Subalgebra generated by a subset, algebras of finite type, and module-finite algebras

Let $R$ be a commutative ring and let $A$ be a commutative $R$-algebra with structure map ${\eta }_A\text{\hspace{0.17em}⁣}:R\to A$ (Algebras over a commutative ring, central structure maps, and algebra homomorphisms).

The subalgebra generated by finitely many elements. Let $n\in \mathbb{N}$ and $a_1,\dots ,a_n\in A$. Iterating the universal property of a polynomial ring (Universal property of $R[x]$: a coefficient homomorphism and the image of $x$ determine a unique ring homomorphism) along the recursion $R[x_1,\dots ,x_{k+1}]=R[x_1,\dots ,x_k][x_{k+1}]$ of Polynomial rings in finitely many commuting indeterminates by iteration gives a unique unital ring homomorphism

$\mathrm{ev}\text{\hspace{0.17em}⁣}:R[x_1,\dots ,x_n]⟶A$

that agrees with ${\eta }_A$ on constants and sends $x_i$ to $a_i$ for each $i$; at each step of the recursion the one-variable universal property supplies existence and uniqueness of the extension, and at $n=0$ the map is ${\eta }_A$ itself. Its image is written $R[a_1,\dots ,a_n]$ and is called the $R$-subalgebra of $A$ generated by $a_1,\dots ,a_n$. It is a subring of $A$ containing ${\eta }_A(R)$, and it is the smallest such subring containing $a_1,\dots ,a_n$, since any subring with those properties is closed under the sums and products that make up a polynomial expression. At $n=0$ it is ${\eta }_A(R)$, the image of $R$ in $A$.

Finite type. $A$ is of finite type over $R$, equivalently a finitely generated $R$-algebra, when $A=R[a_1,\dots ,a_n]$ for some $n\in \mathbb{N}$ and some $a_1,\dots ,a_n\in A$. Equivalently, $A$ is isomorphic as an $R$-algebra to a quotient $R[x_1,\dots ,x_n]\mathrm{/}\mathfrak{a}$ for some $n\in \mathbb{N}$ and some ideal $\mathfrak{a}$: the map $\mathrm{ev}$ above is then surjective, and First isomorphism theorem for rings: $R\mathrm{/}\ker f\cong \mathrm{im}f$ identifies $A$ with $R[x_1,\dots ,x_n]\mathrm{/}\ker \mathrm{ev}$; conversely the composite of the canonical projection with the inclusion of the indeterminates exhibits any such quotient as generated by the residues of $x_1,\dots ,x_n$.

Module-finite. $A$ is module-finite over $R$, equivalently a finite $R$-algebra, when $A$ is finitely generated as an $R$-module (Generated submodule, cyclic and finitely generated modules, module basis and free module) for the action $r\cdot a={\eta }_A(r)a$.

Module-finite implies finite type. If $b_1,\dots ,b_n$ generate $A$ as an $R$-module then $R[b_1,\dots ,b_n]$, being a subring of $A$ that contains ${\eta }_A(R)$ and every $b_i$, contains every $R$-linear combination ${\sum }_i{\eta }_A(r_i)b_i$ and hence all of $A$; so $A=R[b_1,\dots ,b_n]$. The converse fails, and the companion examples page carries a witness.

• Three conditions, three rings. "Finitely generated" is ambiguous on its own: an ideal may be finitely generated as an ideal, a module as a module, and an algebra as an algebra, and the three are different requirements. This page writes "of finite type" for the algebra condition and "module-finite" for the module condition, and always names the ring over which the condition is taken. Sources differ in vocabulary: Totaro and Milne write "finite algebra" for what is called module-finite here, and Altman–Kleiman write "module finite" and "algebra finite".

• The generators need not be algebraically independent. Nothing above asks $\mathrm{ev}$ to be injective. When it is, $A$ is a polynomial ring and the ideal $\mathfrak{a}$ is zero; that is a special case, not the definition.

• The empty list is allowed and is not the same as $A=R$. At $n=0$ the subalgebra generated is ${\eta }_A(R)$, which is a quotient of $R$ rather than a copy of it unless ${\eta }_A$ is injective.