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February 25

Is every PID Noetherian

Hi, I am trying to work out a proof that every principal ideal domain is a Noetherian ring (assuming it is true). I was able to prove that every Euclidean domain was a Noetherian ring, but that was with the assistance of the degree function. I set a~b if a = b*u for some invertible u and it turns out that members of the same equivalence class have the same degree. Since every ideal in a Euclidean domain is principal (which I could prove), the proof that every Euclidean domain is a Noetherian ring follows. Any ideas on whether every PID is Noetherian? Thanks a lot for your help in advance. —Preceding unsigned comment added by 58.161.138.117 (talk) 09:32, 25 February 2009 (UTC)[reply]

Yes. The general proof is (I think) easier than your proof for EDs. Just suppose that you have an ascending chain of ideals, and note that the union of the chain must again be an ideal, and so is principal. More generally, Noetherianity is equivalent to all ideals being finitely generated. Algebraist 10:16, 25 February 2009 (UTC)[reply]

Thanks for your response! But how does what you said show that any ascending chain of ideals terminates? I might just not be getting it... —Preceding unsigned comment added by 58.161.138.117 (talk) 10:29, 25 February 2009 (UTC)[reply]

Algebraist has already given the proof! If it doesn't seem obvious to you, try thinking a little harder. It might just be that you are sleepy!--PS T 11:42, 25 February 2009 (UTC)[reply]

Hey pst!! how are you doing, welcome back! ;) --pma (talk) 11:53, 25 February 2009 (UTC)[reply]

Glad to see the usual folk at the reference desk! ;) --PS T 01:21, 26 February 2009 (UTC)[reply]

Think about where the generator of the union came from. — Emil J. 13:54, 25 February 2009 (UTC)[reply]

By the way, domains where just the f.g. ideals are principal are called Bézout domains. So non-Noetherian ones are sort of non-Noetherian PIDs.John Z (talk)

Simultaneously pyramidal and square

I understand that the only non-trivial natural number to be simultaneously square and pyramidal is 4900. From what I understand, the original proof is non-elementary but there have been some elementary proofs since. I've got through several sheets of paper playing around with congruences and I'm getting nowhere so I'm ready to give up. Strictly speaking, I'm into applied mathematics but I was told about this problem and am genuinely curious. I've even had my old number theory textbook out trying to find inspiration!

Is anyone able to provide an outline of a proof? I haven't found anything on the net so far.

Thanks, Readro (talk) 23:13, 25 February 2009 (UTC)[reply]

The wikipedia article quotes the original result by G.N.Watson (1918). As you are saying there is a more recent and elementary proof: I've found:

Ma, De Gang. An elementary proof of the solutions to the Diophantine equation $6y^{2}=x(x+1)(2x+1)$ . (Chinese) Sichuan Daxue Xuebao 1985, no. 4, 107--116
Anglin, W.S. The square pyramid puzzle. Amer. Math. Monthly 97 (1990), no. 2, 120--124 [1]

It really looks elementary and short. --pma (talk) 12:17, 26 February 2009 (UTC)[reply]

pma, one of your links is in chinese and the other one charges the reader US$4.00 to release the paper. Dauto (talk) 18:58, 1 March 2009 (UTC)[reply]

That reminded me of a problem I worked on when I was a teenager but never managed to find a solution. The questions is: Is it possible to get 24 squares with sides 1, 2, 3, ..., 23, 24 and glue them side by side in such a way to build a square with side 70 ? Dauto (talk) 19:45, 1 March 2009 (UTC) PS: Nowadays I can see that it is not possible...[reply]

See squaring the square for discussion of this problem. --Anonymous, 04:00 UTC, March 4, 2009.

That's true, however I gave the first link because it is the original (the second is mainly a presentation of the first one), and because there are many people who can read chinese. The second link charges you of $4 (very bad) but e.g. from an university you might have it for free (that is, you have already payed for it). --pma (talk) 08:18, 3 March 2009 (UTC)[reply]