Artinian ring without zero divisors is a field Definition: Let $R$ be a commutative algebra with $1$ over the field $k$. $R$ is an Artinian ring over $k$ if $R$ is a finite dimensional vector field over $k$.
Statement: If $R$ is an Artinian ring without zero divisors then it is a field.
I'm a bit stuck with it because as for me it is quite unusual definition of Artinian ring (despite of usual descending chain condition). The only advise that I have is to use the fact that over finite dimensional vector field any injective endomorphism is also surjective.
The usual way to prove that something is a field is to show that any non-zero element has invertible but I have no idea how to do it.

The statement you report is not a definition, but a theorem.

If $R$ is a commutative Artinian ring with no zero divisors, then $R$ is a field.

Take a minimal ideal $I$ of $R$ (which exists by the Artin condition). Let $r\in I$, $r\ne0$. Then $rI$ is an ideal of $R$; moreover $r^2\ne0$ (no zero divisors) and $rI\subset I$. Therefore $rI=I$ by minimality and so there is $s\in I$ with $rs=r$. It follows $s=1$ (no zero divisors) and $I=R$.
