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- Newsgroups: sci.math
- Path: sparky!uunet!zaphod.mps.ohio-state.edu!pacific.mps.ohio-state.edu!linac!mp.cs.niu.edu!rusin
- From: rusin@mp.cs.niu.edu (David Rusin)
- Subject: Re: p prime, p divides ab => pdivides a or b
- Message-ID: <1992Sep2.183341.32017@mp.cs.niu.edu>
- Organization: Northern Illinois University
- References: <Btyo8q.E63@ux1.cso.uiuc.edu>
- Date: Wed, 2 Sep 1992 18:33:41 GMT
- Lines: 19
-
- In article <Btyo8q.E63@ux1.cso.uiuc.edu> ceblair@ux1.cso.uiuc.edu (Charles Blair) writes:
- > My recollection is that a number theory course I took presented
- >this as a difficult result, only proved after doing some stuff
- >with the Euclidean algorithm. Is there a proof which avoids that?
-
- Not really (I just presented my students with this proof an hour ago!).
- More prescisely, what you are proving is the equivalence of what are
- really _irreducible_ elements of the ring (p <> a . b unless a or b = unit)
- and _prime_ elements ( p|ab => p|a or p|b). This is not true in general rings.
- Somehow you have to use something special about the natural integers.
- Being a Principal Ideal Domain is good enough, and you prove that for
- the integers using the Division algorithm.
-
- (Actually I think I remember some quadratic extensions of Z for
- which you can show no division algorithm exists but which are still
- PID's. Readers?)
-
- dave rusin@math.niu.edu
-
-