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Topic: Prime Ideals (Read 305 times) |
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Ben
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Hey.. I have a proof that I can't figure out. I think it's simple, so I'm sure you guys can get it.
In F[x] (a field of polynomials), prove all prime ideals are maximal.
Thanks for any help,
Ben
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Obob
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Re: Prime Ideals
« Reply #1 on: Dec 13th, 2004, 1:15am » |
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Recall that F[x] is a PID. So if I is any ideal, there is a polynomial p for which I=(p). Suppose I is prime. Then it has the property that if the product ab of two polynomials a and b lies in I, then either a or b is in I. Suppose that p is reducible over F, so that p=ab for some polynomials a and b both of positive degree. Since I is prime, this implies that either a or b is in I. But this is impossible, since I, being generated by p, consists entirely of polynomials with degree at least deg p. Therefore p is an irreducible polynomial over F.
We claim that if p is irreducible over F then the ideal (p) is maximal. For suppose that there is an ideal J=(q) such that 0[subset] (p) [subset] (q) [subset] F[x], where all inclusions are strict. Since p is in J and since J is generated by q, this implies that there is a polynomial c for which p=qc. But p is irreducible over F, so this implies that deg q = deg p and therefore that c is a constant. Therefore q=p/c is in (p), and the ideals I and J must be equal as they each contain the generators of each other. This is a contradiction, so (p) is maximal.
Thus we have shown that every prime ideal is generated by an irreducible polynomial and that every ideal generated by an irreducible polynomial is maximal. Hence we are done.
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