Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

As this question suggests, I quite like the notion of permuting the coefficients of polynomials.
And, moreover, I have another question on this direction:If L|F is a finite normal field extension, then it must be normal, then my question is: is there a name for the field extension L|F such that L contains all roots of polynomials obtained by permuting the coefficients of p(x) where p(x) is a polynomial one of whose roots lie in L.
In any case, thanks for paying attention.

share|improve this question
    
Of course this is stronger than the Galois field extensions, and it might be of some subtle interests to me, any way, thanks very much. –  awllower Feb 15 '11 at 11:54
    
I doubt it. This notion is not at all invariant under change of coordinates. –  Qiaochu Yuan Feb 15 '11 at 12:03
2  
Does this condition depend on the particular choice of primitive element? If so, do you want it to hold for some or all primitive elements? In any case, I'm rather sure that the answer to the question is no. –  Matt E Feb 15 '11 at 13:28
    
Thanks, I have edited the question so that it does not depend on primitive element. –  awllower Feb 16 '11 at 0:48
1  
Dear awillower, such a field is algebraically closed! Let $p \in K[x]$ be any polynomial, and consider $xp$; then $xp$ has a root in $K$ (namely, zero); so all the roots of $xp$ lie in $K$, and that includes all roots of $p$. –  Akhil Mathew Feb 16 '11 at 1:01

1 Answer 1

up vote 4 down vote accepted

Such a field is algebraically closed! Let $p \in K[x]$ be any polynomial, and consider $xp$; then $xp$ has a root in $K$ (namely, zero); so all the roots of $xp$ lie in $K$, and that includes all roots of $p$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.