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- From: mkrogers@unix.amherst.edu (MICHAEL K ROGERS)
- Subject: Re: Univariate polynomial equations and the FAQ
- Message-ID: <BxBD94.Irp@unix.amherst.edu>
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- References: <1992Nov6.184527.20793@sun0.urz.uni-heidelberg.de>
- Date: Fri, 6 Nov 1992 21:34:16 GMT
- Lines: 35
-
- Eugen W. Schmidt (gsmith@clio.uucp) wrote:
- : In article <1d72mnINNq2p@mozz.unh.edu> dvf@kepler.unh.edu (David V Feldman) writes:
- :
- : >Fix an integer m. Let K be the extension of Q obtained by adjoining
- : >all roots of all polynomials of the form
- : > n m
- : > x + a x + ... a
- : > m 0
- :
- : >where the coefficients are rational.
- :
- : This sounds like the algebraic closure of Q, Q-bar.
- :
- : Let s(m) be the smallest degree of a
- : >polynomial with rational coefficients which does not have any root in K.
- : >So s(0)=5, by Galois theory. What is known about the function s(m)?
- :
- : This is false for Q-bar. If your original field extension was the
- : maximal solvable extension in Q-bar, then we could try to figure out
- : the answer to your question, if we could figure out the question,
- : which I still can't do!
-
- I'm not sure what the problem is either. It seems like we want
- to consider the field K(m) obtained by adjoining the roots of all
- polynomials over Q of all degrees which have the following property:
-
- The coefficient of x^k is zero if x^k is not the
- leading term and k > m.
-
- The claim s(0)=5 seems false then because the roots of cubic
- polynomials often are of the form sqrt(a + cubert(b)) (or is
- it cubert(a + sqrt(b))?). These roots are not the linear
- combination of square roots and cube roots of rational numbers.
-
- So I suspect the original poster had a different idea in mind.
-