Units in Eisenstein Integers

Detailed Solution

For an element $u=a+b\omega\in\mathbb Z[\omega]$, the norm is \[ N(u)=a^2-ab+b^2. \]An element is a unit if and only if its norm is $1$. Therefore, we need to determine all integer pairs $(a,b)$ satisfying \[ a^2-ab+b^2=1. \]Complete the square: \[ a^2-ab+b^2 =\left(a-\frac b2\right)^2+\frac{3b^2}{4}. \]Since this expression is equal to $1$, we must have \[ \frac{3b^2}{4}\leq 1, \]which gives \[ b^2\leq\frac43. \]Because $b\in\mathbb Z$, we obtain \[ b\in\{-1,0,1\}. \]If $b=0$, then \[ a^2=1, \]so $a=\pm1$. This gives the two units \[ 1,\quad -1. \]If $b=1$, then \[ a^2-a+1=1, \]so \[ a(a-1)=0. \]Hence $a=0$ or $a=1$, giving \[ \omega,\quad 1+\omega. \]Since $\omega$ is a primitive cube root of unity, \[ 1+\omega=-\omega^2. \]Thus these two elements are $\omega$ and $-\omega^2$. If $b=-1$, then \[ a^2+a+1=1, \]so \[ a(a+1)=0. \]Hence $a=0$ or $a=-1$, giving \[ -\omega,\quad -1-\omega. \]Using $1+\omega=-\omega^2$, we have \[ -1-\omega=\omega^2. \]Thus these two elements are $-\omega$ and $\omega^2$. Therefore, the complete set of elements having norm $1$ is \[ \{1,-1,\omega,-\omega,\omega^2,-\omega^2\} = \{\pm1,\pm\omega,\pm\omega^2\}. \]Hence $\mathbb Z[\omega]$ has exactly six units. Option A is incorrect because ${1,-1,i,-i}$ are the units of the Gaussian integers $\mathbb Z[i]$, not the Eisenstein integers. Option C is incorrect because a nonzero Eisenstein integer need not have norm $1$. For example, \[ N(2)=4\neq1, \]so $2$ is not a unit. Option D is incorrect because irrationality of $\omega$ does not prevent it from being a unit. In fact, \[ N(\omega)=0^2-0(1)+1^2=1, \]so $\omega$ is a unit. Thus the correct option is $\boxed{\mathrm{B}}$.