Does perfection of rings commute with products?

Question

The perfection of a ring $A$ of prime characteristic $p$ is the perfect ring $A_{pf}=$ 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 $U\to \Gamma (U,\mathcal{O_\rm{X}})_{pf}$ is a sheaf? This is stated in section 6 in Greenberg: Perfect closures of rings and schemes but I do not understand the proof if perfection does not commute with products.

Answer 1

Colimit perfection does not commute with products. For instance, take $A=\mathbf{F}_p[x_1, x_2, \ldots]/(x_1^p, x_2^{p^2}, x_3^{p^3}, \ldots)$. Then $A_{\mathrm{perf}}=\mathbf{F}_p$. However, the image of the element $(x_1, x_2, \ldots)\in \prod_{i=1}^\infty A$ in $\left(\prod_{i=1}^\infty A\right)_{\mathrm{perf}}$ is non-zero, but its image in $\prod_{i=0}^\infty A_{\mathrm{perf}}$ is zero.

In anycase, if you want to construct the perfection of a scheme you can just take the infinite limit $\lim_{\mathrm{Fr}}X$ along the Frobenius maps. These maps are all affine so the inverse limit is indeed a scheme.