Lemma 15.21.3. Let $R \to S$ be a finite ring map. There exists a finite free ring extension $R \subset R'$ such that $S \otimes _ R R'$ is a quotient of a ring of the form

$R'[T_1, \ldots , T_ n]/(P_1(T_1), \ldots , P_ n(T_ n))$

with $P_ i(T) = \prod _{j = 1, \ldots , d_ i} (T - \alpha _{ij})$ for some $\alpha _{ij} \in R'$.

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 $P(x_ i) = 0$ 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:

$P_ i(T) = \prod \nolimits _{j = 1, \ldots , d_ i} (T - \alpha _{ij})$

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. $\square$

Comment #6603 by Jonas Ehrhard on

In the second sentence of the proof it should be $P_i(x_i) = 0$ (the index $i$ is missing).

There are also:

• 2 comment(s) on Section 15.21: Descent of flatness along integral maps

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).