Consider the polynomial ring of countable variables with coefficients in the real numbers, i.e, $S=\Bbb{R}[x_1,x_2,x_3,...,x_n,...]$. My Question is about the depth of this ring.

Question: Could we find any maximal ideal in $S$ such that it contains an anti-chain with uncountable number of prime ideals?

PS: We recall that a set $\{P_{\alpha}\} _{\alpha \in \mathcal{A}}$ is an anti-chain of prime ideals if for each $u ,v \in T$, neither $P_{u} \subseteq P_{v}$ nor $P_{v} \subseteq P_{u}$.

PPS: By this notation, I am seeking an anti-chain $\{P_{\alpha}\} _{\alpha \in \mathcal{A}}$ of prime ideals in $S$ for which $T$ is uncountable.

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    $\begingroup$ Can't we take $P_u = (x - uy)$ for any real number $u$? $\endgroup$
    – Will Sawin
    Oct 29, 2013 at 16:30

1 Answer 1


In the same spirit as the answer to your earlier question, pick a bijection $\phi$ from the natural numbers to the rationals. For each real number $\alpha$, let $I_\alpha$ be the ideal generated by all $x_n$ such that $\phi(n)<\alpha$ and let $J_\alpha$ be the ideal generated by all $x_n$ such that $\phi(n)>\alpha+1$. Let $P_\alpha=I_\alpha+J_\alpha$. The $P_\alpha$ form an uncountable anti-chain contained in the maximal ideal of $S$ generated by all the $x_n$.

  • $\begingroup$ WoooW!!! what a versatile and basic notion. Thank you very much dear Professor Landsburg. I like the unique and special trick you have used in both of my questions. $\endgroup$
    – Alex
    Oct 29, 2013 at 15:36

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