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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$

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closed as off-topic by Stefan Kohl, Lee Mosher, Charles Rezk, abx, Joe Silverman Feb 9 '14 at 16:55

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – Charles Rezk, abx, Joe Silverman
If this question can be reworded to fit the rules in the help center, please edit the question.

It is impossible to show without $1\ne0$. But this question does not suit this site. – Sasha Anan'in Feb 9 '14 at 16:38
What can I do to make this question more suitable? Can I re-ask if there is a counter example? – Kevin Slote Feb 9 '14 at 16:42
A counter-example is a ring having a unique element $0$. Better to ask at – Sasha Anan'in Feb 9 '14 at 16:45
up vote 0 down vote accepted

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

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