I would never dare posting this here, but the question https://math.stackexchange.com/q/3019853/214353 on math.SE did not receive any feedback (except for 13 views and one upvote) since November 30, so...

I believe I've seen an answer to this somewhere (some Deligne notes?) but cannot recover it.

In the $\mathbb Z/2$-graded formulation, supercommutativity means that two odd degree elements anticommute while all other kinds of homogeneous elements commute.

If 2 is invertible, to name a $\mathbb Z/2$-grading is the same as to name an involution: for $\tau:A\to A$ with $\tau^2$ equal to identity, even elements are those with $\tau e=e$ and the odd ones those with $\tau o=-o$. Then every $a$ is uniquely the sum of the even element $(a+\tau a)/2$ and the odd element $(a-\tau a)/2$.

Now suppose 2 is not invertible, and we still want to formulate supercommutativity in terms of $\tau$ alone, in such a way that it becomes precisely supercommutativity as soon as 2 becomes invertible. How to do it?

My attempts produced ugly and inconclusive things like $$ 2yx=xy+x\tau(y)+\tau(x)y-\tau(x)\tau(y) $$ or $$ yx-\tau(y)x-y\tau(x)+\tau(y)\tau(x)=-xy+x\tau(y)+\tau(x)y-\tau(x)\tau(y). $$

There must be something much better, what is it?

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    $\begingroup$ Why do you expect this to be possible? What if the characteristic is $2$? Note that speaking just in terms of gradings, over any (connected?) base a $\mathbb{Z}/2$-grading is the same as an action of the group scheme $\mu_2$. In characteristic $2$ this group scheme is non-reduced and in particular badly fails to be isomorphic to the constant group scheme $\mathbb{Z}/2$. $\endgroup$ – Qiaochu Yuan Jan 4 at 8:16
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    $\begingroup$ @QiaochuYuan Well I don't even see any rigorous impossibility statement that could be formulated. $\endgroup$ – მამუკა ჯიბლაძე Jan 4 at 9:55
  • $\begingroup$ Could you explain more about $\mu_2$-actions? Do not they just give $\mathbb Z/2$-graded plainly commutative algebras? Or you mean $\mu_2$-equivariant vector bundles over $\mu_2$-schemes? $\endgroup$ – მამუკა ჯიბლაძე Jan 4 at 10:05
  • $\begingroup$ Btw what are tensor powers of $\mu_2$? $\endgroup$ – მამუკა ჯიბლაძე Jan 4 at 10:06
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    $\begingroup$ @HarryGindi Actually I now also remember Deligne working with $\mu_2$ there... $\endgroup$ – მამუკა ჯიბლაძე Jan 4 at 10:13

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