Finitely generated as a module implies f.g. as an algebra

Question

I read some notes on commutative algebra and I got stuck on this proposition. Why finitely generated as an $R$-module implies finitely generated as an $R$-algebra by the same elements? How to deal with non-linear combinations, when we generate an algebra?

Answer 1

In general we have $$a_1R + \cdots + a_nR \subseteq R[a_1,\ldots,a_n]$$ but if $R' = a_1R + \cdots + a_nR$ is a extension ring of $R,$ then also $R'$ contains as a subset the smallest ring containing all the elements of $R$ along with the $a_1,\ldots,a_n.$ That is, $R' \supseteq R[a_1,\ldots,a_n].$

Answer 2

Let $M$ be finitely generated as an $R$-module, then every element in $M$ may be written as an $R$-linear combination of finitely many elements $m_1, \ldots, m_n \in M$. Being finitely generated as an $R$-algebra means (loosely speaking) that every element may be written as a polynomial with coefficients in $R$ and the $m_1, \ldots, m_n$ as variables. I.e. if $M$ is finitely generated as an $R$-module, then every element is a linear polynomial in the variables $m_1, \ldots, m_n$.

Intuitively speaking, "generating as an $R$-algebra" gives one more operations and thus one can generate more, whereas generating as a module only allows to form linear combinations.