TITLE: Is there a finitely generated, algebraic $K$-algebra $A$ that is not a field?
QUESTION [9 upvotes]: There is a well-known theorem that states that if $A$ is a finitely generated $K$-algebra, an integral domain and algebraic over $K$, then $A$ is a field. Is the integral domain condition necesary? I mean, is there an example of an algebraic algebra over $K$, such that is not a field? It may be kind of simple, but I'm a bit confused. Thank you.

REPLY [7 votes]: After all these answers, this is not much more than a comment really. Being finitely generated by algebraic elements over $K$ implies (and is equivalent to) being finite dimensional as a vector space over $K$. And a commutative $K$ algebra that is finite dimesional is a field if and only if it is an integral domain, much in the same way as a finite commutative ring is a field if and only if it is an integral domain. In both cases the finiteness condition ensures that for the operation of multiplication by a fixed element injectivity implies surjectivity, in other words regular elements are invertible. And as the examples show there is no way you can take injectivity for granted: rings with zero divisors are very easy to construct.
I might add that if you drop the condition "integral domain", then your $K$-algebra need not even be commutatitive, and since (in English terminology) fields are supposed to be commutatitive, the quaternions would be another counterexample to your guess.