Is there a finitely generated, algebraic $K$-algebra $A$ that is not a field? 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.
 A: Another example is given by $K\times K$ (the product ring, thought of as a $K$-algebra via the diagonal embedding of $K$).  Any degree two example is either of this form, or of the form $K[x]/(x^2)$ considered in the other answers.
A: 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.
A: For any field $K$, the algebra $A=K[x]/(x^2)$ is a finitely-generated $K$-algebra which is algebraic over $K$, but which is not an integral domain and certainly is not a field.
A: Here is a large class of examples coming from a structure theorem. Suppose we restrict our attention to finite-dimensional commutative algebras over a field $k$ (these are automatically both finitely generated and algebraic). Such algebras are Artinian, and a structure theorem asserts that all Artinian rings are finite direct products of Artinian local rings. Examples which are finite-dimensional commutative over $k$ include finite direct products of the rings of the form $k[x]/x^n$, or more generally $K[x]/x^n$ where $K$ is a finite extension of $k$. 
A: Sure. Consider $A=\mathbb R[x]/(x^2)$, which is generated by $\{1,x\}$ over $\mathbb R$ and is algebraic over $\mathbb R$ since $x^2=0$, yet is clearly not an integral domain hence not a field.
A: You don't need the assumption that $\mathcal A$ is a finitely generated $K$-algebra.
Let $\mathcal A$ be a $K$-algebra that is algebraic over $K$ and an integral domain. Let $a$ be a non-zero element of $\mathcal A$. Since $\mathcal A$ is algebraic, there exists a polynomial $f(x)\in K[x]$ such that $f(a)=0$. Define a homomorphism $$\phi: K[x]\rightarrow K[a]\\g(x)\mapsto g(a)$$ The map is surjective and $(f(x))\subseteq \ker(\phi)\subseteq K[x]$. We have $$K[x]/\ker(\phi)\cong K[a]$$
$K[x]$ is a PID, and $K[a]$ is an integral domain so $\ker(\phi)$ is a maximal ideal. It follows that $a$ is invertible, so $\mathcal A$ is a field.

For completeness:

*

*$\Bbb Q[x]/(x^2)$ is an algebraic $\Bbb Q$-algebra that is neither a integral domain, nor a field. Any element in $\Bbb Q[x]/(x^2)$ may be written $a+bx$, and is a root of the polynomial $y^2-2ay+a^2$.


*$\Bbb R[x]$ is a $\Bbb Z[x]$-algebra that is an integral domain, but neither algebraic nor a field. (I'm not sure if there is a counter-example when $\mathcal A$ is a finitely generated $K$-algebra)
