Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 101 / 6

The Krull dimension of a commutative ring $R$ isThe length counts strict inclusions, so a chain with $n+1$ prime ideals has length $n$. The supremum may be infinite. A field, for example, has just the prime ideal $(0)$ and therefore has Krull dimension zero. For the zero ring, the spectrum of a commutative ring is empty; under the convention $\sup \varnothing =-\infty$, its dimension is $-\infty$.

For a discrete valuation ring $A$,Let $\pi$ be a uniformizer, so its unique nonzero maximal ideal is $\mathfrak{m}=(\pi )$. We use the defining description of a discrete valuation ring: it is a non-field local ring that is a principal ideal domain, and every nonzero element is $u{\pi }^n$ for a unit $u$ and an integer $n\ge 0$.

Every nonzero proper ideal is therefore $({\pi }^n)$ for some $n\ge 1$. Such an ideal is a prime ideal only when $n=1$: for $n\ge 2$, $\pi {\pi }^{n-1}\in ({\pi }^n)$ while neither factor belongs to $({\pi }^n)$. The ideal $(\pi )$ is prime because it is maximal. Also $(0)$ is prime because $A$ is an integral domain. Thusand the longest strict chain has length one.

The original PDF asks about a discrete valuation ring; the local TeX transcription's phrase “discrete variable $X$” is an extraction error.

For a subring $S\subseteq R$ with the same identity, $R$ is an integral extension of $S$ if every $r\in R$ is an integral element over $S$: there exist $n\ge 1$ and $s_0,\dots ,s_{n-1}\in S$ such thatThe requirement that the polynomial be monic is essential. Merely satisfying some polynomial equation over $S$ is insufficient: for instance $1/2\in \mathbb{Q}$ satisfies $2T-1=0$ over $\mathbb{Z}$, but is not integral over $\mathbb{Z}$. To see this directly, a hypothetical monic equation for $1/2$, multiplied by $2^n$, would say that an odd integer $1$ plus a sum of even integers is zero.

The degree and coefficients may depend on $r$. Integrality does not by itself assert that the entire ring $R$ is a finite $S$-module; it does imply that the subalgebra $S[r]$ is generated as an $S$-module by $1,r,\dots ,r^{n-1}$.

We use three theorems for an integral extension $S\subseteq R$, none requiring Noetherian hypotheses. The Lying-over theorem says that each prime ideal $\mathfrak{p}$ of $S$ is $\mathfrak{q}\cap S$ for some prime ideal $\mathfrak{q}$ of $R$. The going-up theorem says that if $\mathfrak{p}\subseteq {\mathfrak{p}}^{\prime }$ in $S$ and $\mathfrak{q}\cap S=\mathfrak{p}$, there is ${\mathfrak{q}}^{\prime }\supseteq \mathfrak{q}$ with ${\mathfrak{q}}^{\prime }\cap S={\mathfrak{p}}^{\prime }$. The incomparability theorem for integral extensions says that comparable prime ideals of $R$ with the same contraction to $S$ are equal.

Take any strict chainin $R$. The contraction of an ideal operation gives ${\mathfrak{p}}_j={\mathfrak{q}}_j\cap S$, which are prime ideals of $S$, by the proof in Question 1(b), and remain strictly increasing by the incomparability theorem for integral extensions. Thus every chain length in $R$ occurs in $S$, giving $\dim R\le \dim S$.

Conversely, take any strict chain ${\mathfrak{p}}_0⊊\cdots ⊊{\mathfrak{p}}_n$ in $S$. Use the Lying-over theorem to choose ${\mathfrak{q}}_0$ over ${\mathfrak{p}}_0$, then repeatedly use the going-up theorem to obtain ${\mathfrak{q}}_0\subseteq \cdots \subseteq {\mathfrak{q}}_n$ with the prescribed contractions. Each inclusion must be strict, since its contractions are distinct. Hence every chain length in $S$ occurs in $R$, giving $\dim S\le \dim R$.

Taking suprema proves integral extensions preserve Krull dimension:The argument works equally well when the dimensions are infinite, since it compares all finite chain lengths.

No. In the Gaussian integers, chooseIndeed, substituting $i=-2$ gives $\mathbb{Z}[i]/(2+i)\cong \mathbb{Z}/5\mathbb{Z}$, so $P$ is a prime ideal and its contraction is $(5)$. Put $\beta =2-i$. Modulo $P$, $\beta$ becomes $4\ne 0$ in ${\mathbb{F}}_5$, so $\beta \notin P$. ThereforeWe show that $\alpha$ is not an integral element over ${\mathbb{Z}}_{(5)}$, using a direct reduction argument. Write $B={\mathbb{Z}}_{(5)}[i]$, viewed inside $\mathbb{Q}(i)$. There is a ring homomorphismwhich reduces rational coefficients modulo $5$. It is well defined because denominators in ${\mathbb{Z}}_{(5)}$ are not divisible by $5$, and $2^2+1=0$ in ${\mathbb{F}}_5$. In particular, $\rho (\beta )=0$ and $\rho (1)=1$.

If $\alpha$ satisfied a monic equation over ${\mathbb{Z}}_{(5)}$, saymultiplication by ${\beta }^n$ would give an equality in $B$:Applying $\rho$ gives $1=0$ in ${\mathbb{F}}_5$, a contradiction. ConsequentlyThis exhibits why localization of an integral extension at a prime need not be integral. Localizing both rings by the same multiplicative subset preserves an integral extension, but $\mathbb{Z}[i{]}_P$ inverts additional elements, including $2-i$, whose image at the other prime ideal $(2-i)$ is zero. The original extension is integral because $i$ satisfies the monic equation $T^2+1=0$: the subalgebra generated by finitely many integral elements property applies to $\mathbb{Z}[i]$.