TITLE: Characterize the commutative rings with trivial group of units
QUESTION [9 upvotes]: 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.)

REPLY [3 votes]: 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 $R$.