# Finitely generated Algebra is a field iff is a domain

Let $K$ be a field and $A$ a finitely generated $K$-algebra (here, $A$ is also commutative and has a unit).

I'm trying to prove the well-known result that $A$ is a domain if and only if $A$ is a field.

I don't know how to prove that and I'm confused by this example: if $A=K[xy]$, we have that $A\subset K[x,y]$, so it is a finitely generated $K$-algebra. It is clearly a domain, but it is not a field, since $xy$ has no inverse.

What am I missing?

• Your statement is wrong! The correct statement is if $A$ is a $K$-algebra which is a finite dimensional vector space over $K$, then it is a domain if and only if it is a field. There are other variations of course. – Mohan May 9 '17 at 15:01
• Furthermore your example is flawed. Of course your $A$ is finitely generated but your reasoning fails...Subrings of polynomial rings are not necessarily finitely generated again. – MooS May 9 '17 at 15:12
• @MooS I've just realized that this reasoning is wrong in deed. – rmdmc89 May 9 '17 at 15:20
• Flawed reasoning aside, it was a little overcomplicated considering that $K[x]$ is already a finitely generated algebra that is a counterexample. – rschwieb May 9 '17 at 16:37
• If you want to stick with finitely generated algebras you can prove that if $A$ is a field then $A$ is a finite extension of $K$. – JJR May 10 '17 at 1:11