Lemma 10.136.14 (03HS)—The Stacks project

Lemma 10.136.14. Suppose that $A$ is a ring, and $P(x) = x^ n + b_1 x^{n-1} + \ldots + b_ n \in A[x]$ is a monic polynomial over $A$. Then there exists a syntomic, finite locally free, faithfully flat ring extension $A \subset A'$ such that $P(x) = \prod _{i = 1, \ldots , n} (x - \beta _ i)$ for certain $\beta _ i \in A'$.

Proof. Take $A' = A \otimes _ R S$, where $R$ and $S$ are as in Example 10.136.8, where $R \to A$ maps $a_ i$ to $b_ i$, and let $\beta _ i = -1 \otimes \alpha _ i$. Observe that $R \to S$ is syntomic (Lemma 10.136.13), $R \to S$ is finite by construction, and $R$ is Noetherian (so any finite $R$-module is finitely presented). Hence $S$ is finite locally free as an $R$-module by Lemma 10.78.2. We omit the verification that $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ is surjective, which shows that $S$ is faithfully flat over $R$ (Lemma 10.39.16). These properties are inherited by the base change $A \to A'$; some details omitted. $\square$

Comments (2)

Comment #7916 by Masugi Kazuki on December 10, 2022 at 11:53

Sorry, I cannot understand why $R \to S$ is locally free, and $A \to A'$ is injection. (c.i.-ness of fiber is okay: because of finiteness, if $R \to S$ is locally free, it is faithfully flat and syntomic.)

Comment #8168 by Aise Johan de Jong on February 28, 2023 at 10:45

OK, thanks for your comment. I have fixed this here by adding a reference to Lemma 10.136.13. I encourage the reader to prove directly that the map $R \to S$ of Example 10.136.8 is finite free of positive rank directly.

There are also:

2 comment(s) on Section 10.136: Syntomic morphisms

Comments (2)

Post a comment