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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.

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  • $\begingroup$ 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/… $\endgroup$
    – tj_
    Feb 28, 2023 at 2:15

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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. $\square$

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  • $\begingroup$ 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. $\endgroup$
    – Jean
    Feb 27, 2023 at 17:03
  • $\begingroup$ 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 $\sum_{\mathfrak q_i\mid \mathfrak p\mathcal O_L} e_{\mathfrak q_i/\mathfrak p}f_{\mathfrak q_i/\mathfrak p} = [L:K]$. $\endgroup$ Feb 27, 2023 at 17:24
  • $\begingroup$ 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! $\endgroup$
    – Jean
    Feb 27, 2023 at 20:29
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    $\begingroup$ 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 $n$. $\endgroup$ Feb 27, 2023 at 21:26
  • $\begingroup$ 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! $\endgroup$
    – Jean
    Feb 28, 2023 at 12:31
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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.

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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.

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  • $\begingroup$ 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. $\endgroup$
    – Jean
    Mar 1, 2023 at 13:19
  • $\begingroup$ @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 $i$. $\endgroup$
    – R.P.
    Mar 1, 2023 at 13:32
  • $\begingroup$ Thanks @R.P. But what if $\gcd(k,n)=1$? Then $p\in K$ as we already know. $\endgroup$
    – Jean
    Mar 1, 2023 at 13:50
  • $\begingroup$ Maybe you meant lcm$(k,n)$? Or am I missing something? Or $d$ must be a multiple of $n$? $\endgroup$
    – Jean
    Mar 1, 2023 at 13:56
  • $\begingroup$ @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. $\endgroup$
    – R.P.
    Mar 1, 2023 at 22:33

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