If $S$ is an $R$-algebra, how does $S$ being a finitely generated $R$-module imply that $S$ is a finite-type $R$-algebra

abstract-algebrafinitely-generatedmodulesring-homomorphismring-theory

Let $R$ be a commutative ring.

If $S$ is a finitely generated $R$-module, then there is an onto $R$-module homomorphism $$R^{\oplus n} \to S.$$

If $S$ is a finite-type $R$-algebra, then there is a onto $R$-algebra homomorphism $$R[x_1, \dots, x_k] \to S.$$


If $S$ is an $R$-algebra and a finitely generated $R$-module, then according this wiki, $S$ is a finite-type $R$-algebra.

I'm having trouble seeing this. How is the onto $R$-algebra hom defined?

Best Answer

Let $x_1, …, x_n$ be a generating system for $S$ as an $R$-module. Then the $R$-algebra morphism $$R[X_1,…, X_n] → S,~X_1 ↦ x_n,~…,~X_n ↦ x_n$$ is surjective, as its image certainly contains $⟨x_1, …, x_n⟩_R = S$.

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