Lemma 16.8.3. Let $R \to A \to \Lambda $ be ring maps with $A$ of finite presentation over $R$. Let $S \subset R$ be a multiplicative set. Let $S^{-1}A \to B' \to S^{-1}\Lambda $ be a factorization with $B'$ smooth over $S^{-1}R$. Then we can find a factorization $A \to B \to \Lambda $ such that some $s \in S$ maps to an elementary standard element (Definition 16.2.3) in $B$ over $R$.

**Proof.**
We first apply Lemma 16.3.4 to $S^{-1}R \to B'$. Thus we may assume $B'$ is standard smooth over $S^{-1}R$. Write $A = R[x_1, \ldots , x_ n]/(g_1, \ldots , g_ t)$ and say $x_ i \mapsto \lambda _ i$ in $\Lambda $. We may write $B' = S^{-1}R[x_1, \ldots , x_{n + m}]/(f_1, \ldots , f_ c)$ for some $c \geq n$ where $\det (\partial f_ j/\partial x_ i)_{i, j = 1, \ldots , c}$ is invertible in $B'$ and such that $A \to B'$ is given by $x_ i \mapsto x_ i$, see Lemma 16.3.6. After multiplying $x_ i$, $i > n$ by an element of $S$ and correspondingly modifying the equations $f_ j$ we may assume $B' \to S^{-1}\Lambda $ maps $x_ i$ to $\lambda _ i/1$ for some $\lambda _ i \in \Lambda $ for $i > n$. Choose a relation

for some $a_ j \in S^{-1}R[x_1, \ldots , x_{n + m}]$. Since each element of $S$ is invertible in $B'$ we may (by clearing denominators) assume that $f_ j, a_ j \in R[x_1, \ldots , x_{n + m}]$ and that

for some $s_0 \in S$. Since $g_ j$ maps to zero in $S^{-1}R[x_1, \ldots , x_{n + m}]/(f_1, \ldots , x_ c)$ we can find elements $s_ j \in S$ such that $s_ j g_ j = 0$ in $R[x_1, \ldots , x_{n + m}]/(f_1, \ldots , f_ c)$. Since $f_ j$ maps to zero in $S^{-1}\Lambda $ we can find $s'_ j \in S$ such that $s'_ j f_ j(\lambda _1, \ldots , \lambda _{n + m}) = 0$ in $\Lambda $. Consider the ring

and the factorization $A \to B \to \Lambda $ with $B \to \Lambda $ given by $x_ i \mapsto \lambda _ i$. We claim that $s = s_0s_1 \ldots s_ ts'_1 \ldots s'_ c$ is elementary standard in $B$ over $R$ which finishes the proof. Namely, $s_ j g_ j \in (f_1, \ldots , f_ c)$ and hence $sg_ j \in (s'_1f_1, \ldots , s'_ cf_ c)$. Finally, we have

which divides $s$ as desired. $\square$

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