**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

Comment #1130 by Johan on