Solvable polynomial definition problem

[jostpuur]

I've now read a definition like this. A polynomial $$f\in K[X]$$ is algebraically solvable if the root field $$N=K(x_1,\ldots, x_n)$$ (here $x_1,\ldots, x_n$ are the roots of the polynomial) is contained in some field extension $$E$$ of $$K$$, for which there exists a sequence of field extension $$K=E_0\subset E_1\subset\cdots\subset E_n=E$$ so that $$E_{k+1} = E_k(a)$$ with $$a^n\in E_k$$ with some $n$.

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

Anything that reduced my confusion would be appreciated.

 

[matt grime]

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.

 

[jostpuur]

With given polynomial $$f=a_n X^n+ \cdots a_1 X + a_0$$, we are supposed to choose as small field K as possible, so that $$f\in K[X]$$, before using the definition of polynomial being algebraically solvable?

 

[matt grime]

You don't "choose" K. K is given to you.

 

[jostpuur]

So instead of speaking about some "polynomial being algebraically solvable", be should more precisely speak about "polynomial being algebraically solvable in some field"?

 

[matt grime]

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