Let $R$ be a commutative ring. Consider $R[X_1,…,X_n]$. Clearly it is a natural $R$-module.
Is it true that $R[X_1,…,X_n]$ is a finitely generated $R$-module? If it is, then every its ideal is a finitely generated $R$-module too right?
For me the answer is yes for both questions since every element $f\in R[X_1,…,X_n]$ has the form $\sum_{i=0}^nr_iX^i$ with $r_0,…,r_n\in R$, so the set of generators is $\{X_1,…,X_n\}$.
Thank you.
Best Answer
It is NOT finitely generated as an $R$-module. It is finitely generated as an algebra, though. Every element is not of the form $\sum_{i=0}^n r_i X^i$. The correct statement would be that every element is of the form $\sum\limits_{I=(i_1,\dots,i_n)} r_I X_1^{i_1}\cdots X_n^{i_n}$. All the monomials $X_1^{i_1} \cdots X_n^{i_n}$ are linearly independent, and so the generating set (as an $R$-module) is infinite.