Let $K$ be a quadratic number field, and $\mu(K)$ the group of roots of unity of $K$. I'm trying to prove that (with the two exceptions of $\mathbb{Q}(\sqrt{-1})$ and $\mathbb{Q}(\sqrt{-3})$)

$$\mu(K) = \{1, -1\}$$

The case where $K$ is imaginary is easier, since by Dirichlet's unit theorem, we have

$$\mathcal{O}_K^* = \mu(K) \times \mathbb{Z}^{1 + 0 - 1} = \mu(K)$$

And it is easy to prove that $\alpha \in \mathcal{O}_K^* \iff N(\alpha) = 1$. Dealing with the $d \equiv 2, 3 \mod 4$ and $d \equiv 1 \mod 4$ separately we get a complete characterization of $\mathcal{O}_K^*$, and therefore of $\mu(K)$, for $K$ imaginary.

But for $K$ real this doesn't work anymore, because we have

$$\mathcal{O}_K^* = \mu(K) \times \mathbb{Z}^{0 + 2 - 1} = \mu(K) \times\mathbb{Z}$$

Instead, and the approach above gives infinite units (solution to Pell's equation), and says nothing about the roots of unity.

I'm guessing that one could use the fact that a real quadratic field has only one embedding, and that embedding is real, together with the fact that the only roots of unity in $\mathbb{R}$ are $\pm 1$.

Is there an elementary way to prove the result without separating the real and imaginary case, and/or without using Dirichlet's unit theorem?


2 Answers 2


The real case is actually quite easy! Consider that the torsion subgroup is $\{\alpha : \alpha^n =1\}$ since the group is multiplicative. But then this is exactly the roots of unity. As such you need only note that since $F\subseteq\Bbb R$ we have that only real roots of unity are in this field, i.e. $\{\pm 1\}$.

If you care about the general case all together, it's also easy, again you note that $\mu(K)$ is exactly the roots of unity since torsion iff finite multiplicative order iff root of unity. But then if you have a root of unity, $\zeta_n$, for some $n\ne 2,3,4,6$ (recall $-\zeta_3$ is a primitive $6^{th}$ root of unity) then you have a proper subfield of degree $\phi(n)$. Since $p-1|\phi(n)$ for all $p|n$ we see $n=2^\alpha3^\beta$ otherwise we have a sub-extension of too large a degree. But we have excluded the (non-trivial) cases

$$\begin{cases} \alpha = \beta = 1\\ \alpha=2, \beta = 0 \\ \alpha = 0, \beta = 1 \end{cases}$$

And if $\alpha >2 $ or $\beta >1$ the degree is again too big since then $\phi(n)>2$, so no other cases can occur.

  • $\begingroup$ @nguyenquangdo I'm not sure I see the intent of your comment. $\endgroup$ Commented Nov 14, 2016 at 23:53
  • $\begingroup$ @nguyenquangdo ah, then you should probably repost that on the original thing and delete these comments here as best practice. $\endgroup$ Commented Nov 15, 2016 at 13:09
  • $\begingroup$ @nguyenquangdo mouse over your comment and a little x should appear to the right of the timestamp after your name. Click the x to delete $\endgroup$ Commented Nov 16, 2016 at 13:57

If $p$ is a prime then $\zeta_p$ satisfies an irreducible equation of degree $p-1$. Since quadratic fields have degree $2$, it remains only to consider the cube roots and the power of $2$ roots.


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