# Factoring polynomials of degree 6 over extension fields.

Let $f(x)$ be a polynomial with integer coefficients that is irreducible over the integers and has degree 6.

Let $L$ be the splitting field of $F$. Then we can ask, whether there exist intermediate fields $K$, $\mathbb{Q}\subseteq K\subseteq L$, such that A) $f(x)$ factors in a non-trivial way in $K[X]$, B) $K$ does not contain any of the zeros of $f(x)$, and C) the degree condition $[K:\mathbb{Q}]=6$ is satisfied.

I wonder if there are $f(x)$ and distinct intermediate fields $K_1$ and $K_2$ such that $f(x) = g_1(x)h_1(x) = g_2(x)h_2(x)$ where $g_1$ and $h_1$ have coefficients in $K_1$ and are of degree 3 and where $g_2$ and $h_2$ have coefficients in $K_2$ and are of degree 3.

That is the main question. However I have some more.

I also wonder if it is possible that $f(x)=q(x^2)$ where $q$ is also an irreducible polynomial.

A more general question is made if we replace $6$ with $2p$ where $p$ is an odd prime and the degrees of the factors are replaced by $p$ ( instead of $3$).

Clearly I considered 2 factorisations , we could also ask how many factorizations and $K_n$ can occur at most for degree $2p$.

A conjecture could be $n = p/3 + O(1)$. Let $t(p)=n$. Then for instance $t(2) = 2$ , $t(3)=2$,$t(5)=2$,$t(7)=3$ which seems imho to weakly suggest $t(p) =$ primecountingfunction$(t)/3 + O(1).$

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possible duplicate of Factoring polynomials of degree 6 in 2 ways. – mick Nov 27 '12 at 22:03

Let the Galois group of $f$ be the dihedral group of order 12. This has a subgroup of order 2 corresponding to rotating a hexagon through $\pi$. The fixed field of this subgroup should be an extension of degree 6 in which $f$ is reducible but has no root.
EDIT: A simple example. $f(x)=x^6-2$ is irreducible over the rationals and has degree 6. Let $K$ be the splitting field of $x^3-2$. Then $$f(x)=(x^2-\root3\of2)(x^2-\omega\root3\of2)(x^2-\omega^2\root3\of2)$$ over $K$, where $\omega=e^{2\pi i/3}$, $K$ contains no zeros of $f$, and $K$ is of degree $6$ over the rationals.