What are the prime ideals of $\mathbb{R}[x_1,x_2,x_3,...]$?

(this is the ring of polynomials over the reals with countably infinite many indeteminates).

My attempt: I think taking the principal ideal generated by an irreducible polynomial gives a prime ideal.

  • $\begingroup$ I would say your attempt is correct, but do notice that $\langle 0 \rangle$ is also prime (and is the only one that you can't reach in this way $\endgroup$ – Belgi Jun 10 '12 at 9:46
  • $\begingroup$ @Belgi: But why are these the only prime ideals? $\endgroup$ – Ola Jun 10 '12 at 9:47
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    $\begingroup$ No, they aren't. For example $(x_1, x_2)$ is prime, as $\mathbb R[x_1, x_2, x_3, \ldots]/(x_1, x_2) \cong \mathbb R[x_3, x_4, \ldots]$ is integral. $\endgroup$ – martini Jun 10 '12 at 10:01
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    $\begingroup$ This is homework? It seems to me that it is quite hard to give a reasonable description of all prime ideals of this ring. Here's one that isn't finitely generated: $(x_1^2 - x_2, x_2^2 - x_3, x_3^2 - x_4, ...)$. Are you sure the question isn't about maximal ideals? $\endgroup$ – Qiaochu Yuan Jun 10 '12 at 10:15
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    $\begingroup$ @Ola: here is an example of easy description. The non-noetherian ring $\mathbb R[x_1,x_2,..]/(x_i\cdot x_j\mid i,j=1,2,3,...)$ has $(x_1,x_2,x_3,...)$ as its only prime ideal. $\endgroup$ – Georges Elencwajg Jun 10 '12 at 10:30

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