This is a problem from Neukirch's book on Algebraic Number Theory.

Let $a$ and $b$ be positive integers that are not perfect squares. How does one show that the fundamental unit of the order $$\mathbb{Z}+\mathbb{Z}\sqrt{a}$$ of the field $\mathbb{Q}(\sqrt{a})$ is also the fundamental unit of the order $$\mathbb{Z} + \mathbb{Z}\sqrt{a}+\mathbb{Z}\sqrt{-b} + \mathbb{Z}\sqrt{a}\sqrt{-b}$$ of $\mathbb{Q}(\sqrt{a},\sqrt{-b})?$ I tried to brute force this, by simply letting $\epsilon$ be the fundamental unit of the order $\mathbb{Z}+\mathbb{Z}\sqrt{a},$ and then trying to show that it is not of the form $\eta^k$ for some $$\eta \in \mathbb{Z} + \mathbb{Z} \sqrt{a} + \mathbb{Z}\sqrt{-b} + \mathbb{Z}\sqrt{a}\sqrt{-b}.$$ This quickly seemed to get out of hand though.


1 Answer 1


I'll assume $\newcommand{\Q}{\Bbb Q}\Q(\sqrt a,\sqrt{-b})$ has only two roots of unity. There will only be many finitely many exceptions to this.

Let $\newcommand{\ep}{\varepsilon}\ep>1$ be the fundamental unit of $R=\newcommand{\Z}{\Bbb Z}\Z[\sqrt a]$ and consider $S=\Z[\sqrt a,\sqrt{-b}]$. The unit group of $S$ is $\{\pm1\}\times\left<\eta\right>$ for some $\eta$. But $N:\alpha\mapsto|\alpha|^2$ is a homomorphism from $S^\times$ to $R^\times$ and takes $\ep$ to $\ep^2$. So the image of $N$ is either $\left<\ep\right>$ or $\left<\ep^2\right>$. In the latter case then $S^\times$ is just $R^\times$. In the former case, we can assume that $\eta^2=\pm\ep$. Either $\eta=\sqrt{\ep}$ or $\eta=\sqrt{-\ep}$.

Consider the fields $K=\Q(\sqrt{\pm\ep})$. If we can show these are not elementary Abelian over $\Q$ we will have succeeded. For a start we need $K$ to be Galois over $\Q$. This means that $K=\Q(\sqrt{\pm\ep'})$ where $\ep'$ is the conjugate of $\ep$ over $\Q$. By Kummer theory $\ep\ep'$ must be a square in $\Q(\sqrt a)$. This is only possible if $\ep\ep'=1$. To get an imaginary extension, we need $K=\Q(\sqrt{-\ep})$.

We must have $\eta^2=-\ep$. But $|\eta|^2=\ep$ and so $\eta$ is totally imaginary: $\eta=u\sqrt{-b}+v\sqrt{-ab}$. Unless $b=1$, $\ep=-\eta^2$ is divisible by $b$.

  • 1
    $\begingroup$ Thank you for your answer. I feel there should be something that does not use Kummer theory though, considering this is in the beginning of Neukirch's book... $\endgroup$
    – Twistediso
    Nov 11, 2017 at 23:06

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