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...
 A: 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.
