Lemma 10.126.1. Let $R \to A$ be a ring map. Consider the category $\mathcal{I}$ of all diagrams of $R$-algebra maps $A' \to A$ with $A'$ finitely presented over $R$. Then $\mathcal{I}$ is filtered, and the colimit of the $A'$ over $\mathcal{I}$ is isomorphic to $A$.

## 10.126 Colimits and maps of finite presentation

In this section we prove some preliminary lemmas which will eventually help us prove result using absolute Noetherian reduction. In Categories, Section 4.19 we discuss filtered colimits in general. Here is an example of this very general notion.

**Proof.**
The category^{1} $\mathcal{I}$ is nonempty as $R \to A$ is an object of it. Consider a pair of objects $A' \to A$, $A'' \to A$ of $\mathcal{I}$. Then $A' \otimes _ R A'' \to A$ is in $\mathcal{I}$ (use Lemmas 10.6.2 and 10.13.2). The ring maps $A' \to A' \otimes _ R A''$ and $A'' \to A' \otimes _ R A''$ define arrows in $\mathcal{I}$ thereby proving the second defining property of a filtered category, see Categories, Definition 4.19.1. Finally, suppose that we have two morphisms $\sigma , \tau : A' \to A''$ in $\mathcal{I}$. If $x_1, \ldots , x_ r \in A'$ are generators of $A'$ as an $R$-algebra, then we can consider $A''' = A''/(\sigma (x_ i) - \tau (x_ i))$. This is a finitely presented $R$-algebra and the given $R$-algebra map $A'' \to A$ factors through the surjection $\nu : A'' \to A'''$. Thus $\nu $ is a morphism in $\mathcal{I}$ equalizing $\sigma $ and $\tau $ as desired.

The fact that our index category is cofiltered means that we may compute the value of $B = \mathop{\mathrm{colim}}\nolimits _{A' \to A} A'$ in the category of sets (some details omitted; compare with the discussion in Categories, Section 4.19). To see that $B \to A$ is surjective, for every $a \in A$ we can use $R[x] \to A$, $x \mapsto a$ to see that $a$ is in the image of $B \to A$. Conversely, if $b \in B$ is mapped to zero in $A$, then we can find $A' \to A$ in $\mathcal{I}$ and $a' \in A'$ which maps to $b$. Then $A'/(a') \to A$ is in $\mathcal{I}$ as well and the map $A' \to B$ factors as $A' \to A'/(a') \to B$ which shows that $b = 0$ as desired. $\square$

Often it is easier to think about colimits over preordered sets. Let $(\Lambda , \geq )$ a preordered set. A system of rings over $\Lambda $ is given by a ring $R_\lambda $ for every $\lambda \in \Lambda $, and a morphism $R_\lambda \to R_\mu $ whenever $\lambda \leq \mu $. These morphisms have to satisfy the rule that $R_\lambda \to R_\mu \to R_\nu $ is equal to the map $R_\lambda \to R_\nu $ for all $\lambda \leq \mu \leq \nu $. See Categories, Section 4.21. We will often assume that $(I, \leq )$ is *directed*, which means that $\Lambda $ is nonempty and given $\lambda , \mu \in \Lambda $ there exists a $\nu \in \Lambda $ with $\lambda \leq \nu $ and $\mu \leq \nu $. Recall that the colimit $\mathop{\mathrm{colim}}\nolimits _\lambda R_\lambda $ is sometimes called a “direct limit” in this case (but we will not use this terminology).

Note that Categories, Lemma 4.21.5 tells us that colimits over filtered index categories are the same thing as colimits over directed sets.

Lemma 10.126.2. Let $R \to A$ be a ring map. There exists a directed system $A_\lambda $ of $R$-algebras of finite presentation such that $A = \mathop{\mathrm{colim}}\nolimits _\lambda A_\lambda $. If $A$ is of finite type over $R$ we may arrange it so that all the transition maps in the system of $A_\lambda $ are surjective.

**Proof.**
The first proof is that this follows from Lemma 10.126.1 and Categories, Lemma 4.21.5.

Second proof. Compare with the proof of Lemma 10.8.12. Consider any finite subset $S \subset A$, and any finite collection of polynomial relations $E$ among the elements of $S$. So each $s \in S$ corresponds to $x_ s \in A$ and each $e \in E$ consists of a polynomial $f_ e \in R[X_ s; s\in S]$ such that $f_ e(x_ s) = 0$. Let $A_{S, E} = R[X_ s; s\in S]/(f_ e; e\in E)$ which is a finitely presented $R$-algebra. There are canonical maps $A_{S, E} \to A$. If $S \subset S'$ and if the elements of $E$ correspond, via the map $R[X_ s; s \in S] \to R[X_ s; s\in S']$, to a subset of $E'$, then there is an obvious map $A_{S, E} \to A_{S', E'}$ commuting with the maps to $A$. Thus, setting $\Lambda $ equal the set of pairs $(S, E)$ with ordering by inclusion as above, we get a directed partially ordered set. It is clear that the colimit of this directed system is $A$.

