Lemma 115.4.3. Let $R \to S$ be a ring map. Let $\mathfrak p \subset R$ be a prime. Let $\mathfrak q \subset S$ be a prime lying over $\mathfrak p$. Assume $S_{\mathfrak q}$ is essentially of finite type over $R_\mathfrak p$. Assume given

1. an integer $n \geq 0$,

2. a prime $\mathfrak a \subset \kappa (\mathfrak p)[x_1, \ldots , x_ n]$,

3. a surjective $\kappa (\mathfrak p)$-homomorphism

$\psi : (\kappa (\mathfrak p)[x_1, \ldots , x_ n])_{\mathfrak a} \longrightarrow S_{\mathfrak q}/\mathfrak p S_{\mathfrak q},$

and

4. elements $\overline{f}_1, \ldots , \overline{f}_ e$ in $\mathop{\mathrm{Ker}}(\psi )$.

Then there exist

1. an integer $m \geq 0$,

2. and element $g \in S$, $g \not\in \mathfrak q$,

3. a map

$\Psi : R[x_1, \ldots , x_ n, x_{n + 1}, \ldots , x_{n + m}] \longrightarrow S_ g,$

and

4. elements $f_1, \ldots , f_ e, f_{e + 1}, \ldots , f_{e + m}$ of $\mathop{\mathrm{Ker}}(\Psi )$

such that

1. the following diagram commutes

$\xymatrix{ R[x_1, \ldots , x_{n + m}] \ar[d]_\Psi \ar[rr]_-{x_{n + j} \mapsto 0} & & (\kappa (\mathfrak p)[x_1, \ldots , x_ n])_{\mathfrak a} \ar[d]^\psi \\ S_ g \ar[rr] & & S_{\mathfrak q}/\mathfrak p S_{\mathfrak q} },$
2. the element $f_ i$, $i \leq n$ maps to a unit times $\overline{f}_ i$ in the local ring

$(\kappa (\mathfrak p)[x_1, \ldots , x_{n + m}])_{ (\mathfrak a, x_{n + 1}, \ldots , x_{n + m})},$
3. the element $f_{e + j}$ maps to a unit times $x_{n + j}$ in the same local ring, and

4. the induced map $R[x_1, \ldots , x_{n + m}]_{\mathfrak b} \to S_{\mathfrak q}$ is surjective, where $\mathfrak b = \Psi ^{-1}(\mathfrak qS_ g)$.

Proof. We claim that it suffices to prove the lemma in case $R$ and $S$ are local with maximal ideals $\mathfrak p$ and $\mathfrak q$. Namely, suppose we have constructed

$\Psi ' : R_{\mathfrak p}[x_1, \ldots , x_{n + m}] \longrightarrow S_{\mathfrak q}$

and $f_1', \ldots , f_{e + m}' \in R_{\mathfrak p}[x_1, \ldots , x_{n + m}]$ with all the required properties. Then there exists an element $f \in R$, $f \not\in \mathfrak p$ such that each $ff_ k'$ comes from an element $f_ k \in R[x_1, \ldots , x_{n + m}]$. Moreover, for a suitable $g \in S$, $g \not\in \mathfrak q$ the elements $\Psi '(x_ i)$ are the image of elements $y_ i \in S_ g$. Let $\Psi$ be the $R$-algebra map defined by the rule $\Psi (x_ i) = y_ i$. Since $\Psi (f_ i)$ is zero in the localization $S_{\mathfrak q}$ we may after possibly replacing $g$ assume that $\Psi (f_ i) = 0$. This proves the claim.

Thus we may assume $R$ and $S$ are local with maximal ideals $\mathfrak p$ and $\mathfrak q$. Pick $y_1, \ldots , y_ n \in S$ such that $y_ i \bmod \mathfrak pS = \psi (x_ i)$. Let $y_{n + 1}, \ldots , y_{n + m} \in S$ be elements which generate an $R$-subalgebra of which $S$ is the localization. These exist by the assumption that $S$ is essentially of finite type over $R$. Since $\psi$ is surjective we may write $y_{n + j} \bmod \mathfrak pS = \psi (h_ j)$ for some $h_ j \in \kappa (\mathfrak p)[x_1, \ldots , x_ n]_{\mathfrak a}$. Write $h_ j = g_ j/d$, $g_ j \in \kappa (\mathfrak p)[x_1, \ldots , x_ n]$ for some common denominator $d \in \kappa (\mathfrak p)[x_1, \ldots , x_ n]$, $d \not\in \mathfrak a$. Choose lifts $G_ j, D \in R[x_1, \ldots , x_ n]$ of $g_ j$ and $d$. Set $y_{n + j}' = D(y_1, \ldots , y_ n) y_{n + j} - G_ j(y_1, \ldots , y_ n)$. By construction $y_{n + j}' \in \mathfrak p S$. It is clear that $y_1, \ldots , y_ n, y_ n', \ldots , y_{n + m}'$ generate an $R$-subalgebra of $S$ whose localization is $S$. We define

$\Psi : R[x_1, \ldots , x_{n + m}] \to S$

to be the map that sends $x_ i$ to $y_ i$ for $i = 1, \ldots , n$ and $x_{n + j}$ to $y'_{n + j}$ for $j = 1, \ldots , m$. Properties (1) and (4) are clear by construction. Moreover the ideal $\mathfrak b$ maps onto the ideal $(\mathfrak a, x_{n + 1}, \ldots , x_{n + m})$ in the polynomial ring $\kappa (\mathfrak p)[x_1, \ldots , x_{n + m}]$.

Denote $J = \mathop{\mathrm{Ker}}(\Psi )$. We have a short exact sequence

$0 \to J_{\mathfrak b} \to R[x_1, \ldots , x_{n + m}]_{\mathfrak b} \to S_{\mathfrak q} \to 0.$

The surjectivity comes from our choice of $y_1, \ldots , y_ n, y_ n', \ldots , y_{n + m}'$ above. This implies that

$J_{\mathfrak b}/ \mathfrak pJ_{\mathfrak b} \to \kappa (\mathfrak p)[x_1, \ldots , x_{n + m}]_{ (\mathfrak a, x_{n + 1}, \ldots , x_{n + m})} \to S_{\mathfrak q}/\mathfrak pS_{\mathfrak q} \to 0$

is exact. By construction $x_ i$ maps to $\psi (x_ i)$ and $x_{n + j}$ maps to zero under the last map. Thus it is easy to choose $f_ i$ as in (2) and (3) of the lemma. $\square$

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