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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).

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  • $\begingroup$ It will depend on whether $k$ is countable or not. $\endgroup$ – Martin Brandenburg May 23 '15 at 23:10
  • $\begingroup$ Yeah so if $k$ is uncountable, I guess the answer will be no. It just occurred to me that it might follow from linear algebra fact: a $k$-vector space $V$ cannot be written as a union of $n$ proper subspaces if $n$ is strictly less than the cardinality of $k$. So that should deal with the case when $k$ is uncountable. $\endgroup$ – Prism May 23 '15 at 23:12
  • $\begingroup$ And when k is countable, then the maximal ideal is countable, and that's it. $\endgroup$ – Martin Brandenburg May 23 '15 at 23:13
  • $\begingroup$ In that case, $R$ is countable and so $\mathfrak{m}$ is countable. And I can just proceed in the same way as "slick solution" in the beginning of the post. Thanks so much Martin! Feel free to post to answer box below, so I can accept your answer. $\endgroup$ – Prism May 23 '15 at 23:15
  • $\begingroup$ @MartinBrandenburg: Wait, the "linear algebra fact" I quoted above is false! See Pete L Clark's answer here. So how do we show that if $k$ is uncountable, then $\mathfrak{m}$ cannot be written as a countable union of strictly smaller prime ideals? $\endgroup$ – Prism May 23 '15 at 23:23
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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 $\kappa < |S| = |k|$.

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  • $\begingroup$ Thank you! This is very nice. I have up-voted your answer for now, and will grant the bounty later (if no other solution gets posted). $\endgroup$ – Prism Jun 4 '15 at 2:32
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This is the case iff $k$ is countable.

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  • $\begingroup$ I just added bounty in order to receive the correct proof of this fact. (My comments show how one natural approach fails in the light of Pete's answer from another thread). $\endgroup$ – Prism Jun 3 '15 at 23:54

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