For the last statement, suppose $A = R[x_1, \ldots , x_ n]/I$. In this case, consider the subset $\Lambda ' \subset \Lambda $ consisting of those systems $(S, E)$ above with $S = \{ x_1, \ldots , x_ n\} $. It is easy to see that still $A = \mathop{\mathrm{colim}}\nolimits _{\lambda ' \in \Lambda '} A_{\lambda '}$. Moreover, the transition maps are clearly surjective. $\square$

It turns out that we can characterize ring maps of finite presentation as follows. This in some sense says that the algebras of finite presentation are the “compact” objects in the category of $R$-algebras.

Lemma 10.126.3. Let $\varphi : R \to S$ be a ring map. The following are equivalent

$\varphi $ is of finite presentation,

for every directed system $A_\lambda $ of $R$-algebras the map

\[ \mathop{\mathrm{colim}}\nolimits _\lambda \mathop{\mathrm{Hom}}\nolimits _ R(S, A_\lambda ) \longrightarrow \mathop{\mathrm{Hom}}\nolimits _ R(S, \mathop{\mathrm{colim}}\nolimits _\lambda A_\lambda ) \]is bijective, and

for every directed system $A_\lambda $ of $R$-algebras the map

\[ \mathop{\mathrm{colim}}\nolimits _\lambda \mathop{\mathrm{Hom}}\nolimits _ R(S, A_\lambda ) \longrightarrow \mathop{\mathrm{Hom}}\nolimits _ R(S, \mathop{\mathrm{colim}}\nolimits _\lambda A_\lambda ) \]is surjective.

**Proof.**
Assume (1) and write $S = R[x_1, \ldots , x_ n] / (f_1, \ldots , f_ m)$. Let $A = \mathop{\mathrm{colim}}\nolimits A_\lambda $. Observe that an $R$-algebra homomorphism $S \to A$ or $S \to A_\lambda $ is determined by the images of $x_1, \ldots , x_ n$. Hence it is clear that $\mathop{\mathrm{colim}}\nolimits _\lambda \mathop{\mathrm{Hom}}\nolimits _ R(S, A_\lambda ) \to \mathop{\mathrm{Hom}}\nolimits _ R(S, A)$ is injective. To see that it is surjective, let $\chi : S \to A$ be an $R$-algebra homomorphism. Then each $x_ i$ maps to some element in the image of some $A_{\lambda _ i}$. We may pick $\mu \geq \lambda _ i$, $i = 1, \ldots , n$ and assume $\chi (x_ i)$ is the image of $y_ i \in A_\mu $ for $i = 1, \ldots , n$. Consider $z_ j = f_ j(y_1, \ldots , y_ n) \in A_\mu $. Since $\chi $ is a homomorphism the image of $z_ j$ in $A = \mathop{\mathrm{colim}}\nolimits _\lambda A_\lambda $ is zero. Hence there exists a $\mu _ j \geq \mu $ such that $z_ j$ maps to zero in $A_{\mu _ j}$. Pick $\nu \geq \mu _ j$, $j = 1, \ldots , m$. Then the images of $z_1, \ldots , z_ m$ are zero in $A_\nu $. This exactly means that the $y_ i$ map to elements $y'_ i \in A_\nu $ which satisfy the relations $f_ j(y'_1, \ldots , y'_ n) = 0$. Thus we obtain a ring map $S \to A_\nu $. This shows that (1) implies (2).

It is clear that (2) implies (3). Assume (3). By Lemma 10.126.2 we may write $S = \mathop{\mathrm{colim}}\nolimits _\lambda S_\lambda $ with $S_\lambda $ of finite presentation over $R$. Then the identity map factors as

for some $\lambda $. This implies that $S$ is finitely presented over $S_\lambda $ by Lemma 10.6.2 part (4) applied to $S \to S_\lambda \to S$. Applying part (2) of the same lemma to $R \to S_\lambda \to S$ we conclude that $S$ is of finite presentation over $R$. $\square$

Using the basic material above we can give a criterion of when an algebra $A$ is a filtered colimit of given type of algebra as follows.

Lemma 10.126.4. Let $R \to \Lambda $ be a ring map. Let $\mathcal{E}$ be a set of $R$-algebras such that each $A \in \mathcal{E}$ is of finite presentation over $R$. Then the following two statements are equivalent

$\Lambda $ is a filtered colimit of elements of $\mathcal{E}$, and

for any $R$ algebra map $A \to \Lambda $ with $A$ of finite presentation over $R$ we can find a factorization $A \to B \to \Lambda $ with $B \in \mathcal{E}$.

**Proof.**
Suppose that $\mathcal{I} \to \mathcal{E}$, $i \mapsto A_ i$ is a filtered diagram such that $\Lambda = \mathop{\mathrm{colim}}\nolimits _ i A_ i$. Let $A \to \Lambda $ be an $R$-algebra map with $A$ of finite presentation over $R$. Then we get a factorization $A \to A_ i \to \Lambda $ by applying Lemma 10.126.3. Thus (1) implies (2).

