Prove that a non-trivial commutative ring $R$ is a field if and only if $0$ and $R$ are the only ideals of $R$.
Proof Let $R$ be a field. Let $a$ be a non-trivial ideal of $R$. Let $x$ be in $a - \{ 0\}$. Then $R = Rx^{−1}x$ in $Rx$ in $a$ in $R$ and $a = R$.
Conversely suppose that $0$ and $R$ are the only ideals of a non-trivial commutative ring $R$. For all $x$ in $R - \{0\}$, $Rx$ is a non-trivial ideal of $R$. Thus $R = Rx$ and there is y in $R$ such that $1 = yx$. Therefore every element of $R -\{0\}$ has an inverse and $R$ is a field.
So this is the proof I have but I do not understand it...
The part I do not understand is "Then $R = Rx^{−1}x$ in $Rx$ in $a$ in $R$ "
Every help is appreciated!
$R$ is a field, then if you take some nontrivial ideal $a$ in $R$ there exists some $x \in a$ such that $x \not= 0$, then as $Rx \subset a \subset R$ and as $R$ is a field, we have that $x^{-1} \in R$, so $rx^{-1} \in R$ for all $r\in R$, ,then $r = rx^{-1}x \in Rx \subset a \implies R \subset Rx \subset a \implies Rx = a = R$