The Stacks project

Proof. Let $\mathfrak p_1, \ldots , \mathfrak p_ t$ be the associated primes of the $R$-module $R/xR$. Since $R/xR$ has no embedded primes we see that each $\mathfrak p_ i$ has height $1$, and is a minimal prime over $(x)$. For each $i$, let $\mathfrak q_{i1}, \ldots , \mathfrak q_{is_ i}$ be the associated primes of the $R^\wedge$-module $R^\wedge /\mathfrak p_ iR^\wedge$. By Lemma 10.162.11 we see that $(R^\wedge )_{\mathfrak q_{ij}}$ is regular. By Lemma 10.65.3 we see that

Let $y \in R^\wedge$ with $y^2 = 0$. As $(R^\wedge )_{\mathfrak q_{ij}}$ is regular, and hence a domain (Lemma 10.106.2) we see that $y$ maps to zero in $(R^\wedge )_{\mathfrak q_{ij}}$. Hence $y$ maps to zero in $R^\wedge /xR^\wedge$ by Lemma 10.63.19. Hence $y = xy'$. Since $x$ is a nonzerodivisor (as $R \to R^\wedge$ is flat) we see that $(y')^2 = 0$. Hence we conclude that $y \in \bigcap x^ nR^\wedge = (0)$ (Lemma 10.51.4). $\square$