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).

  • $\begingroup$ 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

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 $\kappa < |S| = |k|$.

| cite | improve this answer | |
  • $\begingroup$ 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

This is the case iff $k$ is countable.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.