Compact riegs with cancellative addition are trivial

Theorem

Let $R$ be a compact Hausdorff topological rieg. If its additive commutative monoid is cancellative, then $R$ has one element. Connectedness is not required.

Lemma

Every compact Hausdorff commutative cancellative topological monoid $(S,+,0)$ is a group.

Proof. First, every nonempty compact Hausdorff commutative topological semigroup has an idempotent. By compactness and Zorn's lemma, choose a minimal nonempty closed subsemigroup $K$. For $a\in K$, the set $a+K$ is a nonempty closed subsemigroup (using commutativity), hence equals $K$. Thus

is a nonempty closed subsemigroup and equals $K$. In particular, $a+a=a$.

For $s\in S$, the closure of $\{ns:n\geq1\}$ is a compact commutative subsemigroup and contains an idempotent $e$. Cancellativity applied to $e+e=e+0$ gives $e=0$. Choose a net $n_i s\to0$. A subnet of $(n_i-1)s$ converges to some $t$, and continuity gives $s+t=0$. Every element therefore has an additive inverse. $\square$

Proof of the theorem

The lemma gives an additive inverse $-1$ of $1$, and a rieg containing an additive inverse of $1$ is degenerate (Notes on Rieg Theory, Claim 2.3.1: $1=0^0=0^{1+(-1)}=0^1\,0^{-1}=0$). $\square$

What remains

This settles the cancellative case of the connected compact Hausdorff problem. The noncancellative case has a positive answer: the unit interval with $\max$, $\min$ and $x^y=\max(x,1-y)$ is a nontrivial connected compact Hausdorff rieg (Connected compact Hausdorff riegs).