## Show that a given polynomial is irreducible over the rationals with √(-2) adjoined

Prove that $p(x) = x^4 - 4x^2 + 8x + 2$ is irreducible over the quadratic field $\mathbb{Q}(\sqrt{-2})$.

We saw previously that $\mathbb{Z}[\sqrt{-2}]$ is a Euclidean domain with field of fractions $\mathbb{Q}(\sqrt{-2})$. Thus it is a unique factorization domain, and thus by Gauss’ lemma, $p(x)$ is irreducible over $\mathbb{Q}(\sqrt{-2})$ if and only if it is irreducible over $\mathbb{Z}[\sqrt{-2}]$.

Suppose $p(x)$ is reducible in $\mathbb{Z}[\sqrt{-2}][x]$.

Suppose $p(x)$ has a linear factor; say $x - a$. Then we must have that $a$ divides $2$ in $\mathbb{Z}[\sqrt{-2}]$. Recall that the Euclidean norm in this ring is given by $N(a+b\sqrt{-2}) = a^2 + 2b^2$; thus $N(a)$ divides 4, so that $N(a) \in \{ \pm 1, \pm 2, \pm \sqrt{-2}\}$. Evidently, however, we have $p(1) = 7$, $p(-1) = -9$, $p(2) = 18$, $p(-2) = -14$, $p(\sqrt{-2}) = 14 + 8\sqrt{-2}$, and $p(-\sqrt{-2}) = 14 - 8\sqrt{-2}$. So $p(x)$ does not have a linear factor.

Suppose instead that $p(x)$ has a quadratic factor, say $x^2 + ax + b$. Using the long division algorithm for polynomials, we have $p(x) = (x^2 + ax + b)(x^2 - ax + (a^2 - b - 4)) + (8 - a^3 + 2ab +4a)x + (2 - a^2b + b^2 + 4b)$. Since quotients and remainders in the Euclidean algorithm in $F[x]$ are unique, we have $8 - a^3 + 2ab + 4a = 0$ and $2 - a^2b + b^2 + 4b = 0$. Moreover, we have $b \in \{ \pm 1, \pm 2, \pm \sqrt{-2} \}$ as in the linear factor case, and similarly $a^2 - b - 4 \in \{ \pm 1, \pm 2, \pm \sqrt{-2}$. Note that $8 = a(a^2 - 2b - 4)$; computing norms, we see that $N(a)$ divides 64. If $b = 1$, then the second equation gives $2a^2 = -5$. Computing norms, we have that 4 divides 25, a contradiction. If $b = 1$, then we have $-2a^2 = 3$, which is similarly a contradiction. If $b = 2$, then we have $a^2 = -5$. Then $N(a)$ is either 1, 5, or 25; since $N(a)$ divides 64, $N(a) = 1$ and so $a = \pm 1$. Then $a^2 = 1$, a contradiction. If $b = -2$, then we have $a^2 = -1$, a contradiction since no such element exists in $\mathbb{Z}[\sqrt{-2}]$. If $b = \sqrt{-2}$, then the second equation yields $\sqrt{-2}(a^2 + 2) = 1$, a contradiction since the only units in $\mathbb{Z}[\sqrt{-2}]$ are 1 and -1. If $b = -\sqrt{-2}$, then we likewise have $\sqrt{-2}(a^2 + 2) = -1$, again a contradiction. Thus $p(x)$ does not have a quadratic factor.

Thus $p(x)$ is irreducible over $\mathbb{Z}[\sqrt{-2}]$, and thus over $\mathbb{Q}(\sqrt{-2})$.