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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.

asdhfb askldfn
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2 Answers2

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

Sebastian Monnet
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  • Nice! Thanks alot! – asdhfb askldfn Aug 30 '20 at 11:05
  • You're welcome! – Sebastian Monnet Aug 30 '20 at 11:12
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    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. – Douglas Molin Aug 30 '20 at 11:27
  • Just out of interest how did you think of these examples? Any intuition you might be able to give? Many thanks! – asdhfb askldfn Aug 30 '20 at 11:51
  • My original answer was complete wrong! @DouglasMolin's criticism made me look more closely and find that the whole class of counterexamples $\mathbb{Z}_n[X]/(X - a)$ is broken. I have added a completely different argument, proving the claim in the finite case, which I now believe to be true. – Sebastian Monnet Aug 30 '20 at 12:19
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    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. – Sebastian Monnet Aug 30 '20 at 12:24
  • I don't see where your proof uses the assumption that $R$ is finite. Also, I think $\mathbb{Z}[X]/(2X-1)$ is just the ring of dyadic rationals, which does not have a cyclic additive group. – halrankard2 Aug 30 '20 at 12:58
  • @halrankard2 you're right, of course, that was pretty silly of me. So do you think my argument proves the claim for all rings? I've amended the answer under that assumption, but I'd appreciate some external confirmation, because this seems like an easy problem to make unfounded assumptions on. – Sebastian Monnet Aug 30 '20 at 13:01
  • The proof looks right to me. One comment is that a ring with a cyclic additive group has to be commutative, so the statement about noncommutative rings is vacuous. I've not thought about this concept before, but it appears that there is some literature on "cyclic rings". For example, this answer (https://math.stackexchange.com/questions/1914009/classify-all-rings-with-cyclic-additive-group) suggests that a infinite cyclic ring with unity is isomorphic to $\mathbb{Z}$. – halrankard2 Aug 30 '20 at 13:15
  • @halrankard2 also a good point! I've edited the answer again to note that $R$ is commutative. – Sebastian Monnet Aug 30 '20 at 13:22
  • I will take my time to read carefully through all of this. This is very instructive for me. Many thanks! – asdhfb askldfn Aug 30 '20 at 19:21
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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.

Shaun
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rschwieb
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