Lemma 88.28.4. In the category $\widehat{\mathcal{C}}_\Lambda$, if $f_1, f_2 : R \to S$ are formally homotopic and $\mathfrak p \subset R$ is a minimal prime ideal, then $f_1(\mathfrak p)S = f_2(\mathfrak p)S$ as ideals.

Proof. Suppose $(r, h, k)$ is a formal homotopy between $f_1$ and $f_2$. We claim that $\mathfrak pR[[t_1, \ldots , t_ r]] = h(\mathfrak p)R[[t_1, \ldots , t_ r]]$. The claim implies the lemma by further composing with $k$. To prove the claim, observe that the map $\mathfrak p \mapsto \mathfrak pR[[t_1, \ldots , t_ r]]$ is a bijection between the minimal prime ideals of $R$ and the minimal prime ideals of $R[[t_1, \ldots , t_ r]]$. Finally, $h(\mathfrak p)R[[t_1, \ldots , t_ r]]$ is a minimal prime as $h$ is flat, and hence of the form $\mathfrak q R[[t_1, \ldots , t_ r]]$ for some minimal prime $\mathfrak q \subset R$ by what we just said. But since $h \bmod (t_1, \ldots , t_ r) = \text{id}_ R$ by definition of a formal homotopy, we conclude that $\mathfrak q = \mathfrak p$ as desired. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).