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Let $n \ge 2$ and $p,q$ be two different primes. By Eisenstein the polynomials $X^n-p$ and $X^n-q$ are irreducible. My intuition tells me that $\mathbb{Q}(\sqrt[n]{p})$ and $\mathbb{Q}(\sqrt[n]{q})$ are not isomorphic. But for $n>3$ I didn't come up with a proof but the idea of using the discriminant. Is the statement correct and if so how does one prove it?

Is a general formular for the ring of integers known? Even in case $n=2$ this is not always $\mathbb{Z}[\sqrt[n]{p}]$ (in case $p \equiv 1 \pmod 4$ it is bigger, namely $\mathbb{Z}[(1+\sqrt{p})/2]$). Using the knowledge of the ring of the integers one could compute the diskriminants of the number fields and compare them.

While formulating my question I got this older question suggested. It's quite similar but I'd still like to keep my question since it's a bit more generall (in my question $n$ is not required to by prime nor to be different from $p$ and $q$).

principal-ideal-domain
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  • If $\gcd(n,pq)=1$ then $R=\Bbb{Z}[x]/(x^n-p)$ is unramified at $q$ thus so is $O_{Q(p^{1/n})}$ and $R$ is totally ramified at $p$ thus so is $O_{Q(p^{1/n})}$. The opposite holds for $O_{Q(q^{1/n})}$ thus they can't be isomorphic. By the way for this kind of topic nothing is more enlightening than the corresponding extensions of $p$-adics $\Bbb{Q}_p$. – reuns Dec 17 '20 at 08:02
  • @reuns Where can I read more about this ramification theory? – principal-ideal-domain Dec 17 '20 at 08:56
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    A formula for the ring of integers would be tricky, e.g., the ring of integers of $\mathbf Q(\sqrt[n]{2})$ is $\mathbf Z[\sqrt[n]{2}]$ for all $n \leq 1000$ but not for all $n$. The first counterexample is at $n = 1093$ (famous Wieferich prime to base $2$). See https://kconrad.math.uconn.edu/blurbs/gradnumthy/integersradical.pdf. – KCd Dec 30 '20 at 09:39

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Why does your argument for $n \leq 3$ not work for $n > 3$? If such fields were isomorphic for distinct primes $p$ and $q$ then the field $K := \mathbf Q(\sqrt[n]{p})$ is also $\mathbf Q(\sqrt[n]{q})$ since an isomorphism with an abstract $n$th root would work with real $n$th roots by looking at a real embedding of $K$, which does exist. The point is that those $n$th roots both have degree $n$ over $\mathbf Q$.

Here is the key point: if $r \in \mathbf Q$ and $x^n - r$ is irreducible where $n > 1$, then for the field $K = \mathbf Q(\sqrt[n]{r})$ we have ${\rm Tr}_{K/\mathbf Q}(\sqrt[n]{r}) = 0$. If also $K = \mathbf Q(\sqrt[n]{s})$ for another $s \in \mathbf Q$, then ${\rm Tr}_{K/\mathbf Q}(\sqrt[n]{s}) = 0$ too.

If you set up a hypothetical $\mathbf Q$-linear relation $$ \sqrt[n]{q} = a_0 + a_1\sqrt[n]{p} + a_2\sqrt[n]{p}^2 + \cdots + a_{n-1}\sqrt[n]{p}^{n-1} $$ apply ${\rm Tr}_{K/\mathbf Q}$ to both sides. Because $\mathbf Q(\sqrt[n]{p}) = \mathbf Q(\sqrt[n]{q})$, all terms have trace $0$ except the constant term, which has trace $na_0$. This has to be $0$, so $a_0 = 0$. Then divide by $\sqrt[n]{p}$ on both sides and take the trace again (the trace of $\sqrt[n]{q/p}$ from $K$ down to $\mathbf Q$ is again $0$) to see $a_1 = 0$ and so on all the way up, until finally you have $\sqrt[n]{q} = a_{n-1}\sqrt[n]{p}^{n-1}$. Now $a_{n-1}$ is necessarily nonzero. Since $n \geq 2$, take $n$th powers to get $q = a_{n-1}^np^{n-1}$, which is impossible by comparing the power of $q$ on both sides.

KCd
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