Finitely generated algebra

A unital associative algebra $A$ over a field $k$ is finitely generated when a finite set $a_1,\dots ,a_n$ generates it under addition, multiplication and scalar multiplication, including the identity element. Equivalently, every element is a finite linear combination of words in the generators. The words retain their order if multiplication is noncommutative.

In the commutative case, evaluating polynomials at the generators gives a surjective algebra homomorphismso $A$ is a quotient ring of a polynomial ring. This equivalence assumes commutativity. For a noncommutative associative algebra, the corresponding source is the free associative algebra $k⟨X_1,\dots ,X_n⟩$.^[ref]

Finite generation as an algebra differs from being finite-dimensional as a vector space. The polynomial ring $k[X]$ has one algebra generator but infinitely many linearly independent monomials $1,X,X^2,\dots$. By contrast, the quotient ring $k[X]/(X^2)$ has basis $1,\bar{X}$, and its relation truncates every higher power.

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 commutative algebra over a field $k$ has a finite presentation if finitely many generators and finitely many polynomial relations describe it:The polynomial ring supplies freely chosen commuting generators, and the ideal of relations identifies expressions representing the same element. For example, $k[X,Y]/(XY-1)$ is the Laurent polynomial ring in one variable: $Y$ represents the multiplicative inverse of $X$.^[ref]

Every commutative finitely generated algebra over a field admits a finite presentation. Indeed, a field is a Noetherian ring; the Hilbert basis theorem makes its polynomial ring Noetherian, so the kernel of a chosen surjective algebra homomorphism is a finitely generated ideal. This is where commutativity enters the assertion. For a noncommutative associative algebra, a finite presentation instead uses a free associative algebra and finitely many generators for its two-sided ideal of relations; the commutative Hilbert basis theorem does not establish finite presentation there.^[ref]

A finite presentation also organizes differentiation: the module of Kähler differentials is generated by $d{X}_1,\dots ,d{X}_n$, with relations $d{f}_j={\sum }_i(\partial f_j/\partial X_i)d{X}_i$. The Jacobian matrix records those relations; this connects an algebraic description by equations to its infinitesimal geometry.

Let $A$ be a commutative algebra over a field $k$. Suppose $f_1,\dots ,f_r\in A$ generate the unit ideal, and every localization of a ring $A_{f_i}$ is a finitely generated algebra over $k$. Then $A$ is finitely generated over $k$.

Choose $g_i\in A$ with ${\sum }_i{g}_i{f}_i=1$. Let $B\subseteq A$ be the $k$-subalgebra generated by the $f_i,g_i$ and the numerators of finite generating sets of all $A_{f_i}$. Then $B$ is a finitely generated algebra and $B_{f_i}=A_{f_i}$. For $a\in A$, clear the finitely many denominators to choose one $N$ with $f_i^{N}a\in B$ for every $i$. The unit-ideal identity for powers supplies $1={\sum }_i{c}_i{f}_i^N$ with $c_i\in B$, giving $a={\sum }_i{c}_i{f}_i^{N}a\in B$. Thus $A=B$.

Let $A\subseteq B\subseteq C$ be commutative rings. If $A$ is Noetherian, $C$ is a finitely generated algebra over $A$, and $C$ is a finitely generated module over $B$, then $B$ is a finitely generated algebra over $A$.^[ref]

Choose algebra generators $x_1,\dots ,x_r$ for $C$ over $A$ and module generators $c_1=1,c_2,\dots ,c_m$ for $C$ over $B$. Express every $x_i{c}_j$ as a $B$-linear combination of the $c_k$. Let $B_0$ be the $A$-subalgebra generated by the finitely many coefficients in these expressions. The $B_0$-module $M={\sum }_k{B}_0{c}_k$ contains one and is stable under multiplication by every $x_i$, so $M=C$. Thus $C$ is finite as a $B_0$-module.

The Hilbert basis theorem makes $B_0$ Noetherian. Hence its submodule $B\subseteq C$ is finitely generated over $B_0$. Combining those module generators with the finite algebra generators of $B_0$ gives finitely many algebra generators for $B$ over $A$.

Let $L/k$ be a field extension and $A$ a $k$-algebra. If $A{\otimes }_{k}L$ is a finitely generated algebra over $L$, then $A$ is finitely generated over $k$.

Choose finitely many generators of $A{\otimes }_{k}L$. Each is a finite sum of pure tensors, so only finitely many coefficients from $A$ occur. Let $A_0\subseteq A$ be the $k$-subalgebra generated by those coefficients. The image of $A_0{\otimes }_{k}L$ contains every chosen generator and all scalars, hence equals $A{\otimes }_{k}L$. ThereforeHere $A/A_0$ is a quotient vector space, not a quotient algebra: $A_0$ need not be an ideal. The extension $L$ is a faithfully flat module over $k$, so this vanishing forces $A/A_0=0$. Thus $A=A_0$, proving finite generation.