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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?

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    $\begingroup$ 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. $\endgroup$
    – Bernard
    Aug 20, 2019 at 12:15

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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.

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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, $R' \supseteq R[a_1,\ldots,a_n].$

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  • $\begingroup$ 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? $\endgroup$
    – Kirill Losev
    Aug 20, 2019 at 12:28
  • $\begingroup$ Yes. Since $R'$ is a ring, every product $a_ia_j$ can be written as a sum of $R$-multiples of the $a_1, \ldots, a_n$ $\endgroup$
    – Brian Moehring
    Aug 20, 2019 at 12:33

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