# 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.

• Let $K$ be a number field with ring of integers $R$. Since a monic polynomial with rational integer coefficients is irreducible over $K$ iff it's irreducible over $R$, the irreducibility of $X^n-p$ for unramified primes $p$ follows immediately from the elementary Eisenstein irreducibility criterion in $R$: math.stackexchange.com/questions/2758588/…
– tj_
Feb 28, 2023 at 2:15

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. $$\square$$

• Thank you for your help, however I didn't understand some parts. Could you please clarify me? (1) Why $K$ is unramified above $p$? (2) Why $e_{\mathfrak{q}/\mathfrak{p}}=n$ implies $[L: K]\geq n$? Sorry for maybe, simple questions.
– Jean
Feb 27, 2023 at 17:03
• That $K$ is unramified above $p$ is the hypothesis of the lemma. This means that for any given $K$, all but finitely many primes $p$ satisfy the hypothesis. The second follows since $\sum_{\mathfrak q_i\mid \mathfrak p\mathcal O_L} e_{\mathfrak q_i/\mathfrak p}f_{\mathfrak q_i/\mathfrak p} = [L:K]$. Feb 27, 2023 at 17:24
• Thank you again. My two last questions: (1) What means Spec $\mathcal{O}_K$? (2) You said that $\mathcal{O}_L$ contains $\mathbb{Z}[p^{1/n}]$ and so $e_{\mathfrak{q}/\mathfrak{p}}$. What $\mathfrak{q}/\mathfrak{p}$ means (in terms of ideals)? And why this inequality is true? The ramification index in at least $n$? Thank you so much!
– Jean
Feb 27, 2023 at 20:29
• Ok, $\operatorname{Spec} \mathcal O_K$ doesn't really matter (this is algebraic geometry notation for the set of prime ideals in $\mathcal O_K$). Indeed the ramification index is at least $n$, again since ramification index is multiplicative in towers and the ramification index of $\mathbf Z[\sqrt[n\ \ ]{p}]$ is $n$. Feb 27, 2023 at 21:26
• Amazing, thank you very much. Please see if you can help me in the general question mathoverflow.net/questions/441809/… Thank you so much R. van Dobben de Bruyn!
– Jean
Feb 28, 2023 at 12:31

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.

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.

• Thanks for your nice suggestion. I only have two questions: (1) I think $k$ can be $n$ too. Since $g(0)$ can be $p$. Right? (2) Ok, $\epsilon p^{k/n}\in K$, but why this implies that $\zeta p^{1/d}$ belongs to $K$, for some $d$th root of unity $\zeta$, where $d=\gcd(k,n)$? Thanks you again.
– Jean
Mar 1, 2023 at 13:19
• @Jean (1) No, because if $g$ is an irreducible factor of the reducible polynomial $f$, then its degree must be lower than $n$. So its constant term, which is a product of $n$-th power roots of $p$ consisting of $\deg(g)$ factors, can't be equal to $p$ times a root of unity. (2) This is trivial. If you reduce $k/n$ to lowest terms, say you end up with a fraction $t/d$ with $\gcd(t,d)=1$. Then there exists an integer $a$ with $at\equiv 1\pmod{d}$, which means that $p^{ak/n}=p^{at/d}=p^i p^{1/d}$ for some integer $i$.
– R.P.
Mar 1, 2023 at 13:32
• Thanks @R.P. But what if $\gcd(k,n)=1$? Then $p\in K$ as we already know.
– Jean
Mar 1, 2023 at 13:50
• Maybe you meant lcm$(k,n)$? Or am I missing something? Or $d$ must be a multiple of $n$?
– Jean
Mar 1, 2023 at 13:56
• @Jean If $\gcd(k,n)=1$ then the conclusion of my argument is that $p^{1/n}$ is in $K$. The point is that the constant factor of $g$ must really be a non-integral power of $p$ (up to an $n$-th root of unity), from which it follows that $K$ must contain such a power.
– R.P.
Mar 1, 2023 at 22:33