Let $R$ be a commutative unital ring such that $R$ has no non-trivial idempotents, and let $(R^\times,\ \cdot)$ denote the group of units of $R$. Is it true that $(R,+) \ncong (R^\times,\ \cdot)$ ?

See Let $R$ be a commutative unital ring. Is it true that the group of units of $R$ is not isomorphic with the additive group of $R$?. The counter-examples (as far as I can see) do not cover this. Please help. Thanks in advance.

  • $\begingroup$ In this answer it's stated that every element $v$ satisfies $v^2=0,1$, so that ring has no non-trivial idempotents. $\endgroup$ – user26857 Aug 20 '16 at 21:48

Before I present my counterexample, let me point out that this question shows that $R$ cannot be taken to be a field. As you pointed out in the question, the solution is also known in the case when non-trivial idempotents are allowed. My answer to your question is a bit more involved.

My construction depends on a field $K$ and a group $G$. I denote $$\mathcal{H}(K,G)=\left\{\varphi+f\left|\right.\varphi\in KG,\,\,f\in t\,K[t]\right\},$$ that is, $\mathcal{H}(K,G)$ is the set of all formal sums $\varphi+f$ in which $\varphi$ is an element of the group ring $KG$ and $f$ is a polynomial with zero constant term. I define $$(\varphi_1+f_1)(\varphi_2+f_2)=\varphi_1\varphi_2+\sigma(\varphi_1) f_2+\sigma(\varphi_2) f_1+ f_1 f_2,$$ where $\sigma:KG\to K$ is the mapping sending an element of a group ring into the sum of its coefficients.

Let me omit a formal proof that $\mathcal{H}(K,G)$ is indeed a ring. It is also clear that the invertible and idempotent elements in $\mathcal{H}(K,G)$ are precisely those that live in $G$.

To construct the counterexample, I set $K=\mathbb{F}_2$ and $G=\mathbb{Z}_2^\infty$. We get $\varphi^2=\sigma(\varphi)$ for all $\varphi\in KG$, so that there are no non-trivial idempotents, and every invertible element squares to $1$. Therefore, the multiplicative subgroup of $\mathcal{H}(K,G)$ is isomorphic to the direct sum of several copies of $\mathbb{Z}_2$. The same holds for the additive group of $\mathcal{H}(K,G)$ because the underlying field has characteristic two. It remains to note that the cardinalities of the multiplicative and additive groups of $\mathcal{H}(K,G)$ are equal.

  • $\begingroup$ what do you mean by the sum of co-efficients of an element of the group ring ? I thought that the elements were functions .... $\endgroup$ – user228169 Aug 2 '16 at 14:42
  • $\begingroup$ The group ring of $G$ over a field $K$ is the set of all formal sums $s=\sum_{g\in G} k_g g$ with $k_g\in K$ such that only finitely many $k_g$'s are non-zero. (In your terminology, $s$ is a function $g\to k_g$.) In my answer, I define $\sigma(s)=\sum_{g\in G} k_g$, that is, $\sigma$ sends $s$ to the sum of all coefficients $k_g$. $\endgroup$ – heptagon Aug 2 '16 at 18:56
  • $\begingroup$ Nice example, But doesn't $KG$ work just as well as $\mathcal{H}(K,G)$? $\endgroup$ – Jeremy Rickard Aug 4 '16 at 10:36
  • $\begingroup$ Thanks, @JeremyRickard. Of course, this does work. Unfortunately, the time I spent thinking on the question was not enough. Actually, I did not manage to resist reproducing an example, constructed for a different purpose, from my old thesis. I think you can add a separate answer with the group ring, which seems to be a better answer than that of mine. $\endgroup$ – heptagon Aug 9 '16 at 20:00

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