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

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? • Well, if every element is a linear polynomial i $a_1,\dots, a_n$, *a fortiori, it is a ponlynomial in these same elements. He who can do more can do less. Aug 20, 2019 at 12:15

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.
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, $$R' \supseteq R[a_1,\ldots,a_n].$$
• Roughly speaking, all possible powers and products of $a_1,...,a_n$ is in $R^\prime$, hence I can rewrite any polynomial as a linear expression, right? Aug 20, 2019 at 12:28
• Yes. Since $R'$ is a ring, every product $a_ia_j$ can be written as a sum of $R$-multiples of the $a_1, \ldots, a_n$ Aug 20, 2019 at 12:33