TITLE: Show that $x^4-10x^2+1$ is irreducible over $\mathbb{Q}$
QUESTION [13 upvotes]: How do I show that $x^4-10x^2+1$ is irreducible over $\mathbb{Q}$? Someone says I should use the rational root test, but I don't exactly know how that applies. Thanks for any input.

REPLY [2 votes]: Perhaps it is worth noting that $x^{2n} -10 x^n + 1$ is irreducible over $\mathbb{Q}$ for every natural $n\ge 1$. Indeed, for $n=1$ the polynomial 
$$x^2 - 10 x + 1 = (x-(5 + 2\sqrt{6})\,)(x-(5-2\sqrt{6})\,)$$
is irreducible over $\mathbb{Q}$.  Now, it is enough to check that in the decomposition
$$x^{2n}-10 x^n + 1 =(x^n-(5 + 2\sqrt{6})\,)(x^n-(5-2\sqrt{6})\,)$$
both of the factors are irreducible in $\mathbb{Q}(\sqrt{6})[x]$.  
Let's consider the general case of a binomial $x^n - \alpha$ over some field $F$. Decompose into linear factors over some extension 
$x^n - \alpha = (x-\gamma_1)\cdot \ldots \cdot (x-\gamma_n)$. Assume that $x^n -\alpha$ is reducible over $F$. Then we can find $k$ factors ($1\le k \le n-1$) whose product is in $F[x]$. Assume wlog that $$(x-\gamma_1)\cdot \ldots \cdot (x-\gamma_k)\in F[x]$$   We conclude 
$\gamma_1 \cdot \ldots \cdot \gamma_k = \beta \in F$.  Now raise to the power $n$ and get $\beta^n = \gamma_1^n\cdot \ldots \cdot \gamma_k^n  = \alpha^k$.  Conclusion:
If $x^n - \alpha$ is reducible in $F[x]$ then there exists $1\le k \le n-1$ and $\beta \in F$ so that 
$$\alpha^k = \beta^n$$
Back to our problem. It is not hard to check that the group of units of the ring of integers of the field $\mathbb{Q}(\sqrt{6})$ is generated by $5 + 2\sqrt{6}$ and $-1$. Therefore,  for $\alpha = 5\pm 2 \sqrt{6}$ we cannot have an equality as above.