Exercise A0.III.6.15

Proposition

Let $R$ be a commutative ring and let $S$ be a commutative $R$-algebra. Then $S$ is finitely generated as an algebra over $R$ if and only if it is finitely generated as a commutative algebra over $R.$

Proof

Suppose $S$ is finitely generated as an algebra over $R.$ That is, for a finite set $A,$ we have a surjective homomorphism of $R$-algebras $\sigma : R\langle A \rangle \to S$ where $R\langle A \rangle$ is a monoid ring on the free monoid generated by $A.$

Let $C$ be the ideal of $R\langle A \rangle$ generated by all commutators $ab - ba.$ We have a quotient map $\pi : R\langle A \rangle \to R\langle A \rangle / C.$ Now, fix any commutative $R$-algebra $U$ and a set function $f : A \to U.$ By the universal property of $R\langle A \rangle$ this function extends to a morphism $f' : R\langle A \rangle \to U.$ But note that the codomain of $f'$ is commutative, so every commutator of $R\langle A \rangle$ gets killed by $f'.$ The universal property of the quotient gives us the unique morphism $\hat{f} : R\langle A \rangle / C \to U$ such that $\hat{f} \circ \pi \circ i = f,$ but this is just the universal property of a free object in the category of commutative $R$-algebras. Thus, $R\langle A \rangle / C \cong R[A].$

The universal property of quotient lets us factor $\sigma$ through $\sigma' : R\langle A \rangle / C \to S$ as $\sigma = \sigma' \circ \pi.$ Then $\sigma'$ is forced to be surjective and this gives a surjective homomorphism $\sigma' : R[A] \to S.$

Conversely, suppose $S$ is finitely generated as a commutative $R$-algebra. That is, there is a surjective morphism $\sigma' : R[A] \to S.$ We saw earlier that there exists a surjection $\pi : R\langle A \rangle \to R[A],$ so $(\sigma' \circ \pi) : R\langle A \rangle \to S$ gives the required surjective homomorphism.

$\blacksquare$