Is $f(y) = 5 - 4y^2 + y^4$ irreducible over $\mathbb{Q}(\sqrt{5})$?

Question

Using Eisenstein's criterion, we can easily verify that $f(y)$ is irreducible over $\mathbb{Q}$. However, I'm not sure how I'd go about proving whether or not $f$ is irreducible over $\mathbb{Q}(\sqrt{5})$. Does the fact that all of $f$'s roots are imaginary immediately tell us that $f$ is irreducible (I don't think this is the case since $f$ is quartic)?

Answer 1

$f(y) = 0$ has complex roots, which form two conjugate pairs. If $f(y)$ were to be reducible over $\mathbb Q(\sqrt5)$, then the irreducible factor must be of degree 2, with the roots being one of the conjugate pairs.

$f(y) = (y^2 + \sqrt 5)^2 - (4+2\sqrt5)y^2 = (y^2 + \alpha y + \sqrt5)(y^2 - \alpha y + \sqrt5)$, where $\alpha$ is the square root of $4 + 2\sqrt5$.

In $\mathbb Q(\sqrt5)$, you need to figure out whether $4 + 2\sqrt 5$ is a square. But the norm $N(4+2\sqrt5) = 4^2 - 4\times 5 = -4$, so it cannot be a square of something.

Therefore, $f$ is irreducible over $\mathbb Q(\sqrt 5)$.

Answer 2

Having all imaginary roots does not do it. For example, $y^4 + 3 y^2 + 1$ has all imaginary roots, but factors as $(y^2 + (3-\sqrt{5})/2)(y^2 + (3+\sqrt{5})/2))$.

Answer 3

The other answers have already addressed your question about the roots, so let me just add that you can still apply Eisenstein's criterion when testing for irreducibility over $K = \mathbb{Q}(\sqrt{5})$.

Presumably you applied Eisenstein's criterion to $f(x+1) = x^4 + 4 x^3 + 2 x^2 - 4 x + 2$ at the prime $p = 2$, and in fact this same strategy works over $K$. The ring of integers $\DeclareMathOperator{\O}{\mathcal{O}} \O_K$ of $K$ is $\mathbb{Z}[\alpha]$ where $\alpha = \frac{1 + \sqrt{5}}{2}$. The minimal polynomial of $\alpha$ is $x^2 - x - 1$ and since this is irreducible mod $2$, then $(2)$ is a prime ideal of $\O_K$. Applying the general version of Eisenstein's criterion at the prime ideal $(2)$ shows that $f$ is irreducible in $\O_K[x]$ and Gauss's Lemma then implies that $f$ is irreducible in $K[x]$.

I think this idea works more generally, too. Let $L \supseteq K$ be an extension of number fields, let $\mathfrak{p}$ be a prime ideal of $\O_K$ that is unramified in $\O_L$, and let $\mathfrak{P}$ be a prime divisor of $\mathfrak{p} \O_L$. If $f \in K[x]$ is Eisenstein at $\mathfrak{p}$, then $f \in L[x]$ is Eisenstein at $\mathfrak{P}$.