Consider the category $\mathcal{I}$ of Lemma 10.126.1. By Categories, Lemma 4.19.3 the full subcategory $\mathcal{J}$ consisting of those $A \to \Lambda $ with $A \in \mathcal{E}$ is cofinal in $\mathcal{I}$ and is a filtered category. Then $\Lambda $ is also the colimit over $\mathcal{J}$ by Categories, Lemma 4.17.2. $\square$

But more is true. Namely, given $R = \mathop{\mathrm{colim}}\nolimits _\lambda R_\lambda $ we see that the category of finitely presented $R$-modules is equivalent to the limit of the category of finitely presented $R_\lambda $-modules. Similarly for the categories of finitely presented $R$-algebras.

Lemma 10.126.5. Let $A$ be a ring and let $M, N$ be $A$-modules. Suppose that $R = \mathop{\mathrm{colim}}\nolimits _{i \in I} R_ i$ is a directed colimit of $A$-algebras.

If $M$ is a finite $A$-module, and $u, u' : M \to N$ are $A$-module maps such that $u \otimes 1 = u' \otimes 1 : M \otimes _ A R \to N \otimes _ A R$ then for some $i$ we have $u \otimes 1 = u' \otimes 1 : M \otimes _ A R_ i \to N \otimes _ A R_ i$.

If $N$ is a finite $A$-module and $u : M \to N$ is an $A$-module map such that $u \otimes 1 : M \otimes _ A R \to N \otimes _ A R$ is surjective, then for some $i$ the map $u \otimes 1 : M \otimes _ A R_ i \to N \otimes _ A R_ i$ is surjective.

If $N$ is a finitely presented $A$-module, and $v : N \otimes _ A R \to M \otimes _ A R$ is an $R$-module map, then there exists an $i$ and an $R_ i$-module map $v_ i : N \otimes _ A R_ i \to M \otimes _ A R_ i$ such that $v = v_ i \otimes 1$.

If $M$ is a finite $A$-module, $N$ is a finitely presented $A$-module, and $u : M \to N$ is an $A$-module map such that $u \otimes 1 : M \otimes _ A R \to N \otimes _ A R$ is an isomorphism, then for some $i$ the map $u \otimes 1 : M \otimes _ A R_ i \to N \otimes _ A R_ i$ is an isomorphism.

**Proof.**
To prove (1) assume $u$ is as in (1) and let $x_1, \ldots , x_ m \in M$ be generators. Since $N \otimes _ A R = \mathop{\mathrm{colim}}\nolimits _ i N \otimes _ A R_ i$ we may pick an $i \in I$ such that $u(x_ j) \otimes 1 = u'(x_ j) \otimes 1$ in $M \otimes _ A R_ i$, $j = 1, \ldots , m$. For such an $i$ we have $u \otimes 1 = u' \otimes 1 : M \otimes _ A R_ i \to N \otimes _ A R_ i$.

To prove (2) assume $u \otimes 1$ surjective and let $y_1, \ldots , y_ m \in N$ be generators. Since $N \otimes _ A R = \mathop{\mathrm{colim}}\nolimits _ i N \otimes _ A R_ i$ we may pick an $i \in I$ and $z_ j \in M \otimes _ A R_ i$, $j = 1, \ldots , m$ whose images in $N \otimes _ A R$ equal $y_ j \otimes 1$. For such an $i$ the map $u \otimes 1 : M \otimes _ A R_ i \to N \otimes _ A R_ i$ is surjective.

To prove (3) let $y_1, \ldots , y_ m \in N$ be generators. Let $K = \mathop{\mathrm{Ker}}(A^{\oplus m} \to N)$ where the map is given by the rule $(a_1, \ldots , a_ m) \mapsto \sum a_ j x_ j$. Let $k_1, \ldots , k_ t$ be generators for $K$. Say $k_ s = (k_{s1}, \ldots , k_{sm})$. Since $M \otimes _ A R = \mathop{\mathrm{colim}}\nolimits _ i M \otimes _ A R_ i$ we may pick an $i \in I$ and $z_ j \in M \otimes _ A R_ i$, $j = 1, \ldots , m$ whose images in $M \otimes _ A R$ equal $v(y_ j \otimes 1)$. We want to use the $z_ j$ to define the map $v_ i : N \otimes _ A R_ i \to M \otimes _ A R_ i$. Since $K \otimes _ A R_ i \to R_ i^{\oplus m} \to N \otimes _ A R_ i \to 0$ is a presentation, it suffices to check that $\xi _ s = \sum _ j k_{sj}z_ j$ is zero in $M \otimes _ A R_ i$ for each $s = 1, \ldots , t$. This may not be the case, but since the image of $\xi _ s$ in $M \otimes _ A R$ is zero we see that it will be the case after increasing $i$ a bit.

