Any ideas?

f(x) belongs to F[x] and alpha is root of f(x) is some field extension of F. Show irreducible poly. g(x) belonging to F[x] has property that h(x) belonging to F[x] also has alpha as a root then g(x)/h(x) in F[x].

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- July 11th 2007, 07:35 AMPacManisAliveRoots and Irreducible polys
Any ideas?

f(x) belongs to F[x] and alpha is root of f(x) is some field extension of F. Show irreducible poly. g(x) belonging to F[x] has property that h(x) belonging to F[x] also has alpha as a root then g(x)/h(x) in F[x]. - July 11th 2007, 09:39 AMThePerfectHacker
Let be a field and be its extension field. Let be algebraic over (how do we know it is algebraic?). Next define the

*evaluation homomorphism*:

. We know that the is a principal ideal. So for some . This polynomial**must**be of minimial degree because otherwise if is smaller degree then we require that for some because the ideal is principal. But that is a contradiction because . Thus, is of minimal degree. Next we claim that is irreducible because if not and so one of the polynomial has the property that (WLOG) but that cannot be because has a smaller degree.

So we see that all the polynomial having this zero, i.e. are generated by an irreducible polynomial . And so if then for some . This shows that**divides**. - July 12th 2007, 06:46 AMPacManisAliveKernel/Ring homomorphisms
In our abstract class we have not yet been taught about kernels and ring homomorphisms so this has to be proved without using these. Thanks

- July 12th 2007, 07:16 AMThePerfectHacker
1)Okay define .

2)Convince yourself that .

3)Since is a PID it follows for some .

4)Use a similar argument as above.

Note, this is basically the same approach except kernels eliminate the need to prove #1 which is a little lengthy but straightforward checking of the field axioms.