# If $f(x) \Bbb Q[x]$ has splitting field of degree $16$ over $\Bbb Q$ , are the roots of $f(x)$ solvable by radicals.

If $$f(x) \Bbb Q[x]$$ has splitting field of degree $$16$$ over $$\Bbb Q$$ , are the roots of $$f(x)$$ solvable by radicals.

My attemt: Any general equation of degree $$\geq 5$$ is not solvable by radicals.

Now as the splitting field is of degree $$16$$ which is $$2^4$$ so the best candidate should be a poly of degree $$4$$. Now let it's roots are $$a,b,c,d$$ then $$|\Bbb Q(a)/\Bbb Q|=4$$ then for the next root $$b$$ will $$|\Bbb Q(a,b)/\Bbb Q(a)|=2$$or $$3$$?

Help me from here...

Unfortunately, that approach won't work. Clearly we don't want $$[\Bbb Q(a,b):\Bbb Q(a)]=3$$, as $$3$$ is not a factor of $$16$$. So we must have $$2$$. However, if $$b$$ has degree $$2$$ over $$\Bbb Q(a)$$, then $$b$$ has a minimal polynomial $$g$$ of degree $$2$$ over $$\Bbb Q(a)$$. We get $$f(x)=(x-a)g(x)(x-c)$$ for some $$c\in\Bbb C$$. Note that since $$b$$ is a root of $$g$$, the other root of $$g$$ is already contained in $$\Bbb Q(a,b)$$, so $$f$$ factors completely over $$\Bbb Q(a,b)$$ and therefore the extensions stop there.
However, just because general polynomials of degree $$5$$ or higher aren't solvable, that doesn't mean there aren't higher degree solvable polynomials. So it's possible to have a degree $$16$$ solvable polynomial which splits entirely just by adjoining a single root. For instance, take $$\frac{x^{17}-1}{x-1}=x^{16}+x^{15}+\cdots+x+1$$
Finally, I think you have misunderstood the question. I think what the question really asks is "Given that the splitting field of $$f$$ has degree $$16$$ over $$\Bbb Q$$, does it necessarily follow that $$f$$ is solvable by radicals?" And the answer is "yes": the Galois group of the splitting field of $$f$$ has order $$16$$, and any group of order $$16$$ is solvable, since $$A_5$$ is the smallest non-solvable group.
• @Gimgim You mean $c$? Since $(x-a)$ and $g(x)$ are polynomials over $\Bbb Q(a)$, and they are factors of $f$, the division $\frac{f(x)}{(x-a)g(x)}$ can be entirely carried out in $\Bbb Q(a)$. So the result, which is $(x-c)$, must be a polynomial over $\Bbb Q(a)$. So $c\in\Bbb Q(a)$. – Arthur Mar 22 at 5:59
• @Gimgim There are equations of degree $5$ (and any higher degree), like $x^5-x+1$, which do give rise to non-solvable Galois groups. But they all have order at least $60$. So one cannot give a radical formula (like the quadratic formula $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$) for general equations of higher degree. There are, however, always special cases they can be solved, like $x^n-1$. – Arthur Mar 22 at 6:25