To prove (4) assume $u \otimes 1$ is an isomorphism, that $M$ is finite, and that $N$ is finitely presented. Let $v : N \otimes _ A R \to M \otimes _ A R$ be an inverse to $u \otimes 1$. Apply part (3) to get a map $v_ i : N \otimes _ A R_ i \to M \otimes _ A R_ i$ for some $i$. Apply part (1) to see that, after increasing $i$ we have $v_ i \circ (u \otimes 1) = \text{id}_{M \otimes _ R R_ i}$ and $(u \otimes 1) \circ v_ i = \text{id}_{N \otimes _ R R_ i}$. $\square$

Lemma 10.126.6. Suppose that $R = \mathop{\mathrm{colim}}\nolimits _{\lambda \in \Lambda } R_\lambda $ is a directed colimit of rings. Then the category of finitely presented $R$-modules is the colimit of the categories of finitely presented $R_\lambda $-modules. More precisely

Given a finitely presented $R$-module $M$ there exists a $\lambda \in \Lambda $ and a finitely presented $R_\lambda $-module $M_\lambda $ such that $M \cong M_\lambda \otimes _{R_\lambda } R$.

Given a $\lambda \in \Lambda $, finitely presented $R_\lambda $-modules $M_\lambda , N_\lambda $, and an $R$-module map $\varphi : M_\lambda \otimes _{R_\lambda } R \to N_\lambda \otimes _{R_\lambda } R$, then there exists a $\mu \geq \lambda $ and an $R_\mu $-module map $\varphi _\mu : M_\lambda \otimes _{R_\lambda } R_\mu \to N_\lambda \otimes _{R_\lambda } R_\mu $ such that $\varphi = \varphi _\mu \otimes 1_ R$.

Given a $\lambda \in \Lambda $, finitely presented $R_\lambda $-modules $M_\lambda , N_\lambda $, and $R$-module maps $\varphi _\lambda , \psi _\lambda : M_\lambda \to N_\lambda $ such that $\varphi \otimes 1_ R = \psi \otimes 1_ R$, then $\varphi \otimes 1_{R_\mu } = \psi \otimes 1_{R_\mu }$ for some $\mu \geq \lambda $.

**Proof.**
To prove (1) choose a presentation $R^{\oplus m} \to R^{\oplus n} \to M \to 0$. Suppose that the first map is given by the matrix $A = (a_{ij})$. We can choose a $\lambda \in \Lambda $ and a matrix $A_\lambda = (a_{\lambda , ij})$ with coefficients in $R_\lambda $ which maps to $A$ in $R$. Then we simply let $M_\lambda $ be the $R_\lambda $-module with presentation $R_\lambda ^{\oplus m} \to R_\lambda ^{\oplus n} \to M_\lambda \to 0$ where the first arrow is given by $A_\lambda $.

Parts (2) and (3) follow from Lemma 10.126.5. $\square$

Lemma 10.126.7. Let $A$ be a ring and let $B, C$ be $A$-algebras. Suppose that $R = \mathop{\mathrm{colim}}\nolimits _{i \in I} R_ i$ is a directed colimit of $A$-algebras.

If $B$ is a finite type $A$-algebra, and $u, u' : B \to C$ are $A$-algebra maps such that $u \otimes 1 = u' \otimes 1 : B \otimes _ A R \to C \otimes _ A R$ then for some $i$ we have $u \otimes 1 = u' \otimes 1 : B \otimes _ A R_ i \to C \otimes _ A R_ i$.

If $C$ is a finite type $A$-algebra and $u : B \to C$ is an $A$-algebra map such that $u \otimes 1 : B \otimes _ A R \to C \otimes _ A R$ is surjective, then for some $i$ the map $u \otimes 1 : B \otimes _ A R_ i \to C \otimes _ A R_ i$ is surjective.

If $C$ is of finite presentation over $A$ and $v : C \otimes _ A R \to B \otimes _ A R$ is an $R$-algebra map, then there exists an $i$ and an $R_ i$-algebra map $v_ i : C \otimes _ A R_ i \to B \otimes _ A R_ i$ such that $v = v_ i \otimes 1$.

If $B$ is a finite type $A$-algebra, $C$ is a finitely presented $A$-algebra, and $u \otimes 1 : B \otimes _ A R \to C \otimes _ A R$ is an isomorphism, then for some $i$ the map $u \otimes 1 : B \otimes _ A R_ i \to C \otimes _ A R_ i$ is an isomorphism.

**Proof.**
To prove (1) assume $u$ is as in (1) and let $x_1, \ldots , x_ m \in B$ be generators. Since $B \otimes _ A R = \mathop{\mathrm{colim}}\nolimits _ i B \otimes _ A R_ i$ we may pick an $i \in I$ such that $u(x_ j) \otimes 1 = u'(x_ j) \otimes 1$ in $B \otimes _ A R_ i$, $j = 1, \ldots , m$. For such an $i$ we have $u \otimes 1 = u' \otimes 1 : B \otimes _ A R_ i \to C \otimes _ A R_ i$.

