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

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.

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 $$(R, +, 0)$$.

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 $$R = \mathbb{Z}[X]/(2X - 1)$$.

• Nice! Thanks alot! Aug 30, 2020 at 11:05
• You're welcome! Aug 30, 2020 at 11:12
• I don't think the first one is correct. In this ring, writing $x$ for the class of $X$, we have $1+1=2x+2x=4x=0x=0$, so $1=-1$. Hence, $a=-a$ for any $a$, and in particular $1=2x=x+x=x-x=0$. Thus $R=\{0\}$. This happens every time you invert a nilpotent element. Aug 30, 2020 at 11:27
• Just out of interest how did you think of these examples? Any intuition you might be able to give? Many thanks! Aug 30, 2020 at 11:51
• As for thinking of the counterexamples (the intuition is still useful in the infinite case, so I'll answer your question even though I was wrong), we're looking for an element $\alpha$ such that $m\alpha =1$ for some $m$ (i.e. adding $\alpha$ to itself repeatedly gives $1$), but that is not an integer. Seems pretty clear then that we are looking for an element $1/m$ for some integer $m$. Quotienting by the ideal $(mX - 1)$ is just a way of formalising the process of adjoining a reciprocal. However, as shown by my blunder in the finite case, this is risky when you try to invert zero-divisors. Aug 30, 2020 at 12:24

The answer's "yes" when the order of $$1$$ is finite. If $$1$$ has finite order strictly less than $$|R|$$, say $$n$$, then $$n\cdot 1=0$$ would imply that $$n\cdot g=0$$ for every element, but there is supposed to be an element of additive order strictly greater than (potentially even infinite) $$1$$'s order.

The answer is also yes when the order of $$1$$ is infinite (and hence $$R$$ is isomorphic to $$\mathbb Z$$), but the proof is different.

Let $$g$$ additively generate $$R$$. Then there exists some natural number $$n$$ such that $$ng=1$$, and another natural number $$m$$ such that $$mg=g^2$$.

Combining these two, $$nmg=g$$. But since $$R$$ is a free abelian group on $$g$$, this would mean $$mn=1$$, and the only possibilities are $$n=m=1$$ and $$n=m=-1$$, both of which imply $$1$$ is a generator.