# Why is $X^4-16X^2+4$ irreducible in $\mathbb{Q}[X]$?

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?

• The polynomial $(x^2 - x - 1)^2$ has no roots and is reducible over $\mathbb{Q}$. – Andrew Maurer Jun 9 '13 at 22:19
• @GitGud Done. I changed it to be more $\phi$-ish. A good poylnomial, in my opinion. – Andrew Maurer Jun 9 '13 at 22:25
• Thanks for all the answers. I have edited the question, I am wondering whether it is a must to use Gauss Second Lemma. If it does, that's good at least I have learned something new :) – user71346 Jun 9 '13 at 22:29
• @user71346 I don't see how the first lemma can help since it doesn't relate rational coefficients with integer coefficients. – Git Gud Jun 9 '13 at 22:35
• @GitGud. Yes, you are indeed correct! I am just wondering whether it is a must to use the second lemma in this problem. But after I have carefully thought again, I think we indeed have to use the second lemma, otherwise we cannot relate rational coefficient with integer coefficients as you said. – user71346 Jun 9 '13 at 22:41

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.

Why can't we just show that since $x^4−16x^2+4$ has no rational roots hence irreducible?

Because that isn't true. Take for example the polynomial $x^4-5x^2+6$. Now this has no rational roots, but $$x^4-5x^2+6=(x^2-2)(x^2-3)$$ and hence the polynomial isn't irreducible. This trick only works with quadratics and cubics.

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?

The fact that the polynomial has no rational roots guarantees that it has no factors of degree $1$, so it is impossible to factor into polynomials of degree $1$ and degree $3$. The reason why the coefficients have to be integers is just a consequence of Gauss's Lemma: this says that a polynomial with integer coefficients is irreducible over $\mathbb{Q}$ if and only if it is irreducible over $\mathbb{Z}$.

• I think you mean that the polynomial isn't irreducible. – Kortlek Jun 10 '13 at 0:03
• @KimJung-un Thanks for pointing out the typo. :) – Warren Moore Jun 10 '13 at 0:08

You have already answered some of this yourself. You say that the polynomial has no rational root. If (and only if) it had a rational root, it could be factored as a product of a degree 1 polynomial and a degree 3 polynomial, both with rational coefficients. So this possibility has been discounted. But there is still the case to consider that the polynomial factors as a product of two quadratic polynomials with rational coefficients. If it did, you could multiply each of them by a constant to make them monic, so you only need to consider that case. For the rest: Have you done Gauss's Lemma in class yet?

1. A polynomial is irreducible if it cannot be written as product of two polynomials of smaller degree. It is not necessarily the case that one of these factors has degree 1. (Though, if the original polynomial is quadratic or cubic, then it suffices to check against linear factors).

2. If we could find them, we would be perfectly well allowed to factor into linear and cubic. However, there is no linear factor (a fact the author seems to have considerd "trivial", but at least you correctly verified that). This leaves us with quadratic polynomials with rational coefficients. by the Gauss lemma, we may in fact assume that the coefficients are in fact integers - that facilitates the search for as well as any proof of nonexistence of factors. If the leading coefficients of these were anything but $\pm1$, we could not have $1$ as leading coefficient in the product. And of course we can turn a leading $-1$ to a $1$. Therefore, if there is a factor, then these factorse may as well be normalized to be monic integer polynomials.