For commutative rings A in which 2 is invertible, as well as for $A=\mathbb {Z}$ , [Reference Hornbostel and Park3, Proposition 2.3.5] is true. However, for some rings A in which 2 is not invertible, the result is not correct as stated. The reason is the description of the ideal T in the proof of loc. cit. is not correct in this case, and this might imply that the map $\alpha $ in loc. cit. is not an isomorphism. In more detail, let A be a commutative ring. Due to [Reference Dotto, Moi, Patchkoria and Reeh1, Corollary 5.2], $\underline {\pi }_0\mathrm {THR}(A)$ is isomorphic to the Mackey functor
where $\mathrm {res}(x\otimes y)=xy$ for $x,y\in A$ , $\mathrm {tran}(a)=2a\otimes 1$ for $a\in A$ , and $T_A$ is the subgroup generated by $x\otimes a^2 y-a^2 x\otimes y$ and $x\otimes 2ay-2ax\otimes y$ for $a,x,y\in A$ . Let $2A$ denote the ideal $(2)$ in A. We have the monomorphism $2A\to (A\otimes A)/T_A$ given by $2a\mapsto 2a\otimes 1$ for $2a\in 2A$ . Its cokernel is isomorphic to $(A/2\otimes A/2)/T_{A/2}$ , where $A/2:=A/2A$ . Observe that $T_{A/2}$ is the subgroup generated by $x\otimes a^2 y-a^2 x\otimes y$ for $a,x,y\in A/2$ . Hence, we have the short exact sequence
where $\varphi \colon A/2\to A/2$ denotes the Frobenius (i.e., the squaring map). In particular, [Reference Hornbostel and Park3, Proposition 2.3.5] holds if and only if $\varphi $ is surjective.
The only statement in [Reference Hornbostel and Park3] where this proposition is used is the following one in the proof of [Reference Hornbostel and Park3, Proposition 3.2.2], for which we now provide an alternative proof.
Proposition 1. Let $A\to B$ be an étale homomorphism of commutative rings. Then the induced morphism of Mackey functors
is an isomorphism.
Proof. By [Reference Hornbostel and Park2, Lemma 5.1], the Mackey functor $\underline {\pi }_0\mathrm {THR}(\iota A)\square _{\iota A}\iota B$ is isomorphic to
Using the above computations, we have the following commutative diagram where the vertical maps are induced by multiplication:
We only need to show that $\beta $ is an isomorphism. Since B is flat over A, the rows are short exact sequences, and $\alpha $ is an isomorphism. The induced square of commutative rings
is coCartesian by [4, Tag 0EBS] since $A/2\to B/2$ is étale. It follows that $\gamma $ is an isomorphism. Hence, $\beta $ is an isomorphism by the five lemma.