Solvable polynomial definition problem

In summary, the conversation discusses the definition of an algebraically solvable polynomial f in a field K[X], where the root field N is contained in a field extension E of K. The conversation also brings up the confusion about all polynomials over the field of rational numbers being algebraically solvable, and clarifies that this is not the case for polynomials over the field of complex numbers. The conversation also discusses the importance of choosing the smallest field K possible for f to be in, and the necessity of specifying the field in which the polynomial is algebraically solvable.
  • #1
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I've now read a definition like this. A polynomial [tex]f\in K[X][/tex] is algebraically solvable if the root field [tex]N=K(x_1,\ldots, x_n)[/tex] (here [itex]x_1,\ldots, x_n[/itex] are the roots of the polynomial) is contained in some field extension [tex]E[/tex] of [tex]K[/tex], for which there exists a sequence of field extension [tex]K=E_0\subset E_1\subset\cdots\subset E_n=E[/tex] so that [tex]E_{k+1} = E_k(a)[/tex] with [tex]a^n\in E_k[/tex] with some [itex]n[/itex].

I must have understood something wrong, because now it seems that for example all polynomials [tex]f\in\mathbb{Q}[X][/tex] are algebraically solvable. The reason is this. The polynomial is also [tex]f\in\mathbb{C}[X][/tex], the root field is [tex]\mathbb{C}(x_1,\ldots, x_n)=\mathbb{C}[/tex], and it is contained in the trivial field extension where nothing is added.

Anything that reduced my confusion would be appreciated.
 
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  • #2
You've completely forgotten the Q. You can't do that.

Of course, all polys over C[x] are solvable - they are all reducible to linear factors over C, so the Galois theory there is completely vacuous.
 
  • #3
With given polynomial [tex]f=a_n X^n+ \cdots a_1 X + a_0[/tex], we are supposed to choose as small field K as possible, so that [tex]f\in K[X][/tex], before using the definition of polynomial being algebraically solvable?
 
  • #4
You don't "choose" K. K is given to you.
 
  • #5
So instead of speaking about some "polynomial being algebraically solvable", be should more precisely speak about "polynomial being algebraically solvable in some field"?
 
  • #6
Yes, but that was implicit in saying f(x) in K[x].
 

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