= Solution
{c}
The <Krull dimension> of a <commutative ring> $R$ is
$$
\boxed{\dim R=\sup\{d:\text{there exist prime ideals }P_0\subsetneq P_1\subsetneq\cdots\subsetneq P_d\}.}
$$
The 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 $p\ne0$, 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 $(0)\subsetneq\mathfrak m$ is a <prime chain> of length one. Thus the <Krull dimension of a principal ideal domain> that is not a <field> is
$$
\boxed{\dim R=1.}
$$
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