Bourbaki

Exercise 1, § 1

Algebra, VII

Show that if AA is an integral domain then the polynomial ring A[Xi]i∈IA[X_i]_{i \in I} is not a principal ideal domain when one of the following conditions holds : 1) Card⁡(I)≥2\operatorname{Card}(I) \geq 2; 2) AA is not a field (consider the ideals (Xi)+(Xj)(X_i) + (X_j) for a≠βa \neq \beta. and (a)+(Xi)(a) + (X_i) where a≠()a \neq () is a noninvertible element of AA).

Solution

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Transcribed from the printed volume. Not translated.