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Solvable polynomial definition problem

  1. Jun 11, 2008 #1
    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.
    Last edited: Jun 11, 2008
  2. jcsd
  3. Jun 11, 2008 #2

    matt grime

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    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.
  4. Jun 11, 2008 #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?
  5. Jun 11, 2008 #4

    matt grime

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    You don't "choose" K. K is given to you.
  6. Jun 11, 2008 #5
    So instead of speaking about some "polynomial being algebraically solvable", be should more precisely speak about "polynomial being algebraically solvable in some field"?
  7. Jun 12, 2008 #6

    matt grime

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    Yes, but that was implicit in saying f(x) in K[x].
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