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Let $f(x)$ be an irreducible polynomial with integer coefficients, which is irreducible over $\mathbb{Z}$ and has degree $a p^b > p$ with $p,a,b>0$ and $p$ a prime.

It appears that $f(x)$ factors in at least $b$ ways over extensions of degree $p$ that do not belong to the zeros of $f(x)$ or their extensions.

For clarity we count 1 way per extension , even if there are more for that extension.

How to prove or disprove this ?

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    How do you count "ways in which a polynomial factors"? Factorizations into irreducibles over different extensions?2012-09-14
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    Yes different extensions. Thanks for comment.2012-09-14
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    I wrote extensions afterall , not ' an extension ' :) But i admit there are other interpretations possible. For clarity we count 1 way per extension , even if there are more per extension.2012-09-14
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    I have difficulties parsing this question. For example, what is "an extension of degree $p$ that does not belong to the zeros of $f(x)$ or their extension"?2012-09-14
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    @ Jyrki : example : if sqrt(5) is a solution of $f(x)=0$ and $f$ has degree 6 ( multiple of 2 ) then the quadratic ( 2! ) extensions $Z$(sqrt(5)) or $Z$(sqrt(10)) are not allowed.2012-09-14
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    @mick: If an irreducible polynomial has $\sqrt5$ as a root, then that polynomial is quadratic. More precisely, it is $x^2-5$. So I still don't understand. Could you show an example of what you mean by different factorizations of a polynomial? Coefficients in different extension fields?2012-09-16
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    Do you mean factorizations over such extensions that do not contain any zeros of $f$? IOW a zero of a polynomial may belong to an extension field. Not the other way around?2012-09-16
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    Ok another example : x^4 + 1. You can factor it completely by the 4 roots of unity. Or you could write (x^2 + i)(x^2 - i). But there is also a factorization over Q(sqrt(2)). sqrt(2) is not an extension of i or -i nor the zero's of x^4 + 1. Also the zero's of x^4 + 1 are not extensions of Q(sqrt(2)) and neither is + i or - i. We see that x^4 + 1 is degree 2^2 and we have at least 2 ways to factor over extensions namely Q(sqrt(2) and Q(sqrt(-1)) and none of those are extensions of sqrt(i). Exactly as the conjecture in the OP predicts.2012-09-17
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    Ok, I think I understand what you are asking. Thinking...2012-09-17

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I think that the question is asking about the following. I'm prepared for this to be wrong, but I want to open up the discussion with this "guess".

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 $[K:\mathbb{Q}]=p$ is satisfied.

The OP then asks, whether there are always at least the prescribed number of such intermediate fields $K$.

I lead off by proffering a counterexample. Let $$ f(x)=x^4+x^3+x^2+x+1=(x-\zeta)(x-\zeta^2)(x-\overline{\zeta})(x-\overline{\zeta}^2), $$ where $\zeta=e^{2\pi i/5}$.Then the splitting field of $f(x)$ is the fifth cyclotomic field $L=\mathbb{Q}(\zeta)$. We have $\deg f(x)=4=2^2$ as well as $[L:\mathbb{Q}]=4$, so $p=2$, $b=2$, and $a=1$.

But in this case there exists only one intermediate field $K=\mathbb{Q}(\sqrt5).$ This follows from Galois theory. The Galois group $Gal(L/\mathbb{Q})$ is cyclic of order four, so it has a unique subgroup of index two, and the Galois correspondence associates the subfield $K$ with that subgroup.

The corresponding factorization is $$ f(x)=f_1(x)f_2(x), $$ where $$ f_1(x)=(x-\zeta)(x-\overline{\zeta})=x^2-2(\cos\frac{2\pi}5)\,x+1 $$ and $$ f_2(x)=(x-\zeta^2)(x-\overline{\zeta^2})=x^2-2(\cos\frac{4\pi}5)\,x+1. $$ Here the coefficients of the linear term are in the subfield $K$ as $$ \cos\frac{2\pi}5=\frac{\sqrt5 -1}4\qquad\text{and}\qquad\cos\frac{4\pi}5=-\frac{\sqrt5 +1}4 $$ are clearly elements of $K$. Galois theory tells us that the coefficients of other potential factors such as $(x-\zeta)(x-\zeta^2)$ necessarily generate all of $L$.

Cyclotomic polynomials of order $17$, $257$ or $65537$ yield similar polynomials of respective degrees $16$, $256$ and $65536$ that only have a single such field $K$.


Galois theory will say a lot about the problem. An intermediate field $K$ will satisfy condition C) iff the subgroup $H=Gal(L/K)$ is of index $p$ in $G=Gal(L/\mathbb{Q})$. It will satisfy condition B) iff the subgroup $H$ has no fixed points among the roots of $f(x)$. It will satisfy condition A) iff the action of $H$ on the roots of $f(x)$ is not transitive. I dare not suggest the most general case, when all these would hold for a large number of subgroups $H$. When $G$ is elementary $p$-abelian, it seems to be easy to produce examples of several such subgroups $H$. In a comment the OP himself exhibited the example $f(x)=x^4+1$. In this case $G\simeq C_2\times C_2$ is elementary $2$-abelian, and the splitting field $L=\mathbb{Q}(i,\sqrt2)$ has three subfields like $K$ corresponding to the cyclic subgroups generated by the permutations of the roots $\sigma_1=(12)(34)$, $\sigma_2=(13)(24)$ and $\sigma_3=(14)(23)$ respectively. The corresponding subfields are (in some order) $\mathbb{Q}(i)$, $\mathbb{Q}(\sqrt2)$ and $\mathbb{Q}(\sqrt{-2})$.

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    It is still possible that I have misunderstoos something about the question. Do comment!2012-09-17
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    I think you probably got my question right. Assuming I understand your answer somewhat that is. But I don't know much about galois theory unfortunately. Why are the Fermat primes related ? How to find the group of a polynomial ? Does your answer imply that counterexamples are relatively rare ? Only for p-abelian ? In particular this troubles me , quote : Galois theory tells us that the coefficients of other potential factors such as $(x−ζ)(x−ζ_2 )$ necessarily generate all of $L$. Sorry if I'm not good at this and give you a lot of work.2012-09-18
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    @mick: I'm afraid that answering those questions will just cause me to reproduce a largish chunk of Galois theory here, and that simply cannot be done. And you would get better understanding by studying a professional quality exposition of Galois theory anyway. My answer does not really say much about the relative frequency of examples/counterexamples. The cyclotomic polynomials related to Fermat primes were just the first set of counterexamples that occurred to me. The same about the elementary $p$-abelian Galois groups as examples - just something that occurred to me. A good question!2012-09-18
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    Thanks. I think I can modify the question such that it is true or at least harder to give a counterexample. I can imagine other counterexamples to the question stated as of today , but I do not have the skills to check it. Maybe a modification .. or a new thread ? Im not sure. Maybe I should ask some basic galois questions first in a new thread.2012-09-18
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    http://math.stackexchange.com/questions/199105/factoring-polynomials-fgx-over-extension-fiels That is the link for the question.2012-09-19