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.
 A: 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 $I=R$.
A: Let $a  \in R $ \ $\{0\}$ .
Take the sequence of ideals $I_n=(a^n)$ and notice that for every $n$ we have $I_{n+1} \subseteq I_n$.
From the definition of artinian rings $\exists n_0 \in \mathbb{N}$ such that $(a^{n_o})=( a^{n_0+1})=...\Longrightarrow \exists r \in R $ such that $a^{n_0}=ra^{n_0+1} \Longrightarrow a^{n_0}(ar-1)=0$
$R$ has no zero divisors thus $a^{n_0} \neq0$,therefore $ar=1$ proving that $a$ is an invertible elements of $R$
I assumed that $R$ is an artinian integral domain.
If it is not then in the same way you can prove that $a$ has a left and a right inverse,$r_1,r_2 \in R$ and if is that the case , then we know(or you can prove again)  that $r_1=r_2$
A: If your definition of an artinian ring is ‘a finite dimensional algebra $R$ over a field $k$’, just consider, for any $x\in R$, the endomorphism (as a $k$-vector spaceof multiplication by $x$ in $R$: if $x$ is a non-zero divisor, this endomorphism is injective, hence bijective since $R$ is finite-dimensional over $k$. This means $1$ is attained by this endomorphism, i.e. there exists $y\in R$ such that $xy=1$. In other words any $x\ne 0$ has an inverse in $R$.
More generally, one has the following general result:

Let $R,S$ be integral domains and  $R\hookrightarrow S$ an injective ring homomorphism.Suppose $S$ is a finite $R$-module. Then $S$ is a field if and only if $R$ is a field.