To prove (2) assume $u \otimes 1$ surjective and let $y_1, \ldots , y_ m \in C$ be generators. Since $B \otimes _ A R = \mathop{\mathrm{colim}}\nolimits _ i B \otimes _ A R_ i$ we may pick an $i \in I$ and $z_ j \in B \otimes _ A R_ i$, $j = 1, \ldots , m$ whose images in $C \otimes _ A R$ equal $y_ j \otimes 1$. For such an $i$ the map $u \otimes 1 : B \otimes _ A R_ i \to C \otimes _ A R_ i$ is surjective.

To prove (3) let $c_1, \ldots , c_ m \in C$ be generators. Let $K = \mathop{\mathrm{Ker}}(A[x_1, \ldots , x_ m] \to N)$ where the map is given by the rule $x_ j \mapsto \sum c_ j$. Let $f_1, \ldots , f_ t$ be generators for $K$ as an ideal in $A[x_1, \ldots , x_ m]$. We think of $f_ j = f_ j(x_1, \ldots , x_ m)$ as a polynomial. Since $B \otimes _ A R = \mathop{\mathrm{colim}}\nolimits _ i B \otimes _ A R_ i$ we may pick an $i \in I$ and $z_ j \in B \otimes _ A R_ i$, $j = 1, \ldots , m$ whose images in $B \otimes _ A R$ equal $v(c_ j \otimes 1)$. We want to use the $z_ j$ to define a map $v_ i : C \otimes _ A R_ i \to B \otimes _ A R_ i$. Since $K \otimes _ A R_ i \to R_ i[x_1, \ldots , x_ m] \to C \otimes _ A R_ i \to 0$ is a presentation, it suffices to check that $\xi _ s = f_ j(z_1, \ldots , z_ m)$ is zero in $B \otimes _ A R_ i$ for each $s = 1, \ldots , t$. This may not be the case, but since the image of $\xi _ s$ in $B \otimes _ A R$ is zero we see that it will be the case after increasing $i$ a bit.

To prove (4) assume $u \otimes 1$ is an isomorphism, that $B$ is a finite type $A$-algebra, and that $C$ is a finitely presented $A$-algebra. Let $v : B \otimes _ A R \to C \otimes _ A R$ be an inverse to $u \otimes 1$. Let $v_ i : C \otimes _ A R_ i \to B \otimes _ A R_ i$ be as in part (3). Apply part (1) to see that, after increasing $i$ we have $v_ i \circ (u \otimes 1) = \text{id}_{B \otimes _ R R_ i}$ and $(u \otimes 1) \circ v_ i = \text{id}_{C \otimes _ R R_ i}$. $\square$

Lemma 10.126.8. Suppose that $R = \mathop{\mathrm{colim}}\nolimits _{\lambda \in \Lambda } R_\lambda $ is a directed colimit of rings. Then the category of finitely presented $R$-algebras is the colimit of the categories of finitely presented $R_\lambda $-algebras. More precisely

Given a finitely presented $R$-algebra $A$ there exists a $\lambda \in \Lambda $ and a finitely presented $R_\lambda $-algebra $A_\lambda $ such that $A \cong A_\lambda \otimes _{R_\lambda } R$.

Given a $\lambda \in \Lambda $, finitely presented $R_\lambda $-algebras $A_\lambda , B_\lambda $, and an $R$-algebra map $\varphi : A_\lambda \otimes _{R_\lambda } R \to B_\lambda \otimes _{R_\lambda } R$, then there exists a $\mu \geq \lambda $ and an $R_\mu $-algebra map $\varphi _\mu : A_\lambda \otimes _{R_\lambda } R_\mu \to B_\lambda \otimes _{R_\lambda } R_\mu $ such that $\varphi = \varphi _\mu \otimes 1_ R$.

Given a $\lambda \in \Lambda $, finitely presented $R_\lambda $-algebras $A_\lambda , B_\lambda $, and $R_\lambda $-algebra maps $\varphi _\lambda , \psi _\lambda : A_\lambda \to B_\lambda $ such that $\varphi \otimes 1_ R = \psi \otimes 1_ R$, then $\varphi \otimes 1_{R_\mu } = \psi \otimes 1_{R_\mu }$ for some $\mu \geq \lambda $.

**Proof.**
To prove (1) choose a presentation $A = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ m)$. We can choose a $\lambda \in \Lambda $ and elements $f_{\lambda , j} \in R_\lambda [x_1, \ldots , x_ n]$ mapping to $f_ j \in R[x_1, \ldots , x_ n]$. Then we simply let $A_\lambda = R_\lambda [x_1, \ldots , x_ n]/(f_{\lambda , 1}, \ldots , f_{\lambda , m})$.

