I know that assuming axiom of choice or equivalently Zorn's lemma , it can be proved that every non-trivial ring with unity has a maximal ideal (two sided ) . The wiki article on axiom of choice says that this statement regarding existence of maximal ideal in any non-trivial ring with unity is equivalent to axiom of choice , but I am not able to prove this converse implication . Please help . Thanks in advance .
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Do you expect to be able to prove everything that is mentioned in Wikipedia? Some proof are difficult, and require more than a handful of preliminary theorems under your belt. – Asaf Karagila Apr 08 '15 at 13:22
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This has come up before. – Asaf Karagila Apr 08 '15 at 13:23
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See W. Hodges: Krull implies Zorn – martini Apr 08 '15 at 13:27
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@AsafKaragila : Can you at least please give a reference ? – Apr 08 '15 at 13:27
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@martini : I don't have access ... – Apr 08 '15 at 13:28
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I gave three references in my answer, I think that the last one should be accessible without a paywall. – Asaf Karagila Apr 08 '15 at 13:33
1 Answers
The proof is not trivial to come by.
Originally given by Hodges, the proof shows a variant of Zorn's lemma can be proved from the assertion that every commutative ring with a unit has a maximal ideal.
Wilfrid Hodges, Krull implies Zorn, J. London Math. Soc. (2) 19 (1979), no. 2, 285--287.
Some decades later, Banaschewski gave a somewhat different proof of that same fact.
Bernhard Banaschewski, A new proof that “Krull implies Zorn”, Math. Logic Quart. 40 (1994), no. 4, 478--480.
Both paper are not long, and quite readable granted that you're comfortable with reading choice-related papers.
Marcel Erné, A primrose path from Krull to Zorn, Comment. Math. Univ. Carolin. 36 (1995), no. 1, 123–-126.
Is also related to this proof, although I haven't really read that one.
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I said "the last one", and I posted three. You're talking about the second. :-) – Asaf Karagila Apr 08 '15 at 13:39
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I'll add that of the three, I definitely prefer the original paper by Hodges. I find it the easiest to digest. – Asaf Karagila Apr 08 '15 at 13:46
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@Asaf: And now we know: $\text{some}=1.5$, at least when counting decades. – Brian M. Scott Apr 08 '15 at 17:20
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