# Prove that a non-trivial commutative ring $R$ is a field if and only if $0$ and $R$ are the only ideals of $R$.

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!

• Does the first part that you don’t understand actually read $$R=Rx^{-1}x\subseteq Rx\subseteq a\subseteq R\text{ and }a=R\;?$$ You really need to learn how to write mathematics here; there’s a tutorial here. Commented Jun 5, 2020 at 2:28
• yes it reads exaclty that... thank you for the link! Commented Jun 5, 2020 at 2:31

$$R$$ 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 $$r = rx^{-1}x \in Rx \subset a \implies R \subset Rx \subset a \implies Rx = a = R$$