Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 101 / 5

For a prime ideal $\mathfrak{p}$, its height is the supremum of the lengths of strict chains of prime ideals ending at $\mathfrak{p}$. For a proper ideal $I$,This is the height of an ideal.

The Krull height theorem states that if $R$ is Noetherian, $I=(a_1,\dots ,a_m)$, and $\mathfrak{p}$ is minimal over $I$, then

We prove it by induction on $m$. The case $m=1$ is the Krull principal ideal theorem. For the induction step, let ${\mathfrak{q}}_1,\dots ,{\mathfrak{q}}_s$ be the finitely many minimal primes over $(a_2,\dots ,a_m)$ that lie below $\mathfrak{p}$. By induction each has height at most $m-1$; if one equals $\mathfrak{p}$, we are done.

Otherwise, suppose $\mathfrak{p}$ has finite height $d$ and choose a chainwhose first nonminimal term ${\mathfrak{p}}_1$ is contained in none of the ${\mathfrak{q}}_i$. Such a chain is obtained by prime avoidance and the principal ideal theorem: a three-term segment can be replaced by a prime minimal over ${\mathfrak{p}}_0+(b)$ for an element $b$ avoiding the finitely many unwanted primes. Choose

The prime $\mathfrak{p}$ is minimal over $(b,a_2,\dots ,a_m)$. Otherwise a prime strictly between some ${\mathfrak{q}}_i$ and $\mathfrak{p}$ would show that $\mathfrak{p}/(a_2,\dots ,a_m)$ has height at least two, although it is minimal over the principal ideal generated by $a_1$; this contradicts the principal ideal theorem. In $R/(b)$, the prime $\mathfrak{p}/(b)$ is therefore minimal over an ideal generated by $m-1$ elements, so induction bounds its height by $m-1$. The $d-1$ strict inclusions from ${\mathfrak{p}}_1/(b)$ to $\mathfrak{p}/(b)$ giveand hence $d\le m$. If the height were infinite, the same argument applied to arbitrarily long finite chains would give the same fixed bound, which is impossible. This completes the proof.

Let $S=k[x_1,\dots ,x_r]$ with positive degrees $d_i=\mathrm{deg}{x}_i$, and let $M$ be a finitely generated graded $S$-module whose graded pieces are finite-dimensional over $k$. The Hilbert-Serre theorem states thatfor some Laurent polynomial $P(t)\in \mathbb{Z}[t,t^{-1}]$. For the standard grading, the Hilbert function $n\mapsto {\dim }_k{M}_n$ consequently agrees for all sufficiently large $n$ with a polynomial in $n$.

For the proof, induct on $r$. When $r=0$, $M$ is finite-dimensional and its Hilbert series is a Laurent polynomial. For $r>0$, multiplication by $x_r$ gives an exact sequence of graded modulesBoth $K$ and $C$ are annihilated by $x_r$, so they are finitely generated graded modules over $k[x_1,\dots ,x_{r-1}]$. Additivity of the Hilbert series yieldsThe induction hypothesis supplies the required denominator for the right side and proves the rational formula. When all $d_i=1$, expanding $(1-t{)}^{-r}$ shows that its coefficients are binomial polynomials in $n$, which proves eventual polynomiality.

The cases $n=0$ and $n=d$ can indeed be finite: a Noetherian ring has finitely many minimal primes, and a semilocal ring may have finitely many maximal ideals. We prove that every intermediate height occurs infinitely often.

Because $d$ is a finite integer equal to the supremum of prime-chain lengths, there is a chainEach ${\mathfrak{p}}_i$ has height exactly $i$: its displayed lower chain gives height at least $i$, while any longer lower chain could be extended by the remaining displayed primes and would contradict $\mathrm{ht}{\mathfrak{p}}_d=d$.

Suppose $0<n<d$. The three primesfall under prime ideals between a three-prime chain, so infinitely many primes $\mathfrak{q}$ satisfyEvery such $\mathfrak{q}$ has height exactly $n$: the lower inclusion gives height at least $n$, and height at least $n+1$ would, after appending ${\mathfrak{p}}_{n+1}$, contradict its height $n+1$. Therefore there are infinitely many height-$n$ primes. The assumed finiteness forcesThis is infinitude of intermediate-height prime ideals.