The Stacks project

Proof. By induction on the degree of $P$. If $\deg (P) = 1$, then $P(T) = T - \alpha $ and the result is true. If $\deg (P) > 1$, then we can write $P(T) = (T - \alpha )Q(T) + r$ for some polynomial $Q \in R[T]$ of degree $< \deg (P)$ and some $r \in R$ by long division. By assumption $0 = P(\alpha ) = (\alpha - \alpha )Q(\alpha ) + r = r$ and we conclude that $r = 0$ as desired. $\square$

Proof. Write $P(T) = T^ d + a_1T^{d - 1} + \ldots + a_0$. Set $R' = R[x]/(x^ d + a_1x^{d - 1} + \ldots + a_0)$. Set $\alpha $ equal to the congruence class of $x$. Then it is clear that $P(\alpha ) = 0$. Thus we win by Lemma 15.21.1. $\square$

Proof. Let $x_1, \ldots , x_ n \in S$ be generators of $S$ over $R$. For each $i$ we can choose a monic polynomial $P_ i(T) \in R[T]$ such that $P_ i(x_ i) = 0$ in $S$, see Algebra, Lemma 10.36.3. Say $\deg (P_ i) = d_ i$. By Lemma 15.21.2 (applied $\sum d_ i$ times) there exists a finite free ring extension $R \subset R'$ such that each $P_ i$ splits completely:

for certain $\alpha _{ik} \in R'$. Let $R'[T_1, \ldots , T_ n] \to S \otimes _ R R'$ be the $R'$-algebra map which maps $T_ i$ to $x_ i \otimes 1$. As this maps $P_ i(T_ i)$ to zero, this induces the desired surjection. $\square$

Proof. This lemma proves itself. Hint: $V(\bigcap J_{\underline{k}}) = \bigcup V(J_{\underline{k}})$. $\square$

Proof. Let $M$ be an $R$-module such that $M \otimes _ R S$ is flat over $S$. By Algebra, Lemma 10.39.8 in order to prove that $M$ is flat we may replace $R$ by any faithfully flat ring extension. By Lemma 15.21.3 we can find a finite locally free ring extension $R \subset R'$ such that $S' = S \otimes _ R R' = R'[T_1, \ldots , T_ n]/J$ for some ideal $J \subset R'[T_1, \ldots , T_ n]$ which contains the elements of the form $P_ i(T_ i)$ with $P_ i(T) = \prod _{j = 1, \ldots , d_ i} (T - \alpha _{ij})$ for some $\alpha _{ij} \in R'$. Note that $R'$ is Noetherian and that $R' \subset S'$ is a finite extension of rings. Hence we may replace $R$ by $R'$ and assume that $S$ has a presentation as in Lemma 15.21.4. Note that $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ is surjective, see Algebra, Lemma 10.36.17. Thus, using Lemma 15.21.4 we conclude that $I = \bigcap J_{\underline{k}}$ is an ideal such that $V(I) = \mathop{\mathrm{Spec}}(R)$. This means that $I \subset \sqrt{(0)}$, and since $R$ is Noetherian that $I$ is nilpotent. The maps $\Phi _{\underline{k}}$ induce commutative diagrams

from which we conclude that $M/J_{\underline{k}}M$ is flat over $R/J_{\underline{k}}$. By Lemma 15.16.1 we see that $M/IM$ is flat over $R/I$. Finally, applying Algebra, Lemma 10.101.5 we conclude that $M$ is flat over $R$. $\square$

Proof. Choose a presentation

Let's say that the first map is given by the $r \times t$-matrix $T = (f_{ij})$ with $f_{ij} \in R[x_1, \ldots , x_ n]$. Write $f_{ij} = \sum f_{ij, I} x^ I$ with $f_{ij, I} \in R$ (multi-index notation). Consider diagrams

where $R_\lambda $ is a finitely generated $\mathbf{Z}$-subalgebra of $R$ containing all $f_{ij, I}$ and $S_\lambda $ is a finite $R_\lambda $-subalgebra of $S$. Let $M_\lambda $ be the finite $R_\lambda [x_1, \ldots , x_ n]$-module defined by a presentation as above, using the same matrix $T$ but now viewed as a matrix over $R_\lambda [x_1, \ldots , x_ n]$. Note that $S$ is the directed colimit of the $S_\lambda $ (details omitted). By Algebra, Lemma 10.168.1 we see that for some $\lambda $ the module $M_\lambda \otimes _{R_\lambda } S_\lambda $ is flat over $S_\lambda $. By Lemma 15.21.5 we conclude that $M_\lambda $ is flat over $R_\lambda $. Since $M = M_\lambda \otimes _{R_\lambda } R$ we win by Algebra, Lemma 10.39.7. $\square$