Local DVRs at the nonzero primes force dimension one

Let $\mathfrak{p}⊊\mathfrak{q}$ be prime ideals of $R$ with $\mathfrak{p}\mathrm{\ne }(0)$. By hypothesis, $R_{\mathfrak{q}}$ is a discrete valuation ring. Because $R$ is a domain, any nonzero element of $\mathfrak{p}$ remains nonzero in $R_{\mathfrak{q}}$, so $\mathfrak{p}{R}_{\mathfrak{q}}$ is a nonzero prime ideal of $R_{\mathfrak{q}}$. Its contraction is $\mathfrak{p}$, so $\mathfrak{p}{R}_{\mathfrak{q}}⊊\mathfrak{q}{R}_{\mathfrak{q}}$. This contradicts [L3], which allows only $(0)$ and $\mathfrak{q}{R}_{\mathfrak{q}}$ as primes in a DVR. Therefore no nonzero prime ideal of $R$ is properly contained in another prime ideal.

Let $\mathfrak{p}$ be a nonzero prime ideal of $R$. By [L4], choose a maximal ideal $\mathfrak{m}$ containing $\mathfrak{p}$. Since $\mathfrak{p}$ is nonzero, step 1.1 forbids a strict inclusion $\mathfrak{p}⊊\mathfrak{m}$. Hence $\mathfrak{p}=\mathfrak{m}$, so every nonzero prime ideal of $R$ is maximal.

Assume now that $R$ is not a field. Choose a nonzero nonunit $x\in R$. Then $(x)$ is a proper ideal, so [L4] gives a maximal ideal $\mathfrak{m}$ containing it. The ideal $\mathfrak{m}$ is nonzero because it contains $x$, and step 2.1 shows that every nonzero prime ideal is maximal. Thus every strict prime chain has length at most $1$, while $(0)⊊\mathfrak{m}$ has length $1$. Hence $\dim R=1$.