The Stacks project

Proof. The category¹ $\mathcal{I}$ is nonempty as $R \to R$ 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 10.14.2). 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. $\square$

Proof. The first proof is that this follows from Lemma 10.127.1 and Categories, Lemma 4.21.5.

Second proof. Compare with the proof of Lemma 10.11.3. 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. $\square$

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 10.127.2 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 $R$. $\square$

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 10.127.3. Thus (1) implies (2).

Consider the category $\mathcal{I}$ of Lemma 10.127.1. 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. $\square$

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 $(u \otimes 1) \circ v_ i = \text{id}_{N \otimes _ R R_ i}$. $\square$

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 10.127.5. $\square$

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 $(u \otimes 1) \circ v_ i = \text{id}_{C \otimes _ R R_ i}$. $\square$

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 10.127.7. $\square$

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. $\square$

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. $\square$

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

Proof. As in the proof of Lemma 10.127.11 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. $\square$

Proof. This is the non-local version of Lemma 10.127.9. Proof is similar and left to the reader. $\square$

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 10.36.5 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. $\square$

Proof. This is the non-local version of Lemma 10.127.10. Proof is similar and left to the reader. $\square$

Proof. This is the non-local version of Lemma 10.127.11. Proof is similar and left to the reader. $\square$

Proof. This is the non-local version of Lemma 10.127.13. Proof is similar and left to the reader. $\square$