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I'm doing one of the exercises of Stewart and Tall's book on Algebraic Number Theory. The problem concerns finding an expression for the norm in the cyclotomic field $K = \mathbb{Q}(e^{2\pi i / 5})$. The exact problem is the following:

If $\zeta = e^{2 \pi i / 5}$, $K = \mathbb{Q}(e^{2\pi i / 5})$, prove that the norm of $\alpha \in \mathbb{Z}[\zeta]$ is of the form $\frac{1}{4}(A^2 -5B^2)$ where $A, B \in \mathbb{Z}$.

(Hint: In calculating $\textbf{N}(\alpha)$, first calculate $\sigma_1 (\alpha) \sigma_4 (\alpha)$ where $\sigma_i (\zeta) := \zeta^{i}$. Show that this is of the form $q + r\theta + s\phi$ where $q, r, s \in \mathbb{Z}$, $\theta = \zeta + \zeta^{4}$ and $\phi = \zeta^{2} + \zeta^{3}$. In the same way establish $\sigma_2 (\alpha) \sigma_3 (\alpha) = q + s\theta + r\phi$ )

Using Exercise $3$ prove that $\mathbb{Z}[\zeta]$ has an infinite number of units.

Now, I've already done what the hint says and arrived at the following. If we let $\alpha = a +b\zeta^{} + c\zeta^{2} + d\zeta^{3} \in \mathbb{Z}[\zeta]$ then after simplifying I get

$\textbf{N}(\alpha) = \sigma_1 (\alpha) \sigma_4 (\alpha) \sigma_2(\alpha) \sigma_3(\alpha) = ( q + r\theta + s\phi ) ( q + s\theta + r\phi )$

$ = q^2 + (qr + qs)(\theta + \phi) + rs(\theta^2 + \phi^2) + (r^2 + s^2)\theta \phi$

and then it is not that hard to see that $\theta + \phi = -1$, $\theta^2 + \phi^2 = 3$ and $\theta \phi = -1$ so that in the end one obtains

$\textbf{N}(\alpha) = q^2 - (qr + qs) + 3rs - (r^2 + s^2)$

where $q = a^2 + b^2 + c^2 + d^2$, $r = ab + bc + cd$ and $s = ac + ad + bd$.

Now, here I got stuck because I just can't take the last expression for the norm into the form that the exercise wants.

The purpose is to get that nice form for the norm to find units by solving the diophantine equation $\textbf{N}(\alpha) = \pm 1$, which is what the Exercise $3$ mentioned in the statatement of the problem is about.

I already know how to prove the existence of infinitely many units in $\mathbb{Z}[\zeta]$ (without using Dirichlet's Unit Theorem of course), but the exercise also demands a proof that the norm is equal to $\frac{1}{4}(A^2 -5B^2)$.

I even asked my professor about this and we were not able to get the desired form for the norm.

So my question is if anybody knows how to prove that the norm has that form, and if so, how can I show that? Or if it could be that maybe the hint given in the exercise is not that helpful?

Thanks a lot in advance for any help with this.

EDIT

After looking at Derek Jennings' answer below, to get from the expression I had for the norm to the one in Derek's answer is just a matter of taking out a common factor of $1/4$ in the expression and then completing the square,

$\textbf{N}(\alpha) = q^2 - (qr + qs) + 3rs - (r^2 + s^2) = q^2 - q(r+s) + rs - (r-s)^2$

$ = \frac{1}{4}( 4q^2 - 4q(r+s) + 4rs - 4(r-s)^2 ) $ $= \frac{1}{4} ( 4q^2 - 4q(r+s) +\overbrace{(r+s)^2} - \overbrace{(r+s)^2} + 4rs - 4(r-s)^2 )$

$ = \frac{1}{4} ( (2q -(r+s))^2 -(r-s)^2 - 4(r-s)^2 )$

$ = \frac{1}{4}( (2q - r - s)^2 - 5(r-s)^2 ) = \frac{1}{4}(A^2 - 5B^2),$

as desired. Of course it is easier if you already know what to get at =)

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    @Qiaochu Yes, you're right. That is a way to see that it has an infinite number of units because for instance the quadratic field $\mathbb{Q}(\sqrt{5})$ is a subfield of $K$ if I'm not mistaken. So basically, what you're saying is that the hint is not helping me a lot, right?2011-03-03

2 Answers 2

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You're almost there: set $A=2q-r-s$ and $B=r-s.$ Then your expression for $\textbf{N}(\alpha)$ reduces to the desired form. i.e. your

$\textbf{N}(\alpha) = \frac14 \left \lbrace (2q-r-s)^2 - 5(r-s)^2 \right \rbrace = \frac14(A^2-5B^2).$

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    @Adrián: I remembered having done this question many years ago (although I believe it was from another source) and reaching the exact point where you got stuck. That was well before computers were so prominent and already knowing the form of the solution it wasn't too hard to spot.2011-03-03
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I agree completely with Qiaochu. If you understand what he wrote, then you can use this to give a much shorter and easier proof than the hint is suggesting.

In fact I didn't even read the hint. I saw the question, thought "Oh, that looks like the norm from $\mathbb{Q}(\sqrt{5})$". I also know that $\mathbb{Q}(\sqrt{5})$ is the unique quadratic subfield of $\mathbb{Q}(\zeta_5)$ (since $5 \equiv 1 \pmod 4$; there's a little number theory here), and that for any tower of finite field extensions $L/K/F$, we have

$N_{L/F}(x) = N_{K/F}(N_{L/K}(x))$.

Norms also carry algebraic integers to algebraic integers, so this shows that the norm of any element of $\mathbb{Z}[\zeta_5]$ is also the norm of some algebraic integer of $\mathbb{Q}(\sqrt{5})$, i.e., is of the form $(\frac{A+\sqrt{5}B}{2})(\frac{A-\sqrt{5}B}{2})$ for some $A,B \in \mathbb{Z}$. I think we're done.

[I have never read Stewart and Tall, so it may be that they have not assumed or developed this much theory about norm maps at the point they give that exercise. But if you know it, use it!]

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    @Pete: +1, This is a very useful answer! A naive question: How would could write the norm of $\mathbb{Z}[\zeta_{10}]$ in the same way? Would it become $\frac{1}{4}\left(A^2+5B^2\right)$? Just a slight curiosity. I know it is the _same_ ring, but for example $\mathbb{Z}[\frac{1+\sqrt{-3}}{2}]$ has norm $a^2+ab+b^2$ whereas $\mathbb{Z}[\frac{-1+\sqrt{-3}}{2}]$ has norm $a^2-ab+b^2$ even though they are the same.2011-07-05