Bourbaki

Theorem 1

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

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

Let A\mathbf{A} be a (commutative) ring, R\mathbf{R} an algebra over A\mathbf{A} (not necessarily commutative) and xx an element of R\mathbf{R}. The following properties are equivalent:

(E₁) xx is a root of a monic polynomial in the polynomial ring A[X]\mathbf{A}[X]. (E₂) The subalgebra A[x]\mathbf{A}[x] of R\mathbf{R} is a finitely generated A\mathbf{A}-module. (E₃) There exists a faithful module over the ring A[x]\mathbf{A}[x] which is a finitely generated A\mathbf{A}-module.

Let us show first that (E₁) implies (E₂). Let

Xn+a1Xn−1+…+an X^n + a_1 X^{n-1} + \ldots + a_n
be a monic polynomial in A[X]\mathbf{A}[X] with xx as a root; for every integer q≥0q \geq 0 let MqM_q be the sub-A\mathbf{A}-module of R\mathbf{R} generated by 1,x,…,xn+q1, x, \ldots, x^n + q. Then
xn+q=−an+q−1xn+q−1−…−anxq∈M−1 x^{n+q} = -a_{n+q-1} x^{n+q-1} - \ldots - a_n x^q \in M_{-1}
for all q≥1q \geq 1, whence, by induction on qq,
Mq=Mq−1=⋯=M0. M_q = M_{q-1} = \cdots = M_0.

(*) The results of this chapter and the next one depend on no Book other than Books I to VI, nor on Chapter I, § 4, nor on Chapter 111, § 5.

We conclude that A[x]A[x] is equal to M0M_0 and is therefore a finitely generated AA-module.

As the commutative ring A[x]A[x] is a faithful module over itself, (EIIE_{II}) implies (EIIIE_{III}).

Finally, the fact that (EIIIE_{III}) implies (E1E_1) will follow from the following more precise lemma:

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

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