Solvable polynomial definition problem

1. Jun 11, 2008

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.

Last edited: Jun 11, 2008
2. Jun 11, 2008

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.

3. Jun 11, 2008

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?

4. Jun 11, 2008

matt grime

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

5. Jun 11, 2008

jostpuur

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

6. Jun 12, 2008

matt grime

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