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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.

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2 Answers 2

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

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  • $\begingroup$ Nice! Thanks alot! $\endgroup$ Aug 30, 2020 at 11:05
  • $\begingroup$ You're welcome! $\endgroup$ Aug 30, 2020 at 11:12
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    $\begingroup$ 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. $\endgroup$ Aug 30, 2020 at 11:27
  • $\begingroup$ Just out of interest how did you think of these examples? Any intuition you might be able to give? Many thanks! $\endgroup$ Aug 30, 2020 at 11:51
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    $\begingroup$ 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. $\endgroup$ Aug 30, 2020 at 12:24
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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.

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