In Miles Reid's Undergraduate Commutative Algebra he defines a ring $B$ to be finite as an $A$ – algebra if it is finite as an $A$ – module. Now what I don't understand is suppose we look at the polynomial ring $k[x_1,\ldots,x_n]$ where $k$ is a field. Then as a $k$ – algebra it is finitely generated. Is this the same as being a finite $k$ – algebra? For if it is the same this means that $k[x_1,\ldots,x_n]$ is finitely generated as a $k$ – module which is just a $k$ – vector space. However this cannot be possible because the $x_i's$ are not even algebraic over $k$. What am I misunderstanding here?
[Math] Definition of a finitely generated $k$ – algebra
commutative-algebra
Best Answer
An $A$-algebra $B$ is called finite if $B$ is a finitely generated $A$-module, i.e. there are elements $b_1,\dotsc,b_n \in B$ such that $B=A b_1 + \dotsc + A b_n$. It is called finitely generated / of finite type if $B$ is a finitely generated $A$-algebra, i.e. there are elements $b_1,\dotsc,b_n$ such that $B=A[b_1,\dotsc,b_n]$. Clearly every finite algebra is also a finitely generated one. The converse is not true (consider $B=A[T]$). However, there is the following important connection:
For example, $\mathbb{Z}[\sqrt{2}]$ is of finite type over $\mathbb{Z}$ and integral, thus finite. In fact, $1,\sqrt{2}$ is a basis as a module. You can find the proof of the claim above in every introduction to commutative algebra.