Let $R$ be a ring that is Artinian or Noetherian, and $x\in R$. Show that for some $n>0$, the image of $x$ in $R/0:x^n$ is a non zero divisor on that ring.
How should I approach this question? Should I assume that $R$ is Artinian and then try to prove the second statement and do the same when $R$ is Noetherian, or does it not matter?
Also what does it mean by "the image of $x$ in $R/0:x^n$"?
On
I assume that the notation is the same as Intuition for ideal quotient / colon ideal?, that is
$$(I : J) = \{r \in R: rJ \subset I\}$$
In the given question, we consider
$$(0 : x^n) = \{r \in R: rx^n \subset 0\} = \{r \in R: rx^n = 0\}$$
(I'm not going to go further with solving the question until you've had a chance to think about it while knowing the definition).
In answer to your first question, you may have already proved that any Artinian ring is Noetherian, in which case you can just show that $R$ Noetherian implies the property you're considering. However, if not, you will need to show that $R$ Artinian implies that property as well, which might well require a different proof.
I guess that $0:x^n=\{r\in R:rx^n=0\}$, which is an ideal in $R$ (assuming $R$ is commutative). If $I$ is an ideal of $R$ and $x\in R$, then the image of $x$ in $R/I$ is the equivalence class containing $x$, usually written $x+I$.
You need to prove that there exists $n$ such that $$ (a+(0:x^n))(x+(0:x^n))=0+(0:x^n) $$ implies $a+(0:x^n)=0+(0:x^n)$, that is, $a\in(0:x^n)$: this is the explicit translation of “the image of $x$ in $R/(0:x^n)$ is a nonzero divisor”.
Notice that if $rx^p=0$, then also $rx^q=0$, for every $q>p$. Therefore $$ (0:x)\subseteq(0:x^2)\subseteq(0:x^3)\subseteq\dotsb $$ is an ascending chain of ideals. What happens if $R$ is Noetherian?
An Artinian ring is also Noetherian (Hopkins-Levitzki theorem), so examining the Artinian case is not necessary.