The perfection of a ring $A$ of prime characteristic $p$ is the perfect ring $A\_\rm{pf}=$ lim{$A\to A\to ...$} where all maps are Frobenius, so being a filtered colimit in Rings, the perfection functor commutes with **finite** products. But, in general, does perfection commutes with products? If not, why in a scheme $X$, the presheaf associated to $U\to \Gamma (U,\mathcal{O\_\rm{X}})\_\rm{pf}$ is a sheaf? This is stated in section 6 in [Greenberg: Perfect closures of rings and schemes](https://www.ams.org/journals/proc/1965-016-02/S0002-9939-1965-0190169-8/) but I do not understand the proof if perfection does not commute with products.


I asked this question in [math.stackexchange](https://math.stackexchange.com/questions/3575859/does-perfection-of-rings-commute-with-products) a few days ago, but there was no answer.



Does perfection of rings commute with products?
No, even restricting to domains: it doesn't even commute to infinite powers.


Indeed, let $A$ be a non-perfect domain. So $A\_{\mathrm{pf}}$ is an overring of $A$, such that for every $x\in A\_{\mathrm{pf}}$ there exists some minimal $n=N(x)\ge 0$ such that $x^{p^n}\in A$. Then $x\mapsto N(x)$ is surjective onto non-negative integers. Choose $x\_n\in A\_{\mathrm{pf}}$ with $N(x\_n)=n$. Then the sequence $(x\_n)$, viewed as element of $(A\_{\mathrm{pf}})^\mathbf{N}$ has no $p$-power power in $A^\mathbf{N}$. Thus, in this case we see that $(A^\mathbf{N})\_{\mathrm{pf}}$ is a proper subring of $(A\_{\mathrm{pf}})^\mathbf{N}$. 


Note: $A\_{\mathrm{pf}}$ is the initial object in the category of unital commutative rings (of characteristic $p$), endowed with a homomorphism from $A$, and in which $x\mapsto x^p$ is bijective. In particular there's a canonical homomorphism $\big(\prod\_i A\_i\big)\_{\mathrm{pf}}\to \prod\_i ((A\_i)\_{\mathrm{pf}})$. In the above example, it fails to be surjective. ~~I haven't checked carefully but it seems to always be injective.~~




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*Edit:* it's not always injective, even for infinite powers. Denote $\phi=\phi\_A:A\to A$, $x\mapsto x^p$ (for $A$ commutative ring of characteristic $p$). First observe that the kernel of $A\to A\_{\mathrm{pf}}$ is $\bigcup\_n\mathrm{Ker}(\phi\_A^n)$ (indeed, one directly check that for $B=A/\bigcup\_n\mathrm{Ker}(\phi\_A^n)$, we have $\phi\_B$ injective).


Let $A$ be such that $\phi:x\mapsto x^p$ has nontrivial kernel. Hence for every $n\ge 0$ there exists $x\_n$ in $A$ such that $x\_n\in\mathrm{Ker}(\phi^{n+1})\smallsetminus\mathrm{Ker}(\phi^{n})$. Then $(x\_n)\_{n\ge 0}$ is an element of $A^\mathbf{N}$ inducing a nonzero element in the kernel of $(A^\mathbf{N})\_{\mathrm{pf}}\to (A\_{\mathrm{pf}})^\mathbf{N}$.


Note however that if each $A\_i$ is reduced (i.e., has injective $\phi\_{A\_i}$), then $\big(\prod\_i A\_i\big)\_{\mathrm{pf}}\to \prod\_i ((A\_i)\_{\mathrm{pf}})$ is injective.


