# Factoring polynomials of degree $a p^b$ over extension fields.

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? – rschwieb Sep 14 '12 at 12:26
Yes different extensions. Thanks for comment. – mick Sep 14 '12 at 12:30
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. – mick Sep 14 '12 at 12:33
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"? – Jyrki Lahtonen Sep 14 '12 at 12:43
@ 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. – mick Sep 14 '12 at 20:52

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! – Jyrki Lahtonen Sep 17 '12 at 19:38
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. – mick Sep 18 '12 at 18:51
@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! – Jyrki Lahtonen Sep 18 '12 at 19:03
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. – mick Sep 18 '12 at 19:27
math.stackexchange.com/questions/199105/… That is the link for the question. – mick Sep 19 '12 at 12:34