Note that the lemma is equivalent to the statement that the open $U(R \to S, M)$, see Equation (\ref{equation-good-locus}), is dense in $\Spec(R)$. We first prove the lemma for $S = R[x_1, \ldots, x_n]$, and then we deduce the result in general. \medskip\noindent Proof of the case $S = R[x_1, \ldots, x_n]$ and $M$ any finite module over $S$. Note that in this case $S_f = R_f[x_1, \ldots, x_n]$ is free and of finite presentation over $R_f$, so we do not have to worry about the conditions regarding $S$, only those that concern $M$. We will use induction on $n$. \medskip\noindent There exists a finite filtration $$ 0 \subset M_1 \subset M_2 \subset \ldots \subset M_t = M $$ such that $M_i/M_{i - 1} \cong S/J_i$ for some ideal $J_i \subset S$, see Lemma \ref{lemma-trivial-filter-finite-module}. Since a finite intersection of dense opens is dense open, we see from Lemma \ref{lemma-generic-flatness-locus-extension} that it suffices to prove the lemma for each of the modules $R/J_i$. Hence we may assume that $M = S/J$ for some ideal $J$ of $S = R[x_1, \ldots, x_n]$. \medskip\noindent Let $I \subset R$ be the ideal generated by the coefficients of elements of $J$. Let $U_1 = \Spec(R) \setminus V(I)$ and let $$ U_2 = \Spec(R) \setminus \overline{U_1}. $$ Then it is clear that $U = U_1 \cup U_2$ is dense in $\Spec(R)$. Let $f \in R$ be an element such that either (a) $D(f) \subset U_1$ or (b) $D(f) \subset U_2$. If for any such $f$ the lemma holds for the pair $(R_f \to R_f[x_1, \ldots, x_n], M_f)$ then by Lemma \ref{lemma-generic-flatness-locus-reduce} we see that $U(R \to S, M)$ is dense in $\Spec(R)$. Hence we may assume either (a) $I = R$, or (b) $V(I) = \Spec(R)$. \medskip\noindent In case (b) we actually have $I = 0$ as $R$ is reduced! Hence $J = 0$ and $M = S$ and the lemma holds in this case. \medskip\noindent In case (a) we have to do a little bit more work. Note that every element of $I$ is actually the coefficient of a monomial of an element of $J$, because the set of coefficients of elements of $J$ forms an ideal (details omitted). Hence we find an element $$ g = \sum\nolimits_{K \in E} a_K x^K \in J $$ where $E$ is a finite set of multi-indices $K = (k_1, \ldots, k_n)$ with at least one coefficient $a_{K_0}$ a unit in $R$. Actually we can find one which has a coefficient equal to $1$ as $1 \in I$ in case (a). Let $m = \#\{K \in E \mid a_K \text{ is not a unit}\}$. Note that $0 \leq m \leq \# E - 1$. We will argue by induction on $m$. \medskip\noindent The case $m = 0$. In this case all the coefficients $a_K$, $K \in E$ of $g$ are units and $E \not = \emptyset$. If $E = \{K_0\}$ is a singleton and $K_0 = (0, \ldots, 0)$, then $g$ is a unit and $J = S$ so the result holds for sure. (This happens in particular when $n = 0$ and it provides the base case of the induction on $n$.) If not $E = \{(0, \ldots, 0)\}$, then at least one $K$ is not equal to $(0, \ldots, 0)$, i.e., $g \not \in R$. At this point we employ the usual trick of Noether normalization. Namely, we consider $$ G(y_1, \ldots, y_n) = g(y_1 + y_n^{e_1}, y_2 + y_n^{e_2}, \ldots, y_{n - 1} + y_n^{e_{n - 1}}, y_n) $$ with $0 \ll e_{n -1} \ll e_{n - 2} \ll \ldots \ll e_1$. By Lemma \ref{lemma-helper-polynomial} it follows that $G(y_1, \ldots, y_n)$ as a polynomial in $y_n$ looks like $$ a_K y_n^{k_n + \sum_{i = 1, \ldots, n - 1} e_i k_i} + \text{lower order terms in }y_n $$ As $a_K$ is a unit we conclude that $M = R[x_1, \ldots, x_n]/J$ is finite over $R[y_1, \ldots, y_{n - 1}]$. Hence $U(R \to R[x_1, \ldots, x_n], M) = U(R \to R[y_1, \ldots, y_{n - 1}], M)$ and we win by induction on $n$. \medskip\noindent The case $m > 0$. Pick a multi-index $K \in E$ such that $a_K$ is not a unit. As before set $U_1 = \Spec(R_{a_K}) = \Spec(R) \setminus V(a_K)$ and set $$ U_2 = \Spec(R) \setminus \overline{U_1}. $$ Then it is clear that $U = U_1 \cup U_2$ is dense in $\Spec(R)$. Let $f \in R$ be an element such that either (a) $D(f) \subset U_1$ or (b) $D(f) \subset U_2$. If for any such $f$ the lemma holds for the pair $(R_f \to R_f[x_1, \ldots, x_n], M_f)$ then by Lemma \ref{lemma-generic-flatness-locus-reduce} we see that $U(R \to S, M)$ is dense in $\Spec(R)$. Hence we may assume either (a) $a_KR = R$, or (b) $V(a_K) = \Spec(R)$. In case (a) the number $m$ drops, as $a_K$ has turned into a unit. In case (b), since $R$ is reduced, we conclude that $a_K = 0$. Hence the set $E$ decreases so the number $m$ drops as well. In both cases we win by induction on $m$. \medskip\noindent At this point we have proven the lemma in case $S = R[x_1, \ldots, x_n]$. Assume that $(R \to S, M)$ is an arbitrary pair satisfying the conditions of the lemma. Choose a surjection $R[x_1, \ldots, x_n] \to S$. Observe that, with the notation introduced in (\ref{equation-good-locus}), we have $$ U(R \to S, M) = U(R \to R[x_1, \ldots, x_n], S) \cap U(R \to R[x_1, \ldots, x_n], M) $$ Hence as we've just finished proving the right two opens are dense also the open on the left is dense.