The Stacks project

Proof. Let $K$ be the fraction field of $R$. Set $S_ K = K \otimes _ R S$. This is an algebra of finite type over $K$. We will argue by induction on $d = \dim (S_ K)$ (which is finite for example by Noether normalization, see Section 10.115). Fix $d \geq 0$. Assume we know that the lemma holds in all cases where $\dim (S_ K) < d$.

Suppose given $R \to S$ and $M$ as in the lemma with $\dim (S_ K) = d$. By Lemma 10.62.1 there exists a filtration $0 \subset M_1 \subset M_2 \subset \ldots \subset M_ n = M$ so that $M_ i/M_{i - 1}$ is isomorphic to $S/\mathfrak q$ for some prime $\mathfrak q$ of $S$. Note that $\dim ((S/\mathfrak q)_ K) \leq \dim (S_ K)$. Also, note that an extension of free modules is free (see basic notion 50). Thus we may assume $M = S$ and that $S$ is a domain of finite type over $R$.

If $R \to S$ has a nontrivial kernel, then take a nonzero $f \in R$ in this kernel. In this case $S_ f = 0$ and the lemma holds. (This is really the case $d = -1$ and the start of the induction.) Hence we may assume that $R \to S$ is a finite type extension of Noetherian domains.

Apply Lemma 10.115.7 and replace $R$ by $R_ f$ (with $f$ as in the lemma) to get a factorization

where the second extension is finite. Choose $z_1, \ldots , z_ r \in S$ which form a basis for the fraction field of $S$ over the fraction field of $R[y_1, \ldots , y_ d]$. This gives a short exact sequence

By construction $N$ is a finite $R[y_1, \ldots , y_ d]$-module whose support does not contain the generic point $(0)$ of $\mathop{\mathrm{Spec}}(R[y_1, \ldots , y_ d])$. By Lemma 10.40.5 there exists a nonzero $g \in R[y_1, \ldots , y_ d]$ such that $g$ annihilates $N$, so we may view $N$ as a finite module over $S' = R[y_1, \ldots , y_ d]/(g)$. Since $\dim (S'_ K) < d$ by induction there exists a nonzero $f \in R$ such that $N_ f$ is a free $R_ f$-module. Since $(R[y_1, \ldots , y_ d])_ f \cong R_ f[y_1, \ldots , y_ d]$ is free also we conclude by the already mentioned fact that an extension of free modules is free. $\square$

Proof. Write $S = R[x_1, \ldots , x_ n]/(g_1, \ldots , g_ m)$. For $g \in R[x_1, \ldots , x_ n]$ denote $\overline{g}$ its image in $S$. We may write $M = S^{\oplus t}/\sum Sn_ i$ for some $n_ i \in S^{\oplus t}$. Write $n_ i = (\overline{g}_{i1}, \ldots , \overline{g}_{it})$ for some $g_{ij} \in R[x_1, \ldots , x_ n]$. Let $R_0 \subset R$ be the subring generated by all the coefficients of all the elements $g_ i, g_{ij} \in R[x_1, \ldots , x_ n]$. Define $S_0 = R_0[x_1, \ldots , x_ n]/(g_1, \ldots , g_ m)$. Define $M_0 = S_0^{\oplus t}/\sum S_0n_ i$. Then $R_0$ is a domain of finite type over $\mathbf{Z}$ and hence Noetherian (see Lemma 10.31.1). Moreover via the injection $R_0 \to R$ we have $S \cong R \otimes _{R_0} S_0$ and $M \cong R \otimes _{R_0} M_0$. Applying Lemma 10.118.1 we obtain a nonzero $f \in R_0$ such that $(M_0)_ f$ is a free $(R_0)_ f$-module. Hence $M_ f = R_ f \otimes _{(R_0)_ f} (M_0)_ f$ is a free $R_ f$-module. $\square$

Proof. We first prove the lemma for $S = R[x_1, \ldots , x_ n]$, and then we deduce the result in general.

