// Workers AI · dad joke modeWhat did morphism of finite type say? "I'm changing my form.

In commutative algebra, given a homomorphism $A\to B$ of commutative rings, $B$ is called an $A$-algebra of finite type if $B$ can be finitely generated as an $A$-algebra. It is much stronger for $B$ to be a finite $A$-algebra, which means that $B$ is finitely generated as an $A$-module. For example, for any commutative ring $A$ and natural number $n$, the polynomial ring $A[x_{1},\dots ,x_{n}]$ is an $A$-algebra of finite type, but it is not a finite $A$-algebra unless $A$ = 0 or $n$ = 0. Another example of a finite-type homomorphism that is not finite is $\mathbb {C} [t]\to \mathbb {C} [t][x,y]/(y^{2}-x^{3}-t)$.

The analogous notion in terms of schemes is that a morphism $f:X\to Y$ of schemes is of finite type if $Y$ has a covering by affine open subschemes $V_{i}=\operatorname {Spec} (A_{i})$ such that $f^{-1}(V_{i})$ has a finite covering by affine open subschemes $U_{ij}=\operatorname {Spec} (B_{ij})$ of $X$ with $B_{ij}$ an $A_{i}$-algebra of finite type. One also says that $X$ is of finite type over $Y$.

For example, for any natural number $n$ and field $k$, affine $n$-space and projective $n$-space over $k$ are of finite type over $k$ (that is, over $\operatorname {Spec} (k)$), while they are not finite over $k$ unless $n$ = 0. More generally, any quasi-projective scheme over $k$ is of finite type over $k$.

The Noether normalization lemma says, in geometric terms, that every affine scheme $X$ of finite type over a field $k$ has a finite surjective morphism to affine space $\mathbf {A} ^{n}$ over $k$, where $n$ is the dimension of $X$. Likewise, every projective scheme $X$ over a field has a finite surjective morphism to projective space $\mathbf {P} ^{n}$, where $n$ is the dimension of $X$.

References

Bosch, Siegfried (2013). Algebraic Geometry and Commutative Algebra. London: Springer. pp. 360–365. ISBN 9781447148289.