Let $R$ be a local ring where 2 is invertible. Must there exist a faithfully flat $R$-algebra where the squaring map $x\mapsto x^2$ is surjective?


This is certainly true for fields. For DVR's, you can take the strict henselization, and then take the colimit over all extensions taking square roots of the uniformizer.


For a general local ring, I'm a bit lost. Presumably you would start with the strict henselization, then continuously take square roots of stuff in the maximal ideal, though it's unclear to me that the resulting colimit would even range over a set, let alone if the resulting ring is faithfully flat over $R$.



Let $R$ be a local ring where 2 is invertible. Must there exist a faithfully flat $R$-algebra where the squaring map is surjective?
The following works over any ring $R$: Take a family $\underline{X}:=(X\_a)\_{a\in R}$ of indeterminates indexed by $R$, and put $R\_1:=R[\underline{X}]/I$ where $I$ is generated by $(X\_a^2-a)\_{a\in R}$. Then $R\_1$ is free as an $R$-module (you can view it as $\bigotimes\_{a\in R}R[X\_a]/(X\_a^2-a)$) and every element of $R$ becomes a square in $R\_1$. Now just iterate the process to get $R\subset R\_1\subset R\_2\subset\dots$ and put $R\_\infty=\varinjlim\_n R\_n$. This $R\_\infty$ solves the problem.


