Rings with a finite set of units

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?

• Never heard of a name for a ring with finitely many units. Incidentally, a ring with finitely many units must have finitely many nilpotents as well... – rschwieb Oct 9 '18 at 15:53
• Incidentally? Not necessarily? ;-) – Hans-Peter Stricker Oct 9 '18 at 15:54
• Apropos of nothing, that is. I suppose that also means that a commutative Artinian ring with finitely many units is finite. – rschwieb Oct 9 '18 at 16:21
• Doesn't your third example still consist of fourth roots of 1? – pregunton Oct 9 '18 at 16:59
• There are eight elements in the unit group, but $\pm i, \pm j, \pm k$ all have order 4. The number of elements in the unit group and the maximum order of these elements need not be the same, though they are related by Lagrange's theorem. – pregunton Oct 9 '18 at 17:22

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.

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.
• So please allow me to add the requirement that $\mathcal{R}$ is infinite. – Hans-Peter Stricker Oct 9 '18 at 9:23