TITLE: Is the following polynomial reducible over Q?
QUESTION [5 upvotes]: I am looking at an exercise saying that "Demonstrate that x4-22x2+1 is reducible over Q. I have the solution manual and it solves like the following:
If x4-22x2+1 is reducible over Z, then it factors in Z[x], and must therefore
either have a linear factor in Z[x] or factor into two quadratics in Z[x]. The only possibilites for a
linear factor are x ± 1, and clearly neither 1 nor -1 is a zero of the polynomial, so a linear factor is
impossible. Suppose
x4-22x2+1 = (x2 + ax + b)(x2 + cx + d).
Equating coefficients, we see that
x3 coefficient : 0 = a + c
x2 coefficient : −22 = ac + b + d
x coefficient : 0 = bc + ad
constant term : 1 = bd so either b = d = 1 or b = d = −1.
Suppose b = d = 1. Then −22 = ac + 1 + 1 so ac = −24. Because a + c = 0, we have a = −c, so
−c2 = −24 which is impossible for an integer c. Similarly, if b = d = −1, we deduce that −c2 = −20,
which is also impossible. Thus the polynomial is irreducible.
My question is:
1)What does it mean "over Q"? What is the difference between saying over Z and over Q?
2)Why do we factor it as (x2 + ...)(x2 + ...)? Can't it be like (x3 + ...)(x + ...). Also, why don't we factor it like (ax2 + ...)(bx2 + ...), i mean how do we know that the head coefficients are 1?Can someone help with this?
Thanks

REPLY [3 votes]: Over $\mathbb Q$ means that rational coefficients are allowed, not just integers. In fact one can show that they are not needed - but that has to be proved. Suppose $p(x)q(x)=x^4+22x^2+1$ is a rational factorisation. Multiply by $n$ to clear fractions, so that $P(x)Q(x)=n(x^4+22x^2+1)$ is an equation with integer coefficients. Now suppose $p$ is a prime factor of $n$. Then, starting with the highest power of $x$ and the coefficients of the factorisation, it is possible to prove that $p$ is a factor of every coefficient of either $P(x)$ or $Q(x)$. One then cancels the $p$ from both sides. Eventually, every factor of $n$ is cancelled, and we are left with an integer factorisation of the original equation.
If $p(x)=(ax^3+bx^2+cx+d)(x+e)$ then $p(-e)=0$ and $e$ is a root of the equation.