# Algebra of Algebraically Closed Field that is a field but not finitely generated as a module.

I was trying to understand a proof of the Weak Nullstellensatz. In the build-up to the proof it states the following Theorem, without proof.

If K is an algebraically closed field, then any K-algebra cannot be both a field and finitely generated as a K-module.

I'm not sure why this is true. Can somebody help me understand the proof to this?

• If $K\subset L$ is an algebraic field extension, and $K$ is algebraically closed, what can you say? – user26857 Apr 27 at 19:52
• If K is an algebraic field extension, we can say that every element of K is a root of some monic polynomial. If K is algebraically closed, every polynomial has a root. – algaeBurgers Apr 27 at 20:08
• I suggest you to try harder. (Eventually can take a look at this thread, but I won't do this before trying harder.) – user26857 Apr 27 at 20:42
• So I've got that if the K-algebra, L, is a field, then we must have K=L. I just need to make the connection to why this means that L cannot be generated as a K-module. If L is finitely generated as an K-module, then it is an integral extension. – algaeBurgers Apr 27 at 22:56