Does the unit generate the additive group in a unital ring with cyclic additive group

Question

Let $R$ be a unital ring with cyclic additve group $(R, +,0)$. Is it the case that $1$ generates the additive group $(R,+,0)$?

Thoughts:

Maybe classifying the unital rings with cyclic subgroups is possible.

Answer 1

EDIT My original answer was wrong. I've kept it below for completeness, but am writing a new, (hopefully) correct answer at the top.

The statement is true for all rings.

Let $R$ be a unital ring with cyclic additive group generated by $\alpha$. Then $R$ is commutative since $\alpha$ commutes with itself. Then $\alpha^2 = m\alpha$ for some integer $m$, which means that $(m - \alpha)\alpha = 0$. Now, $1 = k\alpha$ for some integer $k$, so $$ 0 = k\cdot 0 = k(m-\alpha)\alpha = (m-\alpha)k\alpha = (m - \alpha)\cdot 1= m - \alpha, $$ so $\alpha = m$, which means that $\alpha$ lies in the additive span of $1$, hence $1$ generates $(R, +, 0)$.

Original wrong answer below

The statement is false for finite and infinite rings.

For the finite case, take $R = \mathbb{Z}_6[X]/(2X - 1)$ and let $\alpha = X + (2X - 1) \in R$. The additive group of $R$ is generated by $\alpha$, but not by $1$.

For the infinite case, do the same thing with $R = \mathbb{Z}[X]/(2X - 1)$.