Determine whether $X^4-16X^2+4$ is irreducible in $\mathbb{Q}[X]$.
To solve this problem, I reasoned that since $X^4-16X^2+4$ has no rational roots hence irreducible.
But there is a hint to this question that uses different approach:
Try supposing it is reducible, then it must factor into a product of two monic quadratic polynomials with integer coefficients. Then show that it is impossible, then conclude the original polynomial is irreducible.
My questions regarding the hints:
1. Why can't we just show that since $X^4-16X^2+4$ has no rational roots hence irreducible?
2. Why do we have to factorise into a product of two monic quadratic polynomials with integer coefficients? Why monic? And why can't we factorise into polynomial with degree 1 and 3? Also, lastly why the coefficients have to be integers?
Thank you for any explanations!
Edit: Thanks for all the answers.
I saw that Second Gauss Lemma is used. But I only learned the first one in class. Which is:
Let $R$ be a Unique Facorisation Domain. If $f,g\in\mathbb{R}[X]$ are primitive, then so too is their product $fg$.
Is it inevitable to use Second Gauss Lemma? Is there any other way around?
Best Answer
Answer to question 1: Consider de polynomial $(x^2+1)^2\in\Bbb Q[x]$. It has no rational roots and it is reducible over $\Bbb Q$. Your reasoning fails.
The absence of roots only guarantees that there are no factors of degree $1$.
Your polynomial has degree $4$, since it has no rational roots, it has no linear factors (that is factors of degree $1$), but it could have two factors of degree $2$.
Assume it does and try to reach a contradiction. If you do, then it is irreducible over $\Bbb Q$.
Answer to question 2: The factors don't need to be monic polynomials, but they can be. If you find a factorization in which the factors aren't monic, you just have to multiply by a certain constant to make the factors monic.
Answer to question 3: If you find a factorization $(x-\alpha )q(x)$, where $q(x)$ is a polynomial with rational coefficients of degree $3$, then the polynomial will have a rational root, namely $\alpha$ and you have estabalished it doesn't.
Answer to question 4: The coefficients need not be integers, but (the second) Gauss's lemma allows us to assume the coefficients are integers, making the calculations much simpler.