0

I want to classify the six units in $\mathbb Z[\zeta_3]$, where $\zeta_3$ is a primitive cube root of unity.

I know the basic idea of this is to show that the norm of $\alpha \in \mathbb Z[\zeta_3]$ can be written $\frac{1}{4}(a^2+3b^2)$ for integers $a$ and $b$ of the same parity, and then use the fact that $\alpha$ is a unit iff $N(\alpha)=\pm1$.

I have proved the condition on $\alpha$ being a unit, and using $N(\alpha)=\frac{1}{4}(a^2+3b^2)=\pm1$ means $a=\pm1, b=\pm1$ or $a=\pm2, b=0$. I am assuming it is the proof that the norm can be written this way that relates the values of $a$ and $b$ to that of $\alpha$.

However, I am not sure how to show that $N(\alpha)=\frac{1}{4}(a^2+3b^2)$. I know the conjugates of $\mathbb Z[\zeta_3]$ are $\{1,\zeta,\zeta^2\}$, and that $N(1-\zeta^j)=p$ for $j=0,1,2,$ but I am not sure if they help. Since $N(\alpha)$ is found by multiplying the conjugates of $\alpha$, would it be a good idea to define $\alpha=a+b\zeta+c\zeta^2$ and try to find a minimal polynomial?

Bill Dubuque
  • 282,220
AccioHogwarts
  • 1,437
  • 1
    Ehh? You do know that $1+\zeta+\zeta^2=0$, don't you? So only $\zeta$ and $\zeta^2=\overline{\zeta}=-1-\zeta$ are conjugates. The elements of $\Bbb{Z}[\zeta]$ are of the form $(a+b\sqrt{-3})/2$ with $a,b$ any integers of the same parity. For many a purpose it might be better to use an integral basis. For example ${1,\zeta}$. The norm does become a bit more complicated then, because $$N(u+v\zeta)=u^2-uv+v^2.$$ But then there are no parity restrictions on $u,v$ - any combination of integers works. – Jyrki Lahtonen Mar 10 '15 at 10:40

3 Answers3

2

You are over-engineering this.

Since $\zeta(\zeta+1)=-1$ you have $\bar\zeta=-1-\zeta$

Now when you note that the norm of some element $\alpha=a+b\zeta$ is equal to $1$ you have $$N(\alpha)=(a+b\zeta)(a+b\bar\zeta)=1$$

It is then clear that $\alpha$ is a unit from the basic definition of a unit, because you've just multiplied it by something else in the original ring and got the answer $1$. (If the Norm were $-1$ you'd just change the sign of the second factor.)

Working through the computation the norm comes out to $$a^2-ab+b^2=\frac{(2a-b)^2+3b^2}4$$ and you can use this (or any other method) to find the relevant $a$ and $b$ - as you have done.

Mark Bennet
  • 101,769
0

Use that $\mathbb Q(\zeta) = \mathbb Q(\sqrt{-3})$, since $(\zeta^2 - \zeta)^2 = -3$, and $\operatorname{Gal}(\mathbb Q(\zeta)|\mathbb Q) = (\mathbb Z/3)^{\times} = \mathbb Z/2$.

Thomas Poguntke
  • 839
0

Here is another approach.

Calculate the norm of $\alpha=a+b\zeta+c\zeta^2$ as $$N(\alpha)=a^2+b^2+c^2-ab-bc-ca=\frac12 ((a-b)^2+(b-c)^2+(c-a)^2).$$ Set $N(\alpha)=1$ (note that $-1$ is not possible). Then the three terms $a-b,b-c,c-a$ must be a permutation of $-1,0,1$. For example let $a-b,b-c,c-a=0,1,-1$. Then $a,b,c=k,k,k-1$ for any $k \in \mathbb{Z}$. Since $1+\zeta+\zeta^2 = 0$, we can pick $c=-1$ as a representative, so $\alpha=-\zeta^2$ is one of the six units.

vuur
  • 1,264
  • But the field of cube roots of unity is not cubic but rather quadratic over $\Bbb Q$. Your basis should have been ${1,\zeta}$. – Lubin Mar 12 '15 at 02:42
  • I don't claim that ${1,\zeta,\zeta^2}$ is a basis. In fact the relation $1+\zeta+\zeta^2 = 0$ is imposed on all $\alpha$, which implies that ${1,\zeta}$ is a basis. The relation is used during the calculation of the norm to pick a representative in $\mathbb{Z}$ (note that the norm is invariant) and later to pick a representative unit. – vuur Mar 12 '15 at 10:14