Past exam of the commutative algebra course of the University of Oxford / 2017 / 3

The Krull dimension of a commutative ring $R$ isThe length of a prime chain counts strict inclusions, rather than the number of prime ideals.

Let $R$ be a principal ideal domain that is not a field. We first show that every nonzero prime ideal is a maximal ideal. Write such a prime ideal as $P=(p)$ with $p\ne 0$, and suppose $P\subseteq (a)$. Then $p=ab$ for some $b\in R$. Since $ab\in P$, either $a\in P$, in which case $(a)=P$, or $b\in P$. In the second case $b=pc$ for some $c$, and $p=apc$ gives $1=ac$ by multiplicative cancellation in an integral domain. Thus $(a)=R$. This proves maximality.

Since $R$ is an integral domain, $(0)$ is a prime ideal, and every other prime ideal is maximal. Consequently every prime chain has length at most one. For the opposite inequality, choose a nonzero nonunit $a$, which exists because $R$ is not a field. The proper principal ideal $(a)$ is contained in a maximal ideal $\mathfrak{m}$; existence follows from Zorn's lemma applied to the proper ideals containing $(a)$. A maximal ideal is a prime ideal, since its quotient ring is a field. Therefore $(0)⊊\mathfrak{m}$ is a prime chain of length one. Thus the Krull dimension of a principal ideal domain that is not a field is

Let $P$ be a prime ideal of $R$ disjoint from the multiplicatively closed set $Y$. In the localization $S=Y^{-1}R$, its extension of an ideal isIndeed, finite sums of such fractions have a common denominator in $Y$ and numerator in $P$, and every generator $p/1$ has this form.

We show that its contraction of an ideal recovers $P$:One inclusion is immediate. Conversely, if $r/1=p/y$ with $p\in P$ and $y\in Y$, equality of fractions in the field of fractions of the integral domain $R$ gives $yr=p\in P$. Since $y\notin P$ and $P$ is a prime ideal, $r\in P$.

If $S{P}_1=S{P}_2$, contracting would therefore give $P_1=P_2$, contrary to the hypothesis. HenceWe will also need the fact that $SP$ is a prime ideal, part of the prime ideal correspondence for localization. It is proper by the contraction identity, since otherwise $1\in P$. If $(a/u)(b/v)\in SP$, multiplication by the denominator gives $ab/1\in SP$, and contraction gives $ab\in P$. Thus $a\in P$ or $b\in P$, so one of the two fractions belongs to $SP$.

Suppose, for a contradiction, that $P_2\cap \mathbb{Z}=(0)$. Then $P_1\cap \mathbb{Z}=(0)$ too. Both prime ideals are disjoint from the multiplicatively closed set $Y=\mathbb{Z}\setminus \{0\}$. The localization iseach fraction on the left has rational coefficients, and every polynomial with rational coefficients has a common nonzero integer denominator.

By b, the extended prime ideals form a strict inclusionThe first is nonzero, because $P_1$ contains a nonzero polynomial and the map $\mathbb{Z}[t]\to \mathbb{Q}[t]$ is injective.

The polynomial ring $\mathbb{Q}[t]$ is a principal ideal domain: in any nonzero ideal, a nonzero polynomial of least polynomial degree generates the ideal, since the polynomial division algorithm gives a remainder of smaller polynomial degree, which must be zero. By a, every nonzero prime ideal of this principal ideal domain is maximal. The strict inclusion of extended prime ideals is therefore impossible. We obtain

For any prime number $p$, there is a prime chainin $\mathbb{Z}[t]$. These are prime ideals because the respective quotient rings are $\mathbb{Z}[t]$, ${\mathbb{F}}_p[t]$, and ${\mathbb{F}}_p$, all integral domains. The inclusions are strict because $p\ne 0$ and $t\notin (p)$. Hence $\dim \mathbb{Z}[t]\ge 2$.

To prove the upper bound, suppose there were a prime chain of length at least three. Since $\mathbb{Z}[t]$ is an integral domain, we could insert $(0)$ at its beginning if necessary and select a chainBy c, $P_2\cap \mathbb{Z}\ne (0)$. This intersection is a proper prime ideal of $\mathbb{Z}$: if a product of integers lies in it, primality of $P_2$ puts one factor in it. Therefore $P_2\cap \mathbb{Z}=(p)$ for a prime number $p$, and $(p)\subseteq P_2$.

We need to check that $P_2\ne (p)$. Here prime ideals below a principal prime give the necessary restriction: no nonzero prime ideal can lie strictly below $(p)$. If $0\ne P⊊(p)$, then $p\notin P$. Choose $0\ne f\in P$ and factor out the largest common power of $p$ from its integer coefficients:Such a finite $k$ exists because $f$ has at least one nonzero integer coefficient. Since $p\notin P$ and $P$ is a prime ideal, repeated primality gives $g\in P\subseteq (p)$, a contradiction. Applied to $P_1⊊P_2$, this rules out $P_2=(p)$.

It follows that $P_2/(p)$ is a nonzero prime ideal ofAs in c, the polynomial division algorithm makes ${\mathbb{F}}_p[t]$ a principal ideal domain, so its nonzero prime ideals are maximal by a. Thus $P_2/(p)$ is maximal. The ideal correspondence theorem makes $P_2$ a maximal ideal of $\mathbb{Z}[t]$, contradicting $P_2⊊P_3$. Consequently