Lemma 10.118.7. Let $R \to S$ be a ring map. Let $M$ be an $S$-module. Assume

1. $R \to S$ is of finite type,

2. $M$ is a finite $S$-module, and

3. $R$ is reduced.

Then there exists a subset $U \subset \mathop{\mathrm{Spec}}(R)$ such that

1. $U$ is open and dense in $\mathop{\mathrm{Spec}}(R)$,

2. for every $u \in U$ there exists an $f \in R$ such that $u \in D(f) \subset U$ and such that we have

1. $M_ f$ and $S_ f$ are free over $R_ f$,

2. $S_ f$ is a finitely presented $R_ f$-algebra, and

3. $M_ f$ is a finitely presented $S_ f$-module.

Proof. Note that the lemma is equivalent to the statement that the open $U(R \to S, M)$, see Equation (10.118.3.2), is dense in $\mathop{\mathrm{Spec}}(R)$. We first prove the lemma for $S = R[x_1, \ldots , x_ n]$, and then we deduce the result in general.

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 $n$.

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 10.5.4. Since a finite intersection of dense opens is dense open, we see from Lemma 10.118.4 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 $S = R[x_1, \ldots , x_ n]$.

Let $I \subset R$ be the ideal generated by the coefficients of elements of $J$. Let $U_1 = \mathop{\mathrm{Spec}}(R) \setminus V(I)$ and let

$U_2 = \mathop{\mathrm{Spec}}(R) \setminus \overline{U_1}.$

Then it is clear that $U = U_1 \cup U_2$ is dense in $\mathop{\mathrm{Spec}}(R)$. 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 10.118.6 we see that $U(R \to S, M)$ is dense in $\mathop{\mathrm{Spec}}(R)$. Hence we may assume either (a) $I = R$, or (b) $V(I) = \mathop{\mathrm{Spec}}(R)$.

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.

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 $m$.

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 10.115.2 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 $n$.

The case $m > 0$. Pick a multi-index $K \in E$ such that $a_ K$ is not a unit. As before set $U_1 = \mathop{\mathrm{Spec}}(R_{a_ K}) = \mathop{\mathrm{Spec}}(R) \setminus V(a_ K)$ and set

$U_2 = \mathop{\mathrm{Spec}}(R) \setminus \overline{U_1}.$

Then it is clear that $U = U_1 \cup U_2$ is dense in $\mathop{\mathrm{Spec}}(R)$. 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 10.118.6 we see that $U(R \to S, M)$ is dense in $\mathop{\mathrm{Spec}}(R)$. Hence we may assume either (a) $a_ KR = R$, or (b) $V(a_ K) = \mathop{\mathrm{Spec}}(R)$. 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 $m$.

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 (10.118.3.2), 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. $\square$

Comment #4642 by Andy on

There's a typo in the last line, the second term should be $U(R \to R[x_1,\ldots,x_n],M)$

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