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I was considering the base field $D$ which is some solvable extenstion of $ \mathbb{Q}$, and a polynomial that isn't solvable in radicals such as $ x^5 - x + 1$. If we let $\zeta$ be a root of this polynomial then something to look at would be the field $D[\zeta]$

Now in traditional fields splitting fields of the form $D[\sqrt[n]{u}]$ all elements are of the form $$ a_0 + a_1 u^{\frac{1}{n}} + a_2 u^{\frac{2}{n}} + ... a_{n-1}u^{\frac{n-1}{n}} , a_i \in D$$

But given that $\zeta$ cannot be expressed in radicals, and so isn't a linear combination of these components, it's not clear that the structure of the elements will be of the form

$$ a_0 + a_1 \zeta + a_2 \zeta^2 + ... a_w \zeta^{w} $$

For any $w \in \mathbb{N}$, $a_i \in D$.

So my question:

What is the actual form of elements in the field for such non solvable extensions, is there a way to derive them?

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This is fairly basic. If your irreducible polynomial $f(x)$ is of degree $n$, you call one root of your polynomial $\zeta$, as you just did, but since $\zeta^n$ is a $D$-linear combination of $\{1,\zeta,\cdots,\zeta^{n-1}\}$, and you show, using various properties of polynomials over fields, that the set of all such linear combinations (at most $n$ terms) is itself a field. Nothing is new, except that $\zeta$ is not an $n$-th root of any kind.

Since we don’t have any (to me) satisfactory way of describing $\zeta$ beyond the known fact that it’s a root of $f$, we mostly have to be satisfied with what we have right there.

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