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.
Is there a specific name for this notion of extensions of fields?
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polynomials
field-theory
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0Of course this is stronger than the Galois field extensions, and it might be of some subtle interests to me, any way, thanks very much. – 2011-02-15
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0I doubt it. This notion is not at all invariant under change of coordinates. – 2011-02-15
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2Does 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. – 2011-02-15
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0Thanks, I have edited the question so that it does not depend on primitive element. – 2011-02-16
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1Dear 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 $p$. – 2011-02-16
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0Thanks, from the point of view of me, you solved this question completely, so if it is not too troublesome to you, could you submit it as an answer?In any case, thanks very much. – 2011-02-16
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0@Akhil: cute! :) – 2011-02-16
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0@awillower: Done. I suspect the answer is, incidentally, the same if we required the polynomial to be *irreducible*(in which case we'd actually have to use the permutability of coefficients, of course). – 2011-02-16
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0What does @Akhil Mathew mean by permutability of coefficients? – 2011-02-16
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0@QiaochuYuan: And what did you mean to say that this is not independent of the coordinates? Is it about the bases changes? Thanks again for the help. – 2011-11-16
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0@awllower: I mean the notion is not invariant under, for example, replacing $p(x)$ with $p(x+a)$ where $a \in F$. – 2011-11-16
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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 $p$.