# Help determine the group of units of the ring $\mathbb{Z}[(1+\sqrt{m})/2]$

let $m$ be an integer with $m\equiv 1 \pmod4$ and $m<-3$.

$U\left(\mathbb{Z}+\mathbb{Z}(\frac{1+\sqrt{m}}{2})\right)=\{\pm1\}$ How can I prove that?

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Hint: If you already have proved some of the properties of norm, this should not be difficult. –  André Nicolas Apr 3 '12 at 20:33
To expand on Andre's comment. If $X = (1+\sqrt{m})/2$, then define $N(a+bX)=(a+bX)(a-bX)$. Notice that $N(AB)=N(A)N(B)$ (the norm is multiplicative). Also, $N(1)=1$. If $AB=1$, what can you say about $N(A),N(B)$? –  Bill Cook Apr 3 '12 at 20:35
@BillCook $(a+bX)(a-bX)$ is not the norm. $(a+bX)(a+b\bar X)$ is the norm. –  Thomas Andrews Apr 3 '12 at 20:38
@ThomasAndrews Oops! Thanks for catching that! –  Bill Cook Apr 3 '12 at 20:48

If you've been introduced to enough machinery of Algebraic Number Theory, you can recognize:

• $\mathbb{Z}[(1 + \sqrt{m})/2]$ is the ring of integers in the number field $\mathbb{Q}(\sqrt{m})$, and so...
• the free part of the unit group is trivial, by Dirichlet's Unit Theorem

So it boils down to looking for roots of unity. However:

• $[ \mathbb{Q}(\zeta_{4}) : \mathbb{Q} ] = 2$
• $[ \mathbb{Q}(\zeta_{2^e}) : \mathbb{Q} ] \geq 4$ for $e \geq 3$
• $[ \mathbb{Q}(\zeta_{3}) : \mathbb{Q} ] = 2$
• $[ \mathbb{Q}(\zeta_{3^e}) : \mathbb{Q} ] \geq 6$ for $e \geq 2$
• $[ \mathbb{Q}(\zeta_{p^e}) : \mathbb{Q} ] \geq p-1$ for $p \geq 3$

Since $[ \mathbb{Q}(\sqrt{m}) : \mathbb{Q} ] = 2$, you just have to check that neither of $\sqrt{-1}$ and $\sqrt{-3}$ are in $\mathbb{Q}(\sqrt{m})$.

In particular, you probably meant to assume $m$ is squarefree, since $\sqrt{-3} \in \mathbb{Q}(\sqrt{-27})$ (and thus so is $\zeta_3$).

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This is a side comment really, but how did you compute did the degree of those field extensions so quickly? –  user38268 Apr 3 '12 at 22:46
@Benjamin: I'm sure that people who do this stuff for a living have those degrees (or the calculation procedure) memorized. For me, really, the only thing I needed for this answer is that, for an odd prime, the roots of $x^p-1$ are $1$ and the $p-1$ primitive $p$-th roots of unity (which are all conjugate), and so $[\mathbb{Q}(\zeta_p) : \mathbb{Q}] = p-1$. The $\zeta_8$ I did from memory (it helps to actually know the complex $8$-th roots of unity). I'm a little iffy on getting $[\mathbb{Q}(\zeta_9) : \mathbb{Q}]$ right, but all I really needed is that it's bigger than $2$. –  Hurkyl Apr 3 '12 at 23:02