I'm confused on how to go in both directions and how to start this proof.
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please try the search function before asking a basic question like this one. You may find your answer right away. Also, try to include your thoughts on the problem to increase your chances of getting good answers. – rschwieb Apr 28 '14 at 02:51
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1While I appreciate your input, this was not helpful in aiding my understanding of the question whatsoever. I understand it might be a 'basic' question to you, but it is not 'basic' to me. – user141447 Apr 28 '14 at 03:02
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really? You followed the duplicates and examined the 10+ solutions with upvotes, one of them with over 10 upvotes, and got no help? Sorry, I thought that was rather a very easy way to connect you with solutions. Come to think of it,you might not have noticed the linked questions list on the right of the question I linked. That's where the dupes are. Regards. – rschwieb Apr 28 '14 at 03:08
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Well, if $R$ is a field, then the only proper ideal is $\{ 0\}$, by virtue of the fact that $0 \ne a \in I$ for an ideal $I$ implies $ba \in I$ for all $b \in R$; then for $x \in R$ we have $x = (xa^{-1})a \in I$, so $R \subseteq I$, so $I = R$, so $I$ is not proper. Going in the other direction, suppose $\{ 0 \} = 0R$ is the only maximal ideal and let $a \in R$ be a non-unit element. Then $a \in aR$ since $1 \in R$, and $1 \notin aR$ since $a$ is not a unit; if $1 \in aR$, then $1 = ba$ for some $b \in R$, contradicting the fact that $a$ is a non-unit element. But then $aR$ is a proper ideal and $0R \subsetneq aR \subsetneq R$, so $0R$ cannot be maximal. Thus every $a \in R$ is a unit and hence $R$ is a field. QED.
Hope this helps. Cheerio,
and as always,
Fiat Lux!!!
Robert Lewis
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@user141447: I surmise you are rather new to abstract algebra, so if you have questions about my answer, if it is confusing to you, feel free to leave me a comment and I'll try to get back to you. Cheers. – Robert Lewis Apr 28 '14 at 03:25