Lemma 10.127.2. 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 10.127.1 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. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)