User lemon sherbet - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T21:40:48Z http://mathoverflow.net/feeds/user/18072 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/76402/q-sqrt2q-sqrt21n/76406#76406 Answer by Lemon Sherbet for $Q(\sqrt{2})=Q((\sqrt{2}+1)^n)$ Lemon Sherbet 2011-09-26T11:52:41Z 2011-09-26T11:52:41Z <p>Choose a prime $p$ that splits completely in $K$, and let $\mathfrak{P}$ denote a prime in $\mathcal{O}_K$ above $p$. There exists an integer $h$ such that $\mathfrak{P}^h$ is principal; write $\mathfrak{P} = (\alpha)$. I claim that $K = \mathbf{Q}(\alpha^n)$ for all $n$. To see this, suppose that $\alpha^n$ lived inside some proper subfield $E$. It's easy to see that the ideal $\alpha^n \mathcal{O}_E$ is divisible by $\mathfrak{p} = \mathfrak{P} \cap \mathcal{O}_E$. But, by construction, the prime $\mathfrak{p}$ splits into $[K:E] > 1$ distinct primes in $K$, which is incompatible with the factorization $(\alpha^n) = \mathfrak{P}^{nh}$ in $\mathcal{O}_K$.</p>