S$ is Jacobson, proving Integral extension of a Jacobson ring.