Let $a$ be an element of ring $R, \ I=\{r \in R: ra=0\}$ and $a$ is an element of every non-zero ideal of $R.$ Could anyone advise me how to show $I$ is a maximal ideal in $R\ ?$
My attempt: Let $J \neq R$ be an non-zero ideal of $R$ where $ \ I \subseteq J$ and $j \in J.$ Then, $ja \in I.$ How do I show $j \in I \ ?$
Hints will suffice, thank you.
Claim. If $a\in R$ belongs to every nonzero left ideal of $R$, then its left annihilator $I = \{r\in R\;:\;ra=0\}$ is a maximal proper left ideal.
Reason: Assume not, and that $J$ is a maximal proper left ideal of $R$ extending $I$. (This means: $I\subsetneq J\subsetneq R$, and there is no left ideal between $J$ and $R$.)
Choose $j\in J-I$. Then $ja\neq 0$, since $j\notin I$, so $Rja$ is a nonzero left ideal of $R$. Now $a\in Rja$, by the assumption on $a$, which means that there is some $s\in R$ so that $a=sja$, or $(1-sj)a=0$. By the definition of $I$ we get $(1-sj)\in I$. But now, since $I\subseteq J$, both $j$ and $1-sj$ belong to $J$. This implies that $1 = (1-sj) + s(j)\in J$, contradicting the fact that $J$ is a maximal proper left ideal. \\\\