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This question already has an answer here:

Obviously, if $F$ is a field, and $I$ is it's nonzero ideal, then it contains an invertible element of $F$(any nonzero element of $F$). Denote this element as $a$. Since $I$ is ideal, $aa^{-1} = 1 \in I$. Hence, $I = F$.

But I'm not sure how to prove that any commutative ring with identity without nonzero proper ideals is a field.

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marked as duplicate by rschwieb ring-theory Feb 15 '16 at 13:42

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  • $\begingroup$ Please try the search feature first, next time. $\endgroup$ – rschwieb Feb 15 '16 at 13:43
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Just think backwards:

If you have a commutative ring $R$ with identity, the only missing property to be a field is, that any element is invertible.

So let's assume $R$ is no field. You have some non-invertible element $r \neq 0$ and thus $rR$ is a proper ideal, since $1 \notin rR$.

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If $\{0\}$ is the only proper ideal, is a maximal ideal and $R/\{0\}\approx R$ is a field.

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  • $\begingroup$ Well, OK, but I I would tend to think of "commutative rings with no nontrivial ideals are fields" as the essential reason why "the quotient of a commutative ring by a maximum ideal is a field", not the other way around... $\endgroup$ – Mike F Feb 15 '16 at 8:19

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