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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.

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I suppose that $K$ is a field: please add that in an edit. The integral domain condition is necessary but the condition that $A$ be finitely generated over $K$ is not necessary. – Georges Elencwajg Jan 5 '13 at 9:28
up vote 10 down vote accepted

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.

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+1, beat me by a few seconds :) – Zev Chonoles Jan 5 '13 at 5:25
+1, yours is more general. Funny how often people here come up with the same counterexamples. Is it because they're natural, or because we've all been taught the same way? – Alex Becker Jan 5 '13 at 5:27
Well, perhaps not natural in the strongest meaning of the word, but I feel that this is a minimal example in a certain sense. – Zev Chonoles Jan 5 '13 at 5:34

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.

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@BenjaLim: Dear Benjamin, It's finite-dimensional, hence algebraic, over $\mathbb R$. (If you go through the proof that finite-dimensional implies algebraic, you can determine the actual minimal polynomials of the elements.) Regards, – Matt E Jan 5 '13 at 6:27

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.

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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 $k$.

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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.

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