A commutative ring whose all proper ideals are prime is a field. [closed]

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.

closed as off-topic by user26857, Dan Rust, mau, TMM, Stefan HansenJan 3 '14 at 17:28

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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 $xx \in x^2 R$.

• 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 $r^2$.

• 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