Let $R$ be a ring. Show that R is a field if and only if $\{ 0 \}$ is the unique maximal ideal of $R$. Thank you
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3Have you tried something? – Patrick Da Silva Feb 28 '14 at 12:59
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Yes, but i couldn't. Help, please. :( – MathStudent Feb 28 '14 at 13:03
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4Can you show one of the directions of the "if and only if"? – Tobias Kildetoft Feb 28 '14 at 13:04
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i would like to learn how can i answer this problem. – MathStudent Feb 28 '14 at 13:06
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The newest edited version does not really make any sense. – Tobias Kildetoft Feb 28 '14 at 13:13
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1@MathStudent It's good to post what you tried even if it didn't work out. With more information, posters can respond with responses tailored to where you started. You'll learn the most this way! – rschwieb Feb 28 '14 at 13:22
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Let $I$ be an ideal properly containing $\{0\}$. That is, there is a nonzero element $a\in I$. By definition of an ideal, also $1=a^{-1}a\in I$, so $I=(1)=F$.
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With a few small pieces, you can make quick work of this proof:
- An ideal is proper if and only if it does not contain a unit.
- Given any $a \in R$, we may define the (left-)ideal "generated by $a$" to be given by $$ \langle a\rangle = \{ra:r \in R\} $$
- If $I$ is an ideal, then $a\in I \implies \langle a \rangle \subset I$.
All of these ideas should be explicitly outlined in your textbook at some point. Note that if $a$ is not a unit, $\langle a \rangle \neq R$ (why)?
Ben Grossmann
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