
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.

1. 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 $u \otimes 1 = u' \otimes 1 : B \otimes _ A R_ i \to C \otimes _ A R_ i$.

2. 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.

3. 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 $v = v_ i \otimes 1$.

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

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).