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Let $P(x)$ be an integer polynomial of degree $6$ that is irreducible over the integers.

$P(x) = x^6 + (A+a) x^5 + (B+ aA+ b) x^4 + (C+aB+bA +c) x^3 + (aC +bB +cA) x^2 + (bC+cB) x + cC = x^6 + (A'+a') x^5 + (B'+ a'A'+ b') x^4 + (C'+a'B'+b'A' +c') x^3 + (a'C' +b'B' +c'A') x^2 + (b'C'+c'B') x + c'C'$

Where $a,b,c,A,B,C,a',b',c',A',B',C'$ are algebraic integers of degree at most $2$.

Notice $P(x)=(x^3+ax^2+bx+c)(x^3+Ax^2+Bx+C))=(x^3+a'x^2+b'x+c')(x^3+A'x^2+B'x+C')$

I am of course intrested in two distinct factorizations of $P(x)$ so trivial solutions such as $a = a' , b = b' , c=c', ...$ are not what I seek.

Do such $P(x)$ exist ? How many exist ?

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Are you aware of the unique factorization theorem for polynomials over a field? Have you looked into its consequences for your situation? – Gerry Myerson Nov 27 '12 at 22:17
Consider $x^4+1$. It factors over the extension sqrt(-1). But another factorization is possible over the extension sqrt(2). So those are 2 dinstinct factorizations of the same polynomial. And both are extensions of degree 2 ; using algebraic integers of degree 2. So I say it might be possible. – mick Nov 27 '12 at 22:25
Yes, but your situation concerns a polynomial of degree 6, not 4. Have you looked into its consequences for your situation? – Gerry Myerson Nov 27 '12 at 23:09
up vote 1 down vote accepted

Here's an attempt to put the problem in a somewhat broader setting, while avoiding Galois Theory.

Let $P$ be a polynomial with integer coefficients, irreducible over the rationals.

Let $P=pq=rs$ where $p,q$ have coefficients in $K$ and are irreducible over $K$, $r,s$ have coefficients in $L$ and are irreducible over $L$, and $K\ne L$.

Renaming the polynomials, if necessary, we may assume there exist $\alpha,\beta$ such that $p(\alpha)=r(\alpha)=0$, $p(\beta)=0\ne r(\beta)$. Then there are polynomials $a,b$ with coefficients in the field $M$ generated by $K$ and $L$ such that $ap+br=0$, with the degree of $a$ less than that of $r$, and the degree of $b$ less than that of $p$. Now if you evaluate at $\beta$ you get $b(\beta)=0$ which gives you further information on the degree of $b$. I think in the setting of the original question, this is enough to prove impossibility; I think that in general, it's enough to give strong conditions on what degeres are possible.

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Thank you. I will look into that. – mick Dec 29 '12 at 20:09

$P$ has six roots in $\mathbb C$, each of the degree 3 factors has three. Hence given different factorization $P=fg=f'g'$, then $f'$ has at least two roots in common with one of $f,g$. Since we want different factorizations, there are also at most two roots in common. Hence $f'$ has exactly one root in common with the other factor. Thus $P$ has a root in an extension of degree a power of two. In fact, we find another root as common factor of $g'$ with $f$ or $g$ and then are left with quadratic factor. Thus if $F$ is the splitting field of $P$, then $[F:\mathbb Q]$ is a power of $2$ and $\operatorname{Gal}_{F/\mathbb Q}$ is a $2$-group. The Galois group for the single degree three factors is a subgroup of this, hence is also a $2$-group. Since it is also a subgroup of the $S_3$ permuting its three roots, we conclude that it is trivial or contains only one two-cycle, contradicting the fact that it should operate transitively on the roots. From this contradiction we learn that the degree three factors cannot be irreducible.

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In a comment to Gerry here I gave the analogue example for degree 4. So are you saying that for any degree the extensions need to be powers of 2. In other words if I asked for say degree 8 there would be a solution ? And there is always a solution for degrees $2^n$ ? – mick Nov 27 '12 at 22:31
Is it possible to answer without galois theory ? – mick Nov 27 '12 at 22:35
I think @Steven has given an answer w/o Galois Theory, though I think it needs some fleshing out. – Gerry Myerson Nov 27 '12 at 23:13

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