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.


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 =)

  • 3
    $\begingroup$ The norm is transitive in towers, and K can be placed in a tower of two quadratic extensions. $\endgroup$ – Qiaochu Yuan Mar 3 '11 at 17:51
  • $\begingroup$ @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? $\endgroup$ – Adrián Barquero Mar 3 '11 at 17:59

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

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

  • $\begingroup$ Thank you very much Derek. I edited my question to reflect on your solution. By the way, is that how you arrived at this expression? $\endgroup$ – Adrián Barquero Mar 3 '11 at 21:21
  • $\begingroup$ @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. $\endgroup$ – Derek Jennings Mar 3 '11 at 22:06

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!]

  • $\begingroup$ Thanks a lot Pete. This is a very nice way of looking at the problem. In fact you're right. Stewart and Tall have not proved the transitivity of norms and traces at that point in the book. In fact I'm not even sure that they prove it at all. But in any case it is proved in most other books I suppose. $\endgroup$ – Adrián Barquero Mar 3 '11 at 21:27
  • $\begingroup$ By the way, how do you know that there's a unique quadratic subfield of $\mathbb{Q}(\zeta_5)$? Is it just using the Galois correspondence and the fact that the Galois group of $\mathbb{Q}(\zeta_5)/\mathbb{Q}$ is isomorphic to $\mathbb{Z}/4\mathbb{Z}$? $\endgroup$ – Adrián Barquero Mar 3 '11 at 21:37
  • $\begingroup$ @Adrián: that's one way to see it. Pete seems to be hinting that since $5$ is the only prime that ramifies in $\mathbb{Q}(\zeta_5)$, the only quadratic subfield it can contain is the unique one with discriminant a power of $5$, but this might be a little too high-brow for where you are. $\endgroup$ – Qiaochu Yuan Mar 3 '11 at 22:34
  • $\begingroup$ @Qiaochu Yes, you're right, I don't know about that yet. We'll probably see some ramification by the end of the semester. But thank you, now I can look this up in a book. $\endgroup$ – Adrián Barquero Mar 3 '11 at 22:48
  • $\begingroup$ @Adrian, @Qiaochu: Qiaochu's explanation is a good way to go. It also follows from my favorite proof of Quadratic Reciprocity, via evaluation of the quadratic Gauss sum: see for instance math.uga.edu/~pete/NT2009qrproof.pdf. (In this one case it would certainly be easy enough to show by hand that $\sqrt{5}$ lies in $\mathbb{Q}(\zeta_5)$...) $\endgroup$ – Pete L. Clark Mar 4 '11 at 0:52

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