= Solution
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 $I$,
$$
\operatorname{ht}(I)=
\inf_{\mathfrak p\supseteq I}\operatorname{ht}(\mathfrak p).
$$
This is the <height of an ideal>.
The <Krull height theorem> states that if $R$ is <Noetherian ring>[Noetherian], $I=(a_1,\ldots,a_m)$, and $\mathfrak p$ is minimal over $I$, then
$$
\boxed{\operatorname{ht}\mathfrak p\leq m}.
$$
We prove it by induction on $m$. The case $m=1$ is the <Krull principal ideal theorem>. For the induction step, let $\mathfrak q_1,\ldots,\mathfrak q_s$ be the finitely many minimal primes over $(a_2,\ldots,a_m)$ 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 chain
$$
\mathfrak p=\mathfrak p_d\supsetneq\cdots\supsetneq\mathfrak p_1\supsetneq\mathfrak p_0
$$
whose 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
$$
b\in\mathfrak p_1\setminus\bigcup_i\mathfrak q_i.
$$
The prime $\mathfrak p$ is minimal over $(b,a_2,\ldots,a_m)$. Otherwise a prime strictly between some $\mathfrak q_i$ and $\mathfrak p$ would show that $\mathfrak p/(a_2,\ldots,a_m)$ 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 $\mathfrak p/(b)$ give
$$
d-1\leq m-1,
$$
and hence $d\leq m$. 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.
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