Parts (2) and (3) follow from Lemma 10.126.7. $\square$

Lemma 10.126.9. Suppose $R \to S$ is a local homomorphism of local rings. There exists a directed set $(\Lambda , \leq )$, and a system of local homomorphisms $R_\lambda \to S_\lambda $ of local rings such that

The colimit of the system $R_\lambda \to S_\lambda $ is equal to $R \to S$.

Each $R_\lambda $ is essentially of finite type over $\mathbf{Z}$.

Each $S_\lambda $ is essentially of finite type over $R_\lambda $.

**Proof.**
Denote $\varphi : R \to S$ the ring map. Let $\mathfrak m \subset R$ be the maximal ideal of $R$ and let $\mathfrak n \subset S$ be the maximal ideal of $S$. Let

As partial ordering we take the inclusion relation. For each $\lambda = (A, B) \in \Lambda $ we let $R'_\lambda $ be the sub $\mathbf{Z}$-algebra generated by $a \in A$, and we let $S'_\lambda $ be the sub $\mathbf{Z}$-algebra generated by $b$, $b \in B$. Let $R_\lambda $ be the localization of $R'_\lambda $ at the prime ideal $R'_\lambda \cap \mathfrak m$ and let $S_\lambda $ be the localization of $S'_\lambda $ at the prime ideal $S'_\lambda \cap \mathfrak n$. In a picture

The transition maps are clear. We leave the proofs of the other assertions to the reader. $\square$

Lemma 10.126.10. Suppose $R \to S$ is a local homomorphism of local rings. Assume that $S$ is essentially of finite type over $R$. Then there exists a directed set $(\Lambda , \leq )$, and a system of local homomorphisms $R_\lambda \to S_\lambda $ of local rings such that

The colimit of the system $R_\lambda \to S_\lambda $ is equal to $R \to S$.

Each $R_\lambda $ is essentially of finite type over $\mathbf{Z}$.

Each $S_\lambda $ is essentially of finite type over $R_\lambda $.

For each $\lambda \leq \mu $ the map $S_\lambda \otimes _{R_\lambda } R_\mu \to S_\mu $ presents $S_\mu $ as the localization of a quotient of $S_\lambda \otimes _{R_\lambda } R_\mu $.

**Proof.**
Denote $\varphi : R \to S$ the ring map. Let $\mathfrak m \subset R$ be the maximal ideal of $R$ and let $\mathfrak n \subset S$ be the maximal ideal of $S$. Let $x_1, \ldots , x_ n \in S$ be elements such that $S$ is a localization of the sub $R$-algebra of $S$ generated by $x_1, \ldots , x_ n$. In other words, $S$ is a quotient of a localization of the polynomial ring $R[x_1, \ldots , x_ n]$.

Let $\Lambda = \{ A \subset R \mid \# A < \infty \} $ be the set of finite subsets of $R$. As partial ordering we take the inclusion relation. For each $\lambda = A \in \Lambda $ we let $R'_\lambda $ be the sub $\mathbf{Z}$-algebra generated by $a \in A$, and we let $S'_\lambda $ be the sub $\mathbf{Z}$-algebra generated by $\varphi (a)$, $a \in A$ and the elements $x_1, \ldots , x_ n$. Let $R_\lambda $ be the localization of $R'_\lambda $ at the prime ideal $R'_\lambda \cap \mathfrak m$ and let $S_\lambda $ be the localization of $S'_\lambda $ at the prime ideal $S'_\lambda \cap \mathfrak n$. In a picture

It is clear that if $A \subset B$ corresponds to $\lambda \leq \mu $ in $\Lambda $, then there are canonical maps $R_\lambda \to R_\mu $, and $S_\lambda \to S_\mu $ and we obtain a system over the directed set $\Lambda $.

The assertion that $R = \mathop{\mathrm{colim}}\nolimits R_\lambda $ is clear because all the maps $R_\lambda \to R$ are injective and any element of $R$ eventually is in the image. The same argument works for $S = \mathop{\mathrm{colim}}\nolimits S_\lambda $. Assertions (2), (3) are true by construction. The final assertion holds because clearly the maps $S'_\lambda \otimes _{R'_\lambda } R'_\mu \to S'_\mu $ are surjective. $\square$

Lemma 10.126.11. Suppose $R \to S$ is a local homomorphism of local rings. Assume that $S$ is essentially of finite presentation over $R$. Then there exists a directed set $(\Lambda , \leq )$, and a system of local homomorphism $R_\lambda \to S_\lambda $ of local rings such that

The colimit of the system $R_\lambda \to S_\lambda $ is equal to $R \to S$.

Each $R_\lambda $ is essentially of finite type over $\mathbf{Z}$.

Each $S_\lambda $ is essentially of finite type over $R_\lambda $.

For each $\lambda \leq \mu $ the map $S_\lambda \otimes _{R_\lambda } R_\mu \to S_\mu $ presents $S_\mu $ as the localization of $S_\lambda \otimes _{R_\lambda } R_\mu $ at a prime ideal.

