Let be the set of polynomials with complex coefficients. For every and , there exist such that True or false?

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- Jun 5th 2010, 06:48 PMNonCommAlgA property of polynomials
Let be the set of polynomials with complex coefficients. For every and , there exist such that True or false?

- Jun 5th 2010, 08:14 PMBruno J.
True!

For every factor of , and every , we multiply by an appropriate number of factors of the form , so as to obtain a polynomial which can be written as a product of powers of polynomials of the form . - Jun 5th 2010, 08:30 PMNonCommAlg
- Jun 5th 2010, 08:54 PMBruno J.
- Jun 5th 2010, 09:25 PMNonCommAlg
- Jun 6th 2010, 02:15 PMBruno J.
Right! And in the case , we can write with ; then we let , so that ...

- Jun 6th 2010, 06:15 PMBruno J.
We write , with . We let be a primitive root of in . Now, we let . A quick check shows that the coefficients of lie in . Now forget that the are polynomials in ; the coefficients of , as a polynomial in , are symmetric polynomials in with coefficients in . But since , only the first and the last symmetric polynomial invariants do not vanish, and hence only powers of remain, i.e. ...

- Jun 8th 2010, 09:04 AMNonCommAlg
- Jun 8th 2010, 09:13 AMBruno J.
- Jun 8th 2010, 09:34 AMNonCommAlg
this problem has a nice result in linear algebra: for a field let be the field of rational functions in the indeterminate i.e.

let then is obviously a subfield of and therefore we can consider as a vector space over using what we just proved, it's easy to see that every element of

can be written uniquely as thus we get the result which is basically the reason that i created this thread! - Jun 8th 2010, 09:59 AMBruno J.
- Jun 8th 2010, 10:53 AMNonCommAlg
- Jun 8th 2010, 06:03 PMBruno J.
Can we generalize Luroth's theorem to rational function fields generated by two or more elements? I suppose...