Rings with a finite set of units

Question

Consider a ring $\mathcal{R}$ with a finite set $\mathcal{R} ^\times$ of units, i.e. divisors of $1$, for example

• $\mathbb{Z}^\times = \{\pm 1\}$

• $\mathbb{Z}[i]^\times = \{\pm 1,\pm i\}$ (Gaussian integers)

• $\mathbb{Z}[i][j][k]^\times = \{\pm 1,\pm i,\pm j,\pm k\}$ (Hamiltonian integers)

Question 1: What's the official name of this property of a ring: having a finite set of units?

For the examples above we have:

• $\mathbb{Z}^\times$ coincides with the square roots of $1$.

• $\mathbb{Z}[i]^\times$ coincides with the $4$th roots of $1$.

• $\mathbb{Z}[i][j][k]^\times$ coincides with the $8$th roots of $1$.

Question 2: Does a finite $\mathcal{R} ^\times$ [added:] for an infinite ring $\mathcal{R}$ always coincide with some $2^k$-th roots of $1$? Of which characteristic property of $\mathcal{R}$ does $k$ depend?

Question 3: If not so [which seems to be the case, see the answers below]: How can the rings for which $\mathcal{R} ^\times$ coincides with some $2^k$-th roots of $1$ be characterized?

Answer 1

Question 1: "A ring whose group of units is finite". I don't think it has a special name.

Question 2: If the group of units of $R$ is finite then by taking the LCM of the (multiplicative) orders of the units, you get an $n$ such that $u^n = 1$ for all units $u$ of $R$. However this $n$ is not necessarily a power of $2$ - consider e.g. $\Bbb Z/7\Bbb Z$ where $n = 6$. Or $(\Bbb Z/7\Bbb Z)[x]$ (which has the same group of units) if you want an infinite example.

An alternative counterexample: $\Bbb Z[x]/(x^3 - 1) = \Bbb Z[\omega]$ where $\omega$ is a cube root of unity.

Answer 2

Answer to question 2:

Every finite field is a ring with 1 and a finite set of unities. These fields have $p^k-1$ elements $\ne0$, where $p$ is a prime number. All these elements are divisors of $1$. So the answer to your question is no.

The ring of polynomials over such a field is still a ring with 1 and its units are again the elements $\ne0$ of the field.