I suspect that this polynomial is irreducible (and therefore separable since $\mathbb{Q}$ has characteristic $0$) because it has no rational roots. Can somebody tell me if I'm correct and explain a more rigorous method to show that the polynomial is irreducible?
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Irreducible over what field? – Pedro Nov 25 '16 at 05:35
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I guess the OP meant $\mathbb{Q}$. ("has no rational roots") – Shraddheya Shendre Nov 25 '16 at 05:37
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If a polynomial has degree $n>3$, then it may be reducible over a field without having any roots in that field. $x^4+1$ over $\mathbb{R}$ is one example.
In your case, however, it turns out that $x^n-2$ is irreducible over $\mathbb{Q}$ for all $n\geq 1$ by Eisenstein's criterion.
carmichael561
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If $\deg f =2,3$, then your method works (and is rigourous). This is because assuming $f = gh$, where $g$ and $h$ are not invertible, we have $\deg g = 1$ or $\deg h = 1$.The conclusion follows since it is well known that $\alpha$ is a root of a polynomial if and only if $X-\alpha$ is a factor.
If $\deg > 3$, the conclusion doesn't follow. All you have to do is take two irreducible polynomials of degree $2$ and multiply them.
Note that you still have to prove separately that $\sqrt[p]2$ is irrational. It is actually easier to do it the other way around: use Eisenstein's criterion, as suggested by carmichael561, to prove that $X^p - 2$ is irreducible and conclude that $\sqrt[p]2$ is irrational.
Ennar
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