Minimal primes over a proper ideal exist

The singleton set $\{1\}$ is multiplicative and is disjoint from $I$ because $I$ is proper. Applying [L1] gives at least one prime ideal containing $I$. Let $\mathrm{\Sigma }$ be the set of all prime ideals containing $I$, ordered by reverse inclusion. Then $\mathrm{\Sigma }\mathrm{\ne }\mathrm{\varnothing }$.

Let $\mathcal{C}\subseteq \mathrm{\Sigma }$ be a chain. Put $\mathfrak{q}={\bigcap }_{\mathfrak{p}\in \mathcal{C}}\mathfrak{p}\mathrm{.}$ Then $I\subseteq \mathfrak{q}$. To see that $\mathfrak{q}$ is prime, let $ab\in \mathfrak{q}$ and assume $a,b\mathrm{\notin }\mathfrak{q}$. Choose ${\mathfrak{p}}_a,{\mathfrak{p}}_b\in \mathcal{C}$ with $a\mathrm{\notin }{\mathfrak{p}}_a$ and $b\mathrm{\notin }{\mathfrak{p}}_b$. Since $\mathcal{C}$ is totally ordered by inclusion, either ${\mathfrak{p}}_a\subseteq {\mathfrak{p}}_b$ or ${\mathfrak{p}}_b\subseteq {\mathfrak{p}}_a$. In the first case $b\mathrm{\notin }{\mathfrak{p}}_a$ because ${\mathfrak{p}}_a\subseteq {\mathfrak{p}}_b$; in the second case $a\mathrm{\notin }{\mathfrak{p}}_b$. Either way one of the primes in the chain contains $ab$ but neither factor, a contradiction. Thus $\mathfrak{q}\in \mathrm{\Sigma }$, and it is an upper bound of $\mathcal{C}$ in the reverse-inclusion order.

By [L2], the poset $\mathrm{\Sigma }$ has a maximal element for reverse inclusion. Such an element is exactly a prime ideal minimal by ordinary inclusion among the primes containing $I$.

Therefore every proper ideal lies under a minimal prime ideal.