Finitely generated algebra

A $k$-algebra is finitely generated when finitely many elements generate it under addition, multiplication, and scalar multiplication. Equivalently, it is a quotient of a polynomial ring $k[x_1,\dots ,x_n]$.

Let $A$ be a commutative algebra over a field $k$, with a homomorphism $A\to k$ restricting to the identity on $k$, and let $I$ be its kernel. If $A$ is generated by finitely many elements, subtracting their images in $k$ gives finitely many generators in $I$. Every polynomial in those generators, modulo $I^2$, is a linear combination of their classes. Thus $I/I^2$, the algebraic cotangent space at this point, is finite-dimensional. For $A=k+xk[x,y]$ and $I=xk[x,y]$, the classes $x{y}^j$, $j\ge 0$, form an infinite basis modulo $I^2=x^{2}k[x,y]$. This is a subalgebra of a polynomial ring which is not a finitely generated algebra.

A presentation of an algebra over a field with finitely many generators and relations. The Hilbert basis theorem ensures that every finitely generated algebra over a field admits such a presentation. Its module of Kähler differentials is presented by the Jacobian matrix of the relations.

Suppose a $k$-algebra $A$ has elements $f_i,g_i$ with ${\sum }_i{g}_i{f}_i=1$, and each $A_{f_i}$ is a finitely generated algebra. Let $B\subseteq A$ contain the $f_i,g_i$ and numerators of finite generating sets of all $A_{f_i}$. Then $B$ is finitely generated and $B_{f_i}=A_{f_i}$. For $a\in A$, clear denominators so that all $f_i^{N}a$ belong to $B$. A unit-ideal identity for powers gives $1={\sum }_i{c}_i{f}_i^N$ in $B$, hence $a={\sum }_i{c}_i{f}_i^{N}a\in B$. Thus $A=B$.

Let $A\subseteq B\subseteq C$ be rings. If $A$ is Noetherian, $C$ is a finitely generated $A$-algebra, and $C$ is a finite $B$-module, then $B$ is a finitely generated $A$-algebra.

Let $L/k$ be a field extension. If $A{\otimes }_{k}L$ is a finitely generated algebra over $L$, then $A$ is finitely generated over $k$. Choose finitely many generators of $A{\otimes }_{k}L$ and collect the finitely many coefficients from $A$ occurring in them. The $k$-subalgebra $A_0$ generated by those coefficients satisfies $(A/A_0){\otimes }_{k}L=0$, and faithful flatness of $L/k$ gives $A=A_0$.