The Stacks project

The second sentence of the statement has an extra "is": it should probably read "…then $a$ has a $p$th root in $A$." In the proof, the phrase "as $B$ is a Nagata" should probably say "…is Nagata" or "…is a Nagata ring".

Yes indeed, thanks. Fixed here.

The proof seems to miss a logical step. Namely, in order to show that $B$ is a domain one needs that $a$ doesn't have a root in the fraction field $K$ of $A$ (or am I overlooking an obvious argument?). For this one needs to use that $A^\hat\cap K=A$ inside $A^\hat\otimes_A K$ and that $A^\hat\otimes_A K$ is reduced (which is clear since $A$ is analytically unramified).

The proof seems to miss a logical step. Namely, in order to show that $B$ is a domain one needs that $a$ doesn't have a root in the fraction field $K$ of $A$ (or am I overlooking an obvious argument?). For this one needs to use that $A^\hat\cap K=A$ inside $A^\hat\otimes_A K$ and that $A^\hat\otimes_A K$ is reduced (which is clear since $A$ is analytically unramified).

Sorry, I meant $A^\wedge\cap K$ and $A^\wedge\otimes_A K$, respectively.