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Mar 2 at 23:07 comment added Todd Trimble @NoahSchweber Thank you so much for the information!
Mar 2 at 20:37 comment added Noah Schweber FWIW the original WKL-based argument isn't silly at all if you're paying attention to computability-theoretic complexity. Indeed, it's basically the key behind the observation that while the existence of principal ideals is equivalent to $\mathsf{ACA}_0$, the existence of prime ideals is strictly weaker; see Downey/Lempp/Mileti, as well as the followup by them + Hirschfeldt/Kach/Montalban. Pardon the nostalgia trip: the DLM paper was the first paper I read!
Mar 2 at 15:16 history edited Todd Trimble CC BY-SA 4.0
scrapped the earlier, ridiculously contorted argument; made CW; Post Made Community Wiki
Mar 2 at 14:23 comment added Todd Trimble @Kapil Coming back to this after 8 years, I would tell my past self to discard this absurdly contorted argument (and I may do just that: delete the post, or completely rewrite it). Asaf was right. Just enumerate the elements $a_1, a_2, \ldots$, and build a maximal ideal $P$ by taking the union of ideals $P_n$ where $P_0 = \{0\}$, defining $P_n = P_{n-1} + \langle a_n\rangle$ if this is proper, else defining $P_n = P_{n-1}$.
Mar 2 at 2:21 comment added Kapil If $a_ia_j$ is in $P$, then all we know is that there is some $w$ in the branch so that $a_ia_j\in P_{w}$. If $k/2 = \binom{i+j+1}{2}+j$, what ensures that $w$ has length $k-1$? I could not complete the argument that $P$ is prime.
Apr 13, 2017 at 12:19 history edited CommunityBot
replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
Jan 23, 2016 at 19:54 comment added Asaf Karagila Yes, of course, but I meant that you can enumerate $R$ and just construct by hand a prime ideal, adding one element after another. There's no need to appeal to WKL, just to the theorem about recursive definitions.
Jan 23, 2016 at 19:20 comment added Todd Trimble @AsafKaragila Isn't the argument I wrote down just an induction, really? The proof of yours of WKL that I cited is an induction argument.
Jan 23, 2016 at 19:16 comment added Asaf Karagila On a countable ring you can use induction to show there is a prime ideal. No need to appeal to anything.
Jan 23, 2016 at 17:57 history answered Todd Trimble CC BY-SA 3.0