Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
possible duplicate of Factoring polynomials of degree 6 in 2 ways. –  mick Nov 27 '12 at 22:03
add comment

1 Answer 1

up vote 1 down vote accepted

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).

share|improve this answer
    
Im not so good with Galois theory. So how is this an answer ? –  mick Nov 23 '12 at 22:34
    
I have added an illustrative example. –  Gerry Myerson Nov 24 '12 at 5:30
    
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. –  mick Nov 24 '12 at 23:02
    
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. –  Gerry Myerson Nov 25 '12 at 3:04
    
I voted for close because I asked a new question basicly the same ... –  mick Nov 27 '12 at 22:03
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.