It seems that it is irreducible over $\mathbb Q$, I tried to apply Eisenstein's criterion to $f(x+a)$ for some $a$, but it didn't work.
Also, I found a root $\alpha = \sqrt[3]{\frac{\sqrt{17}-3}{2}}$ and tried to prove that the degree of $\alpha$ over $\mathbb Q$ is 6, but I stuck in computation.
Is there any approach to prove or disprove the irreducibility of the above polynomial?
Best Answer
Its not hard to show that $[\mathbb{Q}(\alpha):\mathbb{Q}]=6$. We have $$[\mathbb{Q}(\alpha):\mathbb{Q}]=[\mathbb{Q}(\alpha):\mathbb{Q}(\sqrt{17})][\mathbb{Q}(\sqrt{17}):\mathbb{Q}]$$ and $$[\mathbb{Q}(\sqrt{17}):\mathbb{Q}]=2$$ so we must show that $$[\mathbb{Q}(\alpha):\mathbb{Q}(\sqrt{17})]=3.$$
This amounts to showing that $x^3=\frac{-3+\sqrt{17}}{2}$ has no solution in $\mathbb{Q}(\sqrt{17})$ Since $\sqrt{17}\mapsto -\sqrt{17}$ is an automorphism of this field we see that if there is an $x\in \mathbb{Q}(\sqrt{17})$ such that $$x^3=\frac{-3+\sqrt{17}}{2}$$ then there is a $y\in \mathbb{Q}(\sqrt{17})$ such that $$y^3=\frac{-3-\sqrt{17}}{2}$$ multiplying these two equations we get that $$(xy)^3=-2$$ And we know that $\mathbb{Q}(\sqrt[3]{-2}):\mathbb{Q}]=3$ a contradiction so there is no such $x$ and $[\mathbb{Q}(\alpha):\mathbb{Q}(\sqrt{17})]=3.$
Or alternatively you could take the norm to get $(N(x))^3=-2$ an equation in $\mathbb{Q}$ again a contradiction.