Show that a commutative ring with unity having no proper ideals is a
field.
No proper ideals means that $\{0\},R$ are the only ideals of $R$. But then $\{0\}$ is a maximal ideal and thus by applying the first isomorphism theorem we get that $$R/\{0\} \cong R$$ and since $\{0\}$ is maximal we also have that $R/\{0\}$ is a field. Thus $R$ is a field.
Is this reasoning correct?
Best Answer
Yes, that looks fine, although it can be done directly by inspecting the proof of the theorem you used.
Notice that if we pick $a\neq 0$ then the ideal generated by $a$, which is equal to $Ra$ is all of $R$, therefore $1$ is contained in $Ra$.
So there is an element $r\in R$ such that $ra=1$. so $a$ has an inverse.