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Let $R$ be a commutative ring with $1$. Suppose that all ideals $I \neq R$ are prime. Prove that $R$ is a field.

Help me some hints.

Thanks a lot.

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closed as off-topic by YACP, Dan Rust, mau, TMM, Stefan Hansen Jan 3 '14 at 17:28

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The claim is false for $R=0$. – Martin Brandenburg Jan 4 '14 at 8:55
up vote 9 down vote accepted

First show that $R$ is a domain, by a suitable choice of ideal.

Then let $x \ne 0$, and deduce consequences from the fact that $xx \in x^2 R$.

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Can you explain more precise? – chuyenvien94 Jan 3 '14 at 12:15
Did you imply we can choose a noninvertible $x \in R$, take ideal $(x^2)$? – chuyenvien94 Jan 4 '14 at 1:16
Since $(x^2)$ is prime, we have $x \in (x^2)$ – chuyenvien94 Jan 4 '14 at 1:17
That means $x=rx^2,r \in R$ – chuyenvien94 Jan 4 '14 at 1:18
Since $(0)$ is prime, we have $1=rx$, a contradiction – chuyenvien94 Jan 4 '14 at 1:19

Hint: Consider some $r\in R$ and examine the principal ideal generated by $r^2$.

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Can you explain more precise? – chuyenvien94 Jan 3 '14 at 9:26
Well, first you'll need to verify that the cancellation law holds in this ring (how?). Assuming you've done that, suppose for the moment that the principal ideal satisfies $(r^2)=R$. In particular, this says $r$ is a multiple of $r^2$. This implies $r$ is invertible (why?). If $(r^2)\neq R$, then what may be deduced? – Philip Hoskins Jan 3 '14 at 9:36
+1 vote: Thank you – chuyenvien94 Jan 5 '14 at 0:38

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