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$(t)/3 + O(1).$


1 Answer 1


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.

MORE EDIT: Just to clarify, the above is meant as an answer to the question in the second paragraph that goes from "Then we can ask," to conditions A). B). and C).

  • $\begingroup$ Im not so good with Galois theory. So how is this an answer ? $\endgroup$
    – mick
    Nov 23, 2012 at 22:34
  • $\begingroup$ I have added an illustrative example. $\endgroup$ Nov 24, 2012 at 5:30
  • $\begingroup$ Well thanks for the example. But I still do not understand. Is this a yes or a no ? you have 3 factors of degree 2 rather than 2 factors of degree 3. $\endgroup$
    – mick
    Nov 24, 2012 at 23:02
  • $\begingroup$ Mick, you have about 5 questions in the body of your question. I have answered the first one, the one that starts "Then we can ask," and ends with condition C). I don't think it's a good idea to ask half-a-dozen questions in one. If you can work out how to answer your other questions from what I've done for your first question, good --- then you can post the answers. If you can't answer your other questions, then my advice is to post them as new questions, one at a time, with time out to ponder the answers to each before posting the next, and with links back to this question. $\endgroup$ Nov 25, 2012 at 3:04
  • $\begingroup$ I voted for close because I asked a new question basicly the same ... $\endgroup$
    – mick
    Nov 27, 2012 at 22:03

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