Maximal ideal in a local artinian ring.
Matthew Harrington 
I know that an artinian ring $A$ is the union of its units and its zero-divisors.
So every non-zero-divisor is an unit.
I also know that in a local ring every element which is out from the maximal ideal is an unit.
Can I conclude that the set of zero-divisors is the maximal ideal of $A$?
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$\begingroup$Yes. In any commutative ring, the set of zero-divisors is the union of the prime ideals in $\operatorname{Ass}A$. An artinian ring has Krull dimension $0$, hence a local artinian ring has only $1$ prime ideal, and $\operatorname{Ass}A=\operatorname{Max}A=\operatorname{Spec}A$ is the set of zero-divisors in $A$.
$\endgroup$ 1 $\begingroup$I also know that in a local ring every element which is out from the maximal ideal is an unit.
Can I conclude that the set of zero-divisors [of an Artinian local ring] is the maximal ideal of $A$?
Yes: here is an elementary way to see it.
Suppose $M$ is the unique maximal ideal of a commutative Artinian ring $A$. Suppose $a$ is a nonzero element. Then the chain of ideals $aR\supseteq a^2R\supseteq a^3R\supseteq\ldots$ has to stabilize. At some point, $a^nR=a^{n+1}$, so that $a^n=a^{n+1}r$ for some $r\in A$.
Suppose $a$ is not a zero divisor. Then since $a^n(1-ar)=0$, we can cancel $a$'s until $1-ar=0$, and we find $a$ is a unit. Conclusion: $a$ is either a zero divisor or a unit.
Of course, if $M$ contained a unit, it would have to contain all of $A$, so $M$ contains no units. Everything inside must be a zero divisor.
This is actually true more generally for one-sided Artinian rings. Really the property that we used here is much weaker: the descending chain condition on chains of ideals of the form $aR\supseteq a^2R\supseteq a^3R\supseteq\ldots$ A ring that satisfies this chain condition is called strongly $\pi$-regular.
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