Assume $S = R[x_1, \ldots , x_ n]$. Choose elements $m_1, \ldots , m_ t$ which generate $M$. This gives a short exact sequence

Denote $K$ the fraction field of $R$. Denote $S_ K = K \otimes _ R S = K[x_1, \ldots , x_ n]$, and similarly $N_ K = K \otimes _ R N$, $M_ K = K \otimes _ R M$. As $R \to K$ is flat the sequence remains exact after tensoring with $K$. As $S_ K = K[x_1, \ldots , x_ n]$ is a Noetherian ring (see Lemma 10.31.1) we can find finitely many elements $n'_1, \ldots , n'_ s \in N_ K$ which generate it. Choose $n_1, \ldots , n_ r \in N$ such that $n'_ i = \sum a_{ij}n_ j$ for some $a_{ij} \in K$. Set

By construction $M'$ is a finitely presented $S$-module, and there is a surjection $M' \to M$ which induces an isomorphism $M'_ K \cong M_ K$. We may apply Lemma 10.118.2 to $R \to S$ and $M'$ and we find an $f \in R$ such that $M'_ f$ is a free $R_ f$-module. Thus $M'_ f \to M_ f$ is a surjection of modules over the domain $R_ f$ where the source is a free module and which becomes an isomorphism upon tensoring with $K$. Thus it is injective as $M'_ f \subset M'_ K$ because $M'_ f$ is free and $R_ f$ is a domain with fraction field $K$. Hence $M'_ f \to M_ f$ is an isomorphism and the result is proved.

For the general case, choose a surjection $R[x_1, \ldots , x_ n] \to S$. Think of both $S$ and $M$ as finite modules over $R[x_1, \ldots , x_ n]$. By the special case proved above there exists a nonzero $f \in R$ such that both $S_ f$ and $M_ f$ are free as $R_ f$-modules and finitely presented as $R_ f[x_1, \ldots , x_ n]$-modules. Clearly this implies that $S_ f$ is a finitely presented $R_ f$-algebra and that $M_ f$ is a finitely presented $S_ f$-module. $\square$

Proof. Let $u \in U(R \to S, M_1) \cap U(R \to S, M_3)$. Choose $f_1, f_3 \in R$ such that $u \in D(f_1)$, $u \in D(f_3)$ and such that (10.118.3.1) holds for $f_1$ and $M_1$ and for $f_3$ and $M_3$. Then set $f = f_1f_3$. Then $u \in D(f)$ and (10.118.3.1) holds for $f$ and both $M_1$ and $M_3$. An extension of free modules is free, and an extension of finitely presented modules is finitely presented (Lemma 10.5.3). Hence we see that (10.118.3.1) holds for $f$ and $M_2$. Thus $u \in U(R \to S, M_2)$ and we win. $\square$

Proof. Suppose that $u \in U(R_ f \to S_ f, M_ f)$. Then there exists an element $g \in R_ f$ such that $u \in D(g)$ and such that (10.118.3.1) holds for the pair $((R_ f)_ g \to (S_ f)_ g, (M_ f)_ g)$. Write $g = a/f^ n$ for some $a \in R$. Set $h = af$. Then $R_ h = (R_ f)_ g$, $S_ h = (S_ f)_ g$, and $M_ h = (M_ f)_ g$. Moreover $u \in D(h)$. Hence $u \in U(R \to S, M)$. Conversely, suppose that $u \in D(f) \cap U(R \to S, M)$. Then there exists an element $g \in R$ such that $u \in D(g)$ and such that (10.118.3.1) holds for the pair $(R_ g \to S_ g, M_ g)$. Then it is clear that (10.118.3.1) also holds for the pair $(R_{fg} \to S_{fg}, M_{fg}) = ((R_ f)_ g \to (S_ f)_ g, (M_ f)_ g)$. Hence $u \in U(R_ f \to S_ f, M_ f)$ and we win. $\square$

Proof. In view of Lemma 10.118.5 this is a purely topological statement. Namely, by that lemma we see that $U(R \to S, M) \cap D(f_ i)$ is dense in $D(f_ i)$ for each $i \in I$. By Topology, Lemma 5.21.4 we see that $U(R \to S, M) \cap U$ is dense in $U$. Since $U$ is dense in $\mathop{\mathrm{Spec}}(R)$ we conclude that $U(R \to S, M)$ is dense in $\mathop{\mathrm{Spec}}(R)$. $\square$

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

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

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

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

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

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

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

Hence as we've just finished proving the right two opens are dense also the open on the left is dense. $\square$