The Stacks project

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

  1. $\varphi $ is of finite presentation,

  2. 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

  3. 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

\[ S \to S_\lambda \to S \]

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 $R$. $\square$


Comments (2)

Comment #1096 by Antoine Chambert-Loir on

It is not true that a factorization can only exist if . However, it follows from 00F4 (4) that this factorization implies that is finitely presented.

Comment #1130 by on

Oops, yes, that is a mistake. Actually, this also simplifies the thing a bit (hopefully correct this time). Here is the fix.


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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 00QO. Beware of the difference between the letter 'O' and the digit '0'.