$X^4-4X^2-1$ irreducible over $\mathbb{Q}[X]$

Question

I want to show, that $X^4-4X^2-1$ is irreducible over $\mathbb{Q}[X]$. Since there are no roots in $\mathbb{Q}$ it has to be:

$(X^4-4X^2-1)=(X^2+aX+b)(X^2+cX+d)$

Comparision of the coefficients shows that this can not hold over $\mathbb{Q}$. But I am searching for an easier way to show this which uses less calculation.

I tried this:

$X^4-4X^2-1=X^4-4X^2+4-5=(X^2-2)^2-5=(X^2-2-\sqrt{5})(X^2-2+\sqrt{5})$

Which is obviously not in $\mathbb{Q}[X]$ anymore. But would this calculation be enough to show, that $X^4-4X^2-1$ is irreducible. How do I know, that this is the only possible way to factor it?

Thanks in advance.

Answer 1

Your polynomial has no rational roots. Therefore, if it was reducible in $\mathbb{Q}[x]$, then it would have to be the product of two quadratic polynomials $p_1(X),p_2(X)\in\mathbb{Q}[X]$. Each of them either has to real roots or two complex non-real roots.

The roots of your polynomial are $\pm\sqrt{\sqrt5+2}$ and $\pm i\sqrt{\sqrt5-2}$. Of these, only the last two are complex non-real. But$$\left(X-i\sqrt{\sqrt5-2}\,\right)\left(X+i\sqrt{\sqrt5-2}\,\right)=X^2-2+\sqrt5\notin\mathbb{Q}[X].$$Therefore, your polynomial is irreducible.

Answer 2

Let's write $f(x)=x^4-4x^2-1$. Note that since $f(x)=f(-x)$, the roots of $f$ come in positive and negative pairs; say the roots are $a$, $-a$, $b$, and $-b$. Up to renaming the roots, then, there are two cases of factorizations with integer coefficients that $f$ could have. We could have $f(x)=g(x)h(x)$ where $g(x)=(x-a)(x+a)=x^2-a^2$ and $h(x)=(x-b)(x+b)=x^2-b^2$. But that would mean that $x^2-4x-1$ factors as $(x-a^2)(x-b^2)$, and we know $x^2-4x-1$ is irreducible.

The other possibility is that we have a factorization like $f(x)=g(x)h(x)$ with $g(x)=(x-a)(x-b)=x^2-(a+b)x+ab$ and $h(x)=(x+a)(x+b)=x^2+(a+b)x+ab$. In this case, though, $ab$ would be an integer and the constant term of $f(x)$ would be the square of $ab$. Since the constant term is $-1$, this is impossible.

More generally, a similar argument shows that if $e(x)\in\mathbb{Z}[x]$ is irreducible and monic of degree $n$, then $f(x)=e(x^2)$ is irreducible unless its constant term is $(-1)^n$ times a perfect square. See the answers to For which monic irreducible $f(x)\in \mathbb Z[x]$ , is $f(x^2)$ also irreducible in $\mathbb Z[x]$? for more details.

Answer 3

There is also a way to show this by reduction mod $p$:

To see $f$ is irreducible, we use reduction mod 3, which gives $\bar f=x^4-x^2-1$. $\bar f$ clearly has no roots in $\mathbb Z_3$. In $\mathbb Z_3[x]$, the only (monic) irreducible degree 2 polynomials are $$x^2-x+1,x^2-x-1,x^2+x-1,\\x^2+1$$ If $\bar f$ is reducible, then it must be a product of two of them. A few computations show that it can't be the case. Hence $\bar f$ is irreducible and so is $f$.