Example of integral domain with "non-trivial" units

Question

The units of $\mathbb{Z}$ are 1 and -1, this is quite "easy" to see.

Are there any rings $R$ which has "non-trivial" units, in the sense that it takes some work to figure out what are the units of $R$? (I am mainly interested in the case of integral domains.)

Any example will be appreciated.

Thanks a lot.

Answer 1

Here are two:

• In particular, every element of the form $(3+2 \sqrt{2})^n,n \in \Bbb{N}$ is a unit in $\Bbb{Z}[\sqrt{2}]$

  since $$(3+2 \sqrt{2})^n\cdot (3-2 \sqrt{2})^n=\Big[(3+2 \sqrt{2})(3-2 \sqrt{2})\Big]^n=1$$

• $U(\Bbb{Z}[i])=\{\pm 1, \pm i\}$

  everyone for the first time getting this by using $N(a+bi)= a^2+b^2 $

Answer 2

And please don't forget one of my favorites, the Eisenstein integers,

$\Bbb Z[\omega] = \{a + b\omega; \; a, b \in \Bbb Z\} \subset \Bbb C, \; \omega = e^{2\pi i / 3}; \tag 1$

we have

$\omega^2 + \omega + 1 = 0, \tag 2$

whence

$\bar \omega = \omega^2 = -1 - \omega \in \Bbb Z[\omega], \tag 3$

which implies $\Bbb Z[\omega]$ is multiplicatively closed, since it yields

$(a + b\omega)(c + d\omega) = ac + ad\omega + bc\omega + bd\omega^2 = ac + (ad + bc)\omega + bd(-1 - \omega)$ $ = ac - bd + (ad + bc - bd)\omega \in \Bbb Z[\omega]; \tag 4$

(1) implies $\Bbb Z[\omega]$ is an integral domain; $\Bbb Z[\omega]$ has $6$ units, namely

$1 + \omega, (1 + \omega)^2 = \omega, (1 + \omega)^3 = (1 + \omega)^2(1 + \omega) = -1, (1 + \omega)^4 = \omega^2, (1 + \omega)^5 = -\omega, (1 + \omega)^6 = 1, \tag 5$

which together form a group isomorphic to the cyclic group with $6$ elements $\Bbb Z_6$.

Note Added in Edit, Thursday 18 October 2018 8:45 AM PST: What I have actually shown above, cf. (5), is that the group of units $U(\Bbb Z[\omega])$ of $\Bbb Z[\omega]$ contains a subgroup isomorphic to $\Bbb Z_6$; to complete the assertion and show that in fact $U(\Bbb Z[\omega]) \simeq \Bbb Z_6$, we need to prove that the list (4) presents all of $U(\Bbb Z[\omega])$, that is, all of the units in $\Bbb Z[\omega]$. I argue this case as follows: if $0 \ne u \in \Bbb C^\times$, then there exists $0 \ne v \in \Bbb C^\times$ with

$uv = 1; \tag 6$

then

$\bar u \bar v = 1 \tag 7$

as well, and multiplying these two equations yields

$(u \bar u)( v \bar v ) = (uv)( \bar u \bar v) = 1; \tag 8$

now if indeed $0 \ne u, v \in \Bbb Z[\omega]$, we have

$u \bar u, v \bar v \in \Bbb Z_+ = \{ z \in \Bbb Z \mid z \ge 1 \}, \tag 9$

which fact may be seen by setting

$u = a + b\omega, \; a, b \in \Bbb Z; \tag{10}$

then

$u \bar u = (a + b \omega)(a + b \bar \omega) = a^2 + ab \omega + ab \bar \omega + b^2 \omega \bar \omega = a^2 + ab (\omega + \bar \omega) + b^2 \omega \bar \omega; \tag{11}$

since

$\omega \bar \omega = e^{2 \pi i / 3} e^{-2 \pi i / 3} = 1 \tag{12}$

and from (3)

$\omega + \bar \omega = \omega + (-1 - \omega) = -1, \tag{13}$

we obtain

$u \bar u = a^2 - ab + b^2; \tag{14}$

we note that

$u \ne 0 \Longrightarrow a \ne 0 \vee b \ne 0, \tag{15}$

if precisely one of $a, b \ne 0$, then clearly

$a^2 - ab + b^2 \ge 1; \tag{16}$

if both $a, b \ne 0$, then

$\text{sign}(a) \ne \text{sign}(b) \Longrightarrow -ab > 0 \Longrightarrow a^2 - ab + b^2 \ge 1, \tag{17}$

and

$\text{sign}(a) = \text{sign}(b) \Longrightarrow ab > 0$ $\Longrightarrow a^2 - ab + b^2 = a^2 - 2ab + b^2 + ab = (a - b)^2 + ab \ge 1; \tag{18}$

at every branch we find

$u \bar u = a^2 - ab + b^2 \ge 1, \tag{19}$

and likewise

$v \bar v \ge 1; \tag{20}$

it then follows that (8) forces

$u\bar u = v \bar v = 1; \tag{21}$

therefore, to find the elements $u \in U(\Bbb Z[\omega])$ we must discover all pairs $a, b \in \Bbb Z$ with

$a^2 - ab + b^2 = 1; \tag{22}$

a certain symmetry in the solution set to (22) facilitates our search; that is, the pair $(a, b)$ solves (22) if and only if the pair $(b, a)$ does and if and only if the pair $(-a, -b)$ does; thus since $(1, 0)$ satisfies (22) so do $(0, 1)$, $(-1, 0)$ and $(0, -1)$; since $(1, 1)$ is a solution, so is $(-1, -1)$; finally,

$\vert a \vert \ge 2, \vert b \vert \ge 2 \Longrightarrow a^2 - ab + b^2 \ne 1; \tag{23}$

for if say $\vert a \vert \ge 2$ we can by our symmetries assume $a \ge 2$; furthermore we may assume $a \ge b$ since with $b > a \ge 2$ we have

$a^2 - ab + b^2 = a^2 + b(b - a) \ge 4 > 1; \tag{24}$

we rule out the case $b = 0$ since that also leads to (24); if $b = a \ge 2$ then yet again

$a^2 - ab + b^2 = a^2 - b^2 + b^2 = a^2 \ge 4 > 1; \tag{25}$

we are left with

$a \ge 2, a > b \ne 0 \Longrightarrow a^2 - ab + b^2 = a(a - b) + b^2 \ge 2 + b^2 > 2 > 1; \tag{26}$

thus (23) binds and so we must have $\vert a \vert, \vert b \vert \le 1$; switching the roles of $a$ and $b$ completes the argument; there are no pairs $(a, b)$ obeying (22) with either $\vert a \vert \ge 2$ or $\vert b \vert \ge 2$; and we have seen above that every pair with $\vert a \vert, \vert b \vert \le 1$ other than $(0, 0)$ is a solution. In accord with the scheme given above, these $(a, b)$ account for every element of $U(\Bbb Z[\omega])$. End of Note.