Assume that R is a commutative ring and every ideal is either R or 0. Show that R is a field. How can I show this without first being given that $ 1\ne 0 \in R$
closed as offtopic by Stefan Kohl, Lee Mosher, Charles Rezk, abx, Joe Silverman Feb 9 '14 at 16:55This question appears to be offtopic. The users who voted to close gave this specific reason:



If $1=0$ then $R=\{0\}$ since for every $x$ in $R$ one has $x=x1=x0=0$. But $\{0\}$ is not a field, so the result is false without $1\neq 0$. If $1\neq 0$, given $x$ in $R$ with $x\neq 0$, consider the ideal $xR$ generated by $x$. Since $x\neq 0$ and $R$ has a unit, $xR$ is not $\{0\}$. Now $xR$ must be all of $R$, so there exists $y$ in $R$ such that $xy=1$. Of course one also has $yx=1$ so $x$ is invertible, and thus $R$ is a field. This question is nevertheless not appropriate for MO  rather math.stackexchange.com. 

