An ideal contained in a finite union of prime ideals lies in one of them

We argue by induction on $n$. The case $n=1$ is immediate. Assume $n\ge 2$ and that the claim is known for smaller families. Suppose, for contradiction, that $I⊈{\mathfrak{p}}_i$ for every $i$. Then for each $k$, the ideal $I$ is not contained in ${\bigcup }_{i\mathrm{\ne }k}{\mathfrak{p}}_i$ either, because the induction hypothesis would then force $I\subseteq {\mathfrak{p}}_i$ for some $i\mathrm{\ne }k$. Hence for each $k$ we may choose $x_k\in I\setminus {\bigcup }_{i\mathrm{\ne }k}{\mathfrak{p}}_i$. Since $I\subseteq {\bigcup }_i{\mathfrak{p}}_i$, each $x_k$ must lie in ${\mathfrak{p}}_k$.

If $n=2$, then $x_1+x_2\in I$. It is not in ${\mathfrak{p}}_1$, because $x_1\in {\mathfrak{p}}_1$ would then force $x_2=(x_1+x_2)-x_1\in {\mathfrak{p}}_1$, contradicting the choice of $x_2$; the same argument shows $x_1+x_2\mathrm{\notin }{\mathfrak{p}}_2$. This contradicts $I\subseteq {\mathfrak{p}}_1\cup {\mathfrak{p}}_2$.

If $n\ge 3$, put $z=x_n+x_1{x}_2\cdots x_{n-1}\in I\mathrm{.}$ For $i<n$, the product term lies in ${\mathfrak{p}}_i$ because $x_i\in {\mathfrak{p}}_i$, while $x_n\mathrm{\notin }{\mathfrak{p}}_i$ by the choice in step 1.1; hence $z\mathrm{\notin }{\mathfrak{p}}_i$. Also each $x_j$ with $j<n$ lies outside ${\mathfrak{p}}_n$, so [L1] implies $x_1\cdots x_{n-1}\mathrm{\notin }{\mathfrak{p}}_n$; since $x_n\in {\mathfrak{p}}_n$, one gets $z\mathrm{\notin }{\mathfrak{p}}_n$ as well. This again contradicts $I\subseteq {\mathfrak{p}}_1\cup \cdots \cup {\mathfrak{p}}_n$.

The contradictions in steps 2.1 and 2.2 show that the assumption $I⊈{\mathfrak{p}}_i$ for every $i$ is impossible. Therefore $I\subseteq {\mathfrak{p}}_i$ for some $i$.