Here's a problem (Exercise 3.21) from "A Term in Commutative Algebra" by Altman & Kleiman:
Let $k$ be a field, and $R=k[X, Y]$ be polynomial ring in two variables. Let $\mathfrak{m}=\langle X, Y\rangle$ - this is (maximal) ideal generated by $X$ and $Y$. Show that $\mathfrak{m}$ is a union of strictly smaller prime ideals.
And here's a slick solution at the back of the book. For each $f\in \mathfrak{m}$, we know that $f$ has a prime factor $p_{f}$ (because $R$ is UFD), and so $\mathfrak{m} = \bigcup_{f\in \mathfrak{m}} \langle p_{f} \rangle$. Each $\langle p_{f} \rangle$ is a prime ideal, and $\langle p_{f} \rangle\neq \mathfrak{m}$ because $\mathfrak{m}$ is non-principal.
Now my question is: Can we write $\mathfrak{m}$ as a countable union of strictly smaller prime ideals? I guess the answer could potentially depend on whether or not $k$ is infinite. I'd be interested seeing ideas for any field (say $k=\mathbb{C}$ if that makes it simpler).
Consider the set $S = \{X+\alpha Y \mid \alpha\in k\} \subset (X,Y)$.
If an ideal $I$ contains two distinct elements of $S$, $X+\alpha Y$ and $X+\beta Y$, then it contains $\frac{1}{\beta-\alpha}((X-\alpha Y)-(X-\beta Y)) = Y$, and thus contains $X$ as well, so $(X,Y) \subset I$.
Since an ideal properly contained in $(X,Y)$ contains at most one element of $S$, it follows that $(X,Y)$ cannot be the union of $\kappa$ such ideals for $\kappa < |S| = |k|$.