Artin–Tate lemma

The following proof can be found in Atiyah–MacDonald.[3] Let ${\displaystyle x_{1},\ldots ,x_{m}}$ generate ${\displaystyle C}$ as an ${\displaystyle A}$-algebra and let ${\displaystyle y_{1},\ldots ,y_{n}}$ generate ${\displaystyle C}$ as a ${\displaystyle B}$-module. Then we can write

${\displaystyle x_{i}=\sum _{j}b_{ij}y_{j}\quad {\text{and}}\quad y_{i}y_{j}=\sum _{k}b_{ijk}y_{k}}$

with ${\displaystyle b_{ij},b_{ijk}\in B}$. Then ${\displaystyle C}$ is finite over the ${\displaystyle A}$-algebra ${\displaystyle B_{0}}$ generated by the ${\displaystyle b_{ij},b_{ijk}}$. Using that ${\displaystyle A}$ and hence ${\displaystyle B_{0}}$ is Noetherian, also ${\displaystyle B}$ is finite over ${\displaystyle B_{0}}$. Since ${\displaystyle B_{0}}$ is a finitely generated ${\displaystyle A}$-algebra, also ${\displaystyle B}$ is a finitely generated ${\displaystyle A}$-algebra.

Without the assumption that A is Noetherian, the statement of the Artin–Tate lemma is no longer true. Indeed, for any non-Noetherian ring A we can define an A-algebra structure on ${\displaystyle C=A\oplus A}$ by declaring ${\displaystyle (a,x)(b,y)=(ab,bx+ay)}$. Then for any ideal ${\displaystyle I\subset A}$ which is not finitely generated, ${\displaystyle B=A\oplus I\subset C}$ is not of finite type over A, but all conditions as in the lemma are satisfied.

• http://commalg.subwiki.org/wiki/Artin-Tate_lemma