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
Best Answer
$1.$ Being irreducible 'over $\Bbb Q$' means that it can not be factored as the product of polynomials (with degree greater than $0$) with all coefficients in $\Bbb Q$. The difference between irreducibility over $\Bbb Z$ and over $\Bbb Q$ is none due to (the second) Gauss's lemma.
$2.$ If it could be factored as the product of a polynomial of degree $1$ times something else, then it would have a root, namely the root of that same polynomial of degree $1$.