Theorem 2

Tag 051T, permanent. Cite it as [Bourbaki, Tag 051T].

Commutative Algebra, chapter VII, § 3. Factorial domains

Let A be a factorial domain, K its field of fractions, $(p_\lambda)$ a representative system of extremal elements of A and $(P_\lambda)$ a representative system of irreducible polynomials of $K[X]$ , each $P_\lambda$ having 1 as a content. Then: (i) A[X] is a factorial domain; (ii) the set of $p_\lambda$ and $P_\lambda$ is a representative system of extremal elements of A[X].

Let $f$ be a non-zero element of $A[X]$ . In the ring $K[X]$ $f$ can be decomposed uniquely in the form:

f = a \prod_\lambda P_\lambda^{n(\lambda)} \quad (a \in K^*, n(\lambda) \geq 0).

Lemma 1 proves that $a$ is a content off. Hence $a \in A$ . As $A$ is factorial, $a$ can be decomposed uniquely in the form:

a = u \prod p_i^{m(i)} \quad (u \text{ invertible in } A,\ m(i) \geqslant 0).

Whence the existence and uniqueness of the decomposition:

f = u \prod p_i^{m(i)} \bigcap_{\lambda} p_{\lambda}^{n(\lambda)}.

Note that this theorem proves that every element of $A$ admits the same decomposition into extremal elements in $A$ and $A[X]$ . The g.c.d. of a family of elements of $A$ is therefore the same in $A$ and in $A[X]$ .

We may also use Proposition 18 of § 1, no. 10 to show that $A[X]$ is a factorial domain if and only if $A$ is a factorial domain.

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

Theorem 2

Cited by

Cites