**Proof.**
By assumption we may choose an isomorphism $\Phi : (R[x_1, \ldots , x_ n]/I)_{\mathfrak q} \to S$ where $I \subset R[x_1, \ldots , x_ n]$ is a finitely generated ideal, and $\mathfrak q \subset R[x_1, \ldots , x_ n]/I$ is a prime. (Note that $R \cap \mathfrak q$ is equal to the maximal ideal $\mathfrak m$ of $R$.) We also choose generators $f_1, \ldots , f_ m \in I$ for the ideal $I$. Write $R$ in any way as a colimit $R = \mathop{\mathrm{colim}}\nolimits R_\lambda $ over a directed set $(\Lambda , \leq )$, with each $R_\lambda $ local and essentially of finite type over $\mathbf{Z}$. There exists some $\lambda _0 \in \Lambda $ such that $f_ j$ is the image of some $f_{j, \lambda _0} \in R_{\lambda _0}[x_1, \ldots , x_ n]$. For all $\lambda \geq \lambda _0$ denote $f_{j, \lambda } \in R_{\lambda }[x_1, \ldots , x_ n]$ the image of $f_{j, \lambda _0}$. Thus we obtain a system of ring maps

Set $\mathfrak q_\lambda $ the inverse image of $\mathfrak q$. Set $S_\lambda = (R_\lambda [x_1, \ldots , x_ n]/ (f_{1, \lambda }, \ldots , f_{n, \lambda }))_{\mathfrak q_\lambda }$. We leave it to the reader to see that this works. $\square$

Remark 10.126.12. Suppose that $R \to S$ is a local homomorphism of local rings, which is essentially of finite presentation. Take any system $(\Lambda , \leq )$, $R_\lambda \to S_\lambda $ with the properties listed in Lemma 10.126.10. What may happen is that this is the “wrong” system, namely, it may happen that property (4) of Lemma 10.126.11 is not satisfied. Here is an example. Let $k$ be a field. Consider the ring

Set $S = R/zR$. As system take $\Lambda = \mathbf{N}$ and $R_ n = k[[z, y_1, \ldots , y_ n]]/(\{ y_ i^2 - zy_{i + 1}\} _{i \leq n-1})$ and $S_ n = R_ n/(z, y_ n^2)$. All the maps $S_ n \otimes _{R_ n} R_{n + 1} \to S_{n + 1}$ are not localizations (i.e., isomorphisms in this case) since $1 \otimes y_{n + 1}^2$ maps to zero. If we take instead $S_ n' = R_ n/zR_ n$ then the maps $S'_ n \otimes _{R_ n} R_{n + 1} \to S'_{n + 1}$ are isomorphisms. The moral of this remark is that we do have to be a little careful in choosing the systems.

Lemma 10.126.13. Suppose $R \to S$ is a local homomorphism of local rings. Assume that $S$ is essentially of finite presentation over $R$. Let $M$ be a finitely presented $S$-module. Then there exists a directed set $(\Lambda , \leq )$, and a system of local homomorphisms $R_\lambda \to S_\lambda $ of local rings together with $S_\lambda $-modules $M_\lambda $, such that

The colimit of the system $R_\lambda \to S_\lambda $ is equal to $R \to S$. The colimit of the system $M_\lambda $ is $M$.

Each $R_\lambda $ is essentially of finite type over $\mathbf{Z}$.

Each $S_\lambda $ is essentially of finite type over $R_\lambda $.

Each $M_\lambda $ is finite over $S_\lambda $.

For each $\lambda \leq \mu $ the map $S_\lambda \otimes _{R_\lambda } R_\mu \to S_\mu $ presents $S_\mu $ as the localization of $S_\lambda \otimes _{R_\lambda } R_\mu $ at a prime ideal.

For each $\lambda \leq \mu $ the map $M_\lambda \otimes _{S_\lambda } S_\mu \to M_\mu $ is an isomorphism.

**Proof.**
As in the proof of Lemma 10.126.11 we may first write $R = \mathop{\mathrm{colim}}\nolimits R_\lambda $ as a directed colimit of local $\mathbf{Z}$-algebras which are essentially of finite type. Next, we may assume that for some $\lambda _1 \in \Lambda $ there exist $f_{j, \lambda _1} \in R_{\lambda _1}[x_1, \ldots , x_ n]$ such that

Choose a presentation

of $M$ over $S$. Let $A \in \text{Mat}(t \times s, S)$ be the matrix of the presentation. For some $\lambda _2 \in \Lambda $, $\lambda _2 \geq \lambda _1$ we can find a matrix $A_{\lambda _2} \in \text{Mat}(t \times s, S_{\lambda _2})$ which maps to $A$. For all $\lambda \geq \lambda _2$ we let $M_\lambda = \mathop{\mathrm{Coker}}(S_\lambda ^{\oplus s} \xrightarrow {A_\lambda } S_\lambda ^{\oplus t})$. We leave it to the reader to see that this works. $\square$

Lemma 10.126.14. Suppose $R \to S$ is a ring map. Then there exists a directed set $(\Lambda , \leq )$, and a system of ring maps $R_\lambda \to S_\lambda $ such that

The colimit of the system $R_\lambda \to S_\lambda $ is equal to $R \to S$.

