Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 101 / 6

The height $\mathrm{ht}\mathfrak{p}$ is the supremum of lengths of strict chains of primes ending at $\mathfrak{p}$. The Krull principal ideal theorem says that in a Noetherian ring every prime minimal over a principal proper ideal has height at most one.

Induct on $n$. If $\mathfrak{p}$ is minimal over $I=(x_1,\dots ,x_n)$, choose a prime $\mathfrak{q}\subseteq \mathfrak{p}$ minimal over $(x_1,\dots ,x_{n-1})$ and localize appropriately. In $R/\mathfrak{q}$, the prime $\mathfrak{p}/\mathfrak{q}$ is minimal over the principal ideal generated by $x_n$, so its relative height is at most one. Induction gives $\mathrm{ht}\mathfrak{q}\le n-1$, hence

Write $\mathfrak{p}=(x)$. The principal ideal theorem and the hypothesis give $\mathrm{ht}\mathfrak{p}=1$. We use the standard principal prime in a Noetherian local ring lemma: a principal prime of positive height in a Noetherian local ring is generated by a nonzerodivisor and is the unique minimal prime above zero. The lemma follows by applying the associated-prime description of zero divisors and Nakayama's lemma to $\mathrm{ann}(x)$; if a nonzero annihilator or another minimal component existed, the principal prime would have height zero.

Here is the needed argument directly. For every $n\ge 1$,Indeed, if $y{x}^n=0$ with $y\notin \mathfrak{p}$, then $y$ is a unit in $R_{\mathfrak{p}}$, so $x^n=0$ there. The maximal ideal $\mathfrak{p}{R}_{\mathfrak{p}}=(x)$ would then be nilpotent, making $R_{\mathfrak{p}}$ zero-dimensional, contrary to $\mathrm{ht}\mathfrak{p}\ge 1$.

Now suppose $xy=0$. The displayed containment gives $y=x{y}_1$, then $x^2{y}_1=0$ gives $y_1=x{y}_2$, and inductivelyBecause $R$ is Noetherian, the ascending chain $\mathrm{ann}(x)\subseteq \mathrm{ann}(x^2)\subseteq \cdots$ stabilizes, say at $n=N$. Then $y_N\in \mathrm{ann}(x^{N+1})=\mathrm{ann}(x^N)$, and hence $y=x^N{y}_N=0$. Thus $x$ is a nonzerodivisor.

Let $N={\bigcap }_{n\ge 0}{x}^{n}R$. It is a finitely generated ideal. If $y=xz\in N$, then $xz\in x^{n+1}R$ for every $n$; cancellation of the nonzerodivisor $x$ gives $z\in x^{n}R$ for every $n$. Hence $N=xN$, and Nakayama lemma gives $N=0$ because $x$ lies in the maximal ideal.

Every nonzero element consequently has a finite $x$-adic order. If nonzero $a,b$ satisfied $ab=0$, write $a=x^r{a}^{\prime }$ and $b=x^s{b}^{\prime }$ with $a^{\prime },b^{\prime }\notin (x)$. Cancelling $x^{r+s}$ gives $a^{\prime }{b}^{\prime }=0$, contradicting primality of $(x)$. Equivalently, $(0)$ is prime, so

The formal power series ring $P=k[[X]]$ is Noetherian, so the finite product $R=P\times P$ is Noetherian. Its maximal ideals areso there are exactly two.

The idealis principal and prime because $R/\mathfrak{p}\simeq k$. The prime chainshows that it has height one, and no longer chain exists because $\dim P=1$. Yetwith both factors nonzero, so $R$ is not a domain. This shows why locality is essential in part i.

A stable $I$-filtration of $M$ is a descending sequence $M=M_0\supseteq M_1\supseteq \cdots$ with $I{M}_n\subseteq M_{n+1}$ for all $n$ and equality for all sufficiently large $n$. The Rees ring and associated Rees module areThe filtration condition makes multiplication by $I^r{t}^r$ send $M_n{t}^n$ into $M_{n+r}{t}^{n+r}$, so $M^{\ast }$ is a graded $R^{\ast }$-module.

If the filtration is stable from degree $\ell$, then $M^{\ast }$ is generated over $R^{\ast }$ by finite generating sets for $M_0,\dots ,M_{\ell }$. Conversely, let homogeneous elements of degrees at most $\ell$ generate $M^{\ast }$. In every degree $n>\ell$, each expression for an element of $M_n$ uses a positive-degree coefficient from $R^{\ast }$, so $M_n=I{M}_{n-1}$. Hence finite generation is equivalent to stability.

For $M^{\prime }\subseteq M$, takeThen $M^{\ast }$ is a graded submodule of the finite Rees module $⨁I^{n}M$. Since $R$ is Noetherian and $I$ is finitely generated, $R^{\ast }$ is Noetherian; hence $M^{\ast }$ is finite and the filtration is stable. Therefore, for some $\ell$ and all $n\ge \ell$,This is the Artin-Rees lemma.

Let $N={\bigcap }_{n\ge 0}{I}^{n}M$. Apply Artin--Rees to $N\subseteq M$. Since $N\subseteq I^{n}M$ for every $n$, stability givesfor all sufficiently large $n$, and in particular $N=IN$. The module $N$ is finitely generated because $M$ is Noetherian. The determinant trick applied to a finite generating set of $N=IN$ produces $a\in I$ withTaking $x=-a\in I$ yieldswhich is Krull intersection theorem.