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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 $R$.

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  • $\begingroup$ If $R$ is a local Noetherian ring of characteristic $2$, then adding all square roots amounts to passing to the perfection (direct limit over $x \mapsto x^2$) which is flat if and only if $R$ is regular. $\endgroup$
    – Pol van Hoften
    Jul 16, 2020 at 10:29
  • $\begingroup$ Now let $R$ be a local excellent domain of positive characteristic. Then you could ask for the (presumably stronger) condition that $R \to R^{+}$ is flat, where $R^{+}$ is the absolute integral closure (the integral closure of $R$ in an algebraic closure of its field of fractions). Theorem 4.13.(2) of arxiv.org/pdf/1803.03229.pdf says that this happens if and only if $R$ is regular. $\endgroup$
    – Pol van Hoften
    Jul 16, 2020 at 10:32

1 Answer 1

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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.

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  • $\begingroup$ Okay, I can see that $R_1$ is faithfully flat, since $R_1$ can be viewed as a colimit of $R_S$, where $S\subset R$ is a finite subset, and $R_S := R[(X_s)_{s\in S}]/(X_s^2-s)$. This is a directed colimit of faithfully flat $R$-algebras, and hence is also faithfully flat by stacks.math.columbia.edu/tag/090N. Though, out of curiosity, how are you seeing that $R_1$ is free? How do you define the infinite tensor product? Is a directed colimit of free modules free? $\endgroup$
    – stupid_question_bot
    Jul 16, 2020 at 20:44
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    $\begingroup$ An infinite tensor product (of algebras, not modules!) can be defined as a coproduct, and constructed as a colimit of finite tensor products. If $(A_i)$ is a family of $R$-algebras, then $A=\bigotimes_i A_i$ is generated as $R$-module by tensors $\otimes_i x_i$ with finite support ($x_i=1$ for almost all $i$). If $B_i$ is an $R$-basis of $A_i$ for each $i$, it is easy to check that taking $x_i$ in $B_i$ gives rise to a basis of $A$. In the above example, a basis of $R_1$ is given by the finite products $\prod_{a\in S}x_a$ where $S$ runs through finite subsets of $R$ and $x_a=$ class of $X_a$. $\endgroup$
    – Laurent Moret-Bailly
    Jul 17, 2020 at 7:33
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    $\begingroup$ For the last question, the $\mathbb{Z}$-module $\mathbb{Q}$ is the colimit of the free modules $\frac{1}{n}\mathbb{Z}$ ($n\geq1$) but is not free. In fact, every flat modules is a directed colimit of free modules (Lazard's theorem). $\endgroup$
    – Laurent Moret-Bailly
    Jul 17, 2020 at 7:38

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