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.

  • $\begingroup$ What about the 4th root of f? $\endgroup$ – Gimgim Mar 22 at 5:56
  • $\begingroup$ @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 $c\in\Bbb Q(a)$. $\endgroup$ – Arthur Mar 22 at 5:59
  • $\begingroup$ @Gimgim I added a small bit at the end. $\endgroup$ – Arthur Mar 22 at 6:07
  • $\begingroup$ Then what does the statement involving general eqn of degree 5 mean? $\endgroup$ – Gimgim Mar 22 at 6:14
  • $\begingroup$ @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 $x^n-1$. $\endgroup$ – Arthur Mar 22 at 6:25

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