Each $R_\lambda $ is of finite type over $\mathbf{Z}$.

Each $S_\lambda $ is of finite type over $R_\lambda $.

**Proof.**
This is the non-local version of Lemma 10.126.9. Proof is similar and left to the reader.
$\square$

Lemma 10.126.15. Suppose $R \to S$ is a ring map. Assume that $S$ is integral over $R$. Then there exists a directed set $(\Lambda , \leq )$, and a system of ring maps $R_\lambda \to S_\lambda $ such that

The colimit of the system $R_\lambda \to S_\lambda $ is equal to $R \to S$.

Each $R_\lambda $ is of finite type over $\mathbf{Z}$.

Each $S_\lambda $ is of finite over $R_\lambda $.

**Proof.**
Consider the set $\Lambda $ of pairs $(E, F)$ where $E \subset R$ is a finite subset, $F \subset S$ is a finite subset, and every element $f \in F$ is the root of a monic $P(X) \in R[X]$ whose coefficients are in $E$. Say $(E, F) \leq (E', F')$ if $E \subset E'$ and $F \subset F'$. Given $\lambda = (E, F) \in \Lambda $ set $R_\lambda \subset R$ equal to the $\mathbf{Z}$-subalgebra of $R$ generated by $E$ and $S_\lambda \subset S$ equal to the $\mathbf{Z}$-subalgebra generated by $F$ and the image of $E$ in $S$. It is clear that $R = \mathop{\mathrm{colim}}\nolimits R_\lambda $. We have $S = \mathop{\mathrm{colim}}\nolimits S_\lambda $ as every element of $S$ is integral over $S$. The ring maps $R_\lambda \to S_\lambda $ are finite by Lemma 10.35.5 and the fact that $S_\lambda $ is generated over $R_\lambda $ by the elements of $F$ which are integral over $R_\lambda $ by our condition on the pairs $(E, F)$. The lemma follows.
$\square$

Lemma 10.126.16. Suppose $R \to S$ is a ring map. Assume that $S$ is of finite type over $R$. Then there exists a directed set $(\Lambda , \leq )$, and a system of ring maps $R_\lambda \to S_\lambda $ such that

The colimit of the system $R_\lambda \to S_\lambda $ is equal to $R \to S$.

Each $R_\lambda $ is of finite type over $\mathbf{Z}$.

Each $S_\lambda $ is of finite type over $R_\lambda $.

For each $\lambda \leq \mu $ the map $S_\lambda \otimes _{R_\lambda } R_\mu \to S_\mu $ presents $S_\mu $ as a quotient of $S_\lambda \otimes _{R_\lambda } R_\mu $.

**Proof.**
This is the non-local version of Lemma 10.126.10. Proof is similar and left to the reader.
$\square$

Lemma 10.126.17. Suppose $R \to S$ is a ring map. Assume that $S$ is of finite presentation over $R$. Then there exists a directed set $(\Lambda , \leq )$, and a system of ring maps $R_\lambda \to S_\lambda $ such that

The colimit of the system $R_\lambda \to S_\lambda $ is equal to $R \to S$.

Each $R_\lambda $ is of finite type over $\mathbf{Z}$.

Each $S_\lambda $ is of finite type over $R_\lambda $.

For each $\lambda \leq \mu $ the map $S_\lambda \otimes _{R_\lambda } R_\mu \to S_\mu $ is an isomorphism.

**Proof.**
This is the non-local version of Lemma 10.126.11. Proof is similar and left to the reader.
$\square$

Lemma 10.126.18. Suppose $R \to S$ is a ring map. Assume that $S$ is of finite presentation over $R$. Let $M$ be a finitely presented $S$-module. Then there exists a directed set $(\Lambda , \leq )$, and a system of ring maps $R_\lambda \to S_\lambda $ together with $S_\lambda $-modules $M_\lambda $, such that

The colimit of the system $R_\lambda \to S_\lambda $ is equal to $R \to S$. The colimit of the system $M_\lambda $ is $M$.

Each $R_\lambda $ is of finite type over $\mathbf{Z}$.

Each $S_\lambda $ is of finite type over $R_\lambda $.

Each $M_\lambda $ is finite over $S_\lambda $.

For each $\lambda \leq \mu $ the map $S_\lambda \otimes _{R_\lambda } R_\mu \to S_\mu $ is an isomorphism.

For each $\lambda \leq \mu $ the map $M_\lambda \otimes _{S_\lambda } S_\mu \to M_\mu $ is an isomorphism.

In particular, for every $\lambda \in \Lambda $ we have

**Proof.**
This is the non-local version of Lemma 10.126.13. Proof is similar and left to the reader.
$\square$

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