Irreducibility of polynomials over some number fields Recently, a problem in my research has appeared and now I need to construct some algebraic numbers with special properties (related to its degree and some other fields extensions).
Now, in order to help me, I would like to prove the following result:
Proposition. Let $\alpha$ be a real algebraic number and let $n>4$ be a positive integer, then the polynomial $X^n-p$ is irreducible over $K:=\mathbb{Q}(\alpha)$, for all large enough prime number $p$.
I tried to use some splitting fields, discriminantes properties, (un)ramified primes to prove it, but I was not able to do it.
Any suggestion is very welcomed.
Thanks in advance.
 A: Here is the Vahlen-Capelli irreducibility criterion for binomial polynomials.
Theorem. For a field $F$, nonzero $a \in F$, and integer $n \geq 2$, the polynomial $x^n − a$ is irreducible in $F[x]$ if and only if
the following two conditions are satisfied:
$(i)$ for each prime $\ell$ dividing $n$, $a$ is not an $\ell$th power in $F$,
and $(ii)$ if $4 \mid n$, then $a$ is not $-4b^4$ for some $b \in F$.
This is proved in Lang's Algebra, Theorem 9.1 of Section 9 of Chapter VI. Lang only proves that under those two conditions, $x^n - a$ is irreducible. The converse is easy (you'll need to know $x^4 + 4$ has a universal factorization: it's $(x^2 + 2x + 2)(x^2-2x+2)$).
Example. Let $F$ be an arbitrary number field, $n \geq 2$, and $p$ be a sufficiently large prime number. Then $p$ is unramified in $F$, so $p$ is not a $k$-th power in $F$ when $k \geq 2$: if $p = \gamma^k$ for some $\gamma \in F$ then $\gamma \in \mathcal O_F$ and $(p) = (\gamma)^k$ as ideals, but an  unramified prime has a squarefree prime ideal factorization. Thus we get a contradiction since $k \geq 2$.  So $p$ is not an $\ell$th power in $F$ for each prime $\ell$ dividing $n$. If $4 \mid n$ and $p = -4b^4$ for some $b \in F$ then $(p) = (2b^2)^2$ is the square of an ideal, which is false for large $p$ since $(p)$ is squarefree. Thus the conditions of the theorem are met, so $x^n - p$ is irreducible.
That argument did not need $p$ to be a large prime: all the reasoning works for the polynomial $x^n - a$ where $a$ is a sufficiently large squarefree integer.
A: Lemma. Let $K$ be any number field, and $p$ a prime unramified in $K$. Then $X^n-p$ is irreducible over $K$.
Proof. It suffices to show that the field $L = K(\sqrt[n\ \ ]{p})$ has degree $n$ over $K$. Let $\mathfrak q \subseteq \mathcal O_L$ be a prime above $p$, and let $\mathfrak p = \mathcal O_K \cap \mathfrak q$ be its image in $\operatorname{Spec} \mathcal O_K$. Since $\mathcal O_L$ contains $\mathbf Z[\sqrt[n\ \ ]{p}]$, we have $e_{\mathfrak q/p} \geq n$. But $K$ is unramified above $p$, so $e_{\mathfrak p/p} = 1$. We conclude that $e_{\mathfrak q/\mathfrak p} = n$, so $[L:K] \geq n$. The reverse inequality is clear. $\square$
A: Here is a different approach, which is arguably a bit more elementary. If $f=X^n-p$ splits in $K$, and $g$ is one of its factors, then the constant term of $g$, being a product of zeros of $f$, must be of the form $\epsilon p^{k/n}$, where $1 \leq k < n=\deg(f)$ and $\epsilon$ is a root of unity, and therefore $K$ contains a zero of the polynomial $X^d - p$, where $d=\gcd(k,n)$ is a divisor of $n$. (This then also shows that $p$ is ramified in the field extension $K/\mathbb{Q}$, but we will not use this.)
Now it is a standard exercise in Galois theory to show that $\mathbb{Q}(\sqrt[d]{p_1},\ldots,\sqrt[d]{p_m})$ has degree $d^m$ over $\mathbb{Q}$, if $p_1,\ldots,p_m$ are prime numbers. So the numbers of primes $p$ such that $K$ contains a zero of $X^d-p$ is finite. Since the number of possible $d$ is also finite, this concludes the proof.
