Characterize the commutative rings with trivial group of units

Question

This question suggested me the following:

Characterize the commutative unitary rings $R$ with trivial group of units, that is, $R^{\times}=\{1\}$.

The local case was solved here long time ago and it's very simple. In general, such a ring must have characteristic $2$ and the Jacobson radical $J(R)=(0)$. At present I know two classes of examples: direct products of $\mathbb Z/2\mathbb Z$ and polynomial rings over $\mathbb Z/2\mathbb Z$.

(One can also ask for such a characterization in the non-commutative case, but I'm primarily interested in the commutative setting.)

Answer 1

This is only a partial answer, to record some thoughts:

If $R$ is semilocal, i.e. has only finitely many maximal ideals (in particular, if $R$ is finite), then one can characterize these rings as precisely the finite products of copies of $\mathbb{F}_2$. If $R$ is semilocal, then any surjection $R \twoheadrightarrow R/I$ induces a surjection on unit groups $R^\times \twoheadrightarrow (R/I)^\times$ (see here for proof). In particular, for any maximal ideal $m$ of $R$, $R/m$ is a field with only one unit, hence must be $\mathbb{F}_2$. Since the Jacobson radical of $R$ is $0$, there is an isomorphism $R \cong \prod_{m \in \text{mSpec}(R)} R/m = \prod \mathbb{F}_2$ by Chinese Remainder, and conversely any finite product of $\mathbb{F}_2$'s does indeed have only one unit.

If there are infinitely many maximal ideals, then it is not clear if every residue field at a maximal ideal is $\mathbb{F}_2$. If this is the case though, then although Chinese remainder fails, one can still realize $R$ as a subring of a product of copies of $\mathbb{F}_2$, so we get a characterization in this case. Thus:

If $R$ is a subring of a direct product of copies of $\mathbb{F}_2$, then $R$ has trivial unit group. The converse holds if every maximal ideal of $R$ has index $2$; in particular it holds if $R$ is semilocal.

Update: There are more examples of such rings than products of $\mathbb{F}_2$ or polynomial rings over $\mathbb{F}_2$ though. If $S = \mathbb{F}_2[x_1, \ldots]$ is a polynomial ring over $\mathbb{F}_2$ (in any number of variables) then for any homogeneous prime ideal $P \subseteq S$ (necessarily contained in the irrelevant ideal $(x_1, \ldots)$), the ring $S/P$ has trivial unit group. Since the property of having trivial unit group passes to products and subrings, the same holds if $P$ is only assumed to be radical (and still homogeneous).

Conversely, any ring $R$ with trivial unit group is a reduced $\mathbb{F}_2$-algebra, hence has a presentation $R \cong \mathbb{F}_2[x_1, \ldots]/I$, where $I$ is radical. We can even realize it as a demohogenization $R \cong (\mathbb{F}_2[t, x_1, \ldots]/J)/(t-1)$, where $J$ is a homogeneous radical ideal. Thus if any dehomogenization (at a variable) of a ring of the form $S/I$, where $I$ is a homogeneous radical ideal, has trivial unit group, this would yield a characterization. This in turn is equivalent to asking whether or not the multiplicative set $1 + (t - 1)$ is saturated in $S/I$ (at this point, I must leave this line of reasoning as is, but would welcome any feedback).

Update 2: Upon reflection, it's easy to see that not every dehomogenization of a graded reduced $\mathbb{F}_2$-algebra will have trivial unit group, i.e. for $\mathbb{F}_2[t,x,y]/(xy - t^2)$, setting $t = 1$ gives $\mathbb{F}_2[x,y]/(xy - 1)$ which has nontrivial units. I'll have to think a little more about the right strengthening of the condition on $I$.

Answer 2

I don't know if this actually contributes more than what has been already said, but in Rings of zero-divisors (1958, Theorem 3), Cohn shows that if $R$ is a commutative ring without nontrivial units, then every element $x\in R$ is either idempotent or transcendental (over $\mathbb{F}_2$). Moreover $R$ is a subdirect product of extension fields of $\mathbb{F}_2$.

These assertions are easy to prove: the subdirect product part is due to $J(R)=0$. If $x\in R$ is algebraic, then $\mathbb{F}_2[x]$ is finite-dimensional over $\mathbb{F}_2$, hence a product of extension fields without nontrivial units, hence a product of copies of $\mathbb{F}_2$, hence $x$ is idempotent.

In fact, the same argument is done by Cohn for any $F$-algebra $R$, $F$ a field, $R$ without units outside of $F$.

EDITED: For the noncommutative case, what we can say with the same arguments is very similar: If $R$ is a ring without nontrivial units, then $R$ is an $\mathbb{F}_2$-algebra in which every element is either idempotent of transcendental. Moreover, $R$ is a subdirect product of domains.

The statement about the elements has the same proof as in the commutative case (since $\mathbb{F}_2(x)$ is commutative). Let us prove the last claim: Since $x^2=0$ implies $(1+x^2)=(1+x)^2=1$, we have $1+x=1$, so $x=0$; hence $R$ is reduced (in particular semiprime). In a semiprime ring, the intersection of prime ideals is $0$; since every prime ideal contains a minimal prime ideal, the intersection of minimal prime ideals is also $0$. Now, in a reduced ring, every minimal prime ideal is completely prime, so that its quotient ring is a domain. Therefore $R$ is a subdirect product of the domains $R/P_i$, where $\{P_i\}_{i\in I}$ is the family of completely prime ideals of $R$.

Answer 3

If the ring is finite, then the ring must be boolean (and hence commutative). I have handled this problem in an article, which you can find on the web: Rodney Coleman: Some properties of finite rings.