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
Comments (4)
Comment #1096 by Antoine Chambert-Loir on
Comment #1130 by Johan on
Comment #9336 by Manolis Tsakiris on
Comment #9430 by Stacks project on