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## Main Question or Discussion Point

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.

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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