Proposition 2

Tag 04N0, permanent. Cite it as [Bourbaki, Tag 04N0].

Commutative Algebra, chapter V, § 1. Notion of an integral element

*Let $A, A'$ be two rings, $R$ an $A$ -algebra, $R$ an $A'$ -algebra (not necessarily commutative) and $f : A \to A'$ and $g : R \to R'$ two ring homomorphisms such that the diagram

\begin{array}{ccc} A & \xrightarrow{f} & A' \\ | & & | \\ \mathbf{R} & \xrightarrow{g} & \mathbf{R}' \end{array}

is commutative. If an element $x \in R$ is integral over $A$ , then $g(x)$ is integral over $A'$ .

If $x^n + a_1 x^{n-1} + \cdots + a_n = 0$ where $a_i \in A$ for $1 \leq i \leq n$ , we deduce that

(g(x))^n + f(a_1)(g(x))^{n-1} + \cdots + f(a_n) = 0.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

Cited by

Exercise 15, § 1
Corollary, § 1
Proposition 16, § 1
Corollary 1, § 1
Proposition 19, § 1
Proposition 20, § 1
Proposition 3, § 1
Proposition 5, § 1
Proposition 8, § 1
Proposition 4, § 2
Theorem 1 (Normalization Lemma), § 3
Corollary 1, § 3

Cites

Nothing.