Noether Normalization and Lie Algebra of derivations on a commutative algebra $A$

41 Views Asked by At

In my Lie algebra course, the professor asked us to use Noether's Normalization Theorem and use it to say something about the Lie algebra of derivations over a commutative, associative algebra. Now, Noether Normalization theorem goes like this :

Let $A$ be a commutative finitely generated algebra over a field $k$. Then there exists algebraically independent elements $z_1, \dots, z_d\in A$ (i.e. $B=k[z_1, \dots, z_d]$ is isomorphic to a polynomial ring) such that $A$ is a finitely generated module over $B$.

Now the question posed by our professor is certainly open-ended, but I ask this question because what I'm thinking seems too strong and feels like it classifies a big class of derivation algebras, which seems unlikely. So far what I have got is that, $\text{Der}(k[x_1, \dots, x_n])$ can be thought of as a free module over $k[x_1, \dots, x_n]$ with generators given by $\frac{\partial}{\partial x_i}$. So for a finitely generated commutative algebra $A$, $\text{Der}(A)$ is the ring of derivations of $A$ which is a module over $k[x_1, \dots, x_n]$. Intuitively I want to say that $$\text{Der}(A)\cong \text{Der}(k[x_1, \dots, x_n])\otimes_{k[x_1, \dots, x_n]} A?$$ I know this is true when the module is a localization, i.e. if $A=k[x_1, \dots, x_n][f^{-1}]$ then my claim holds. I expect it fails generally, but does the Noether Normalization theorem make life simpler in any ways at all?

Any comments and suggestions welcomed, I don't want a full solution but maybe something that might suggest I am moving in the right direction or in the wrong direction. Thanks in advance.