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For any commutative ring $R$ and an ideal $I$ of $R$, $I \neq R$, show that $I$ is a maximal ideal iff $R/I$ is a field.

I write my own proof and it checks with the 'traditional' proof which makes use of an element that is not in $I$. When I check the notes, my teacher gives this proof:

$R/I$ is a field

$\Leftrightarrow$ the only ideals in $R/I$ are zero ideal and itself

$\Leftrightarrow$ the only ideals in $R$ containing $I$ are $I$ and $R$

$\Leftrightarrow$ $I$ is maximal

The first and second equivalence relations are not so obvious for me. Can anyone provide some hints? Thanks in advance.

Martin Sleziak
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Nighty
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2 Answers2

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Suppose that $F$ is a field, and $I$ is a non-zero ideal in $F$; show that $1_F\in I$ and then that $I=F$. The second equivalence is from the fourth isomorphism theorem for rings.

Brian M. Scott
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  • Thanks for replying. I claim this for the first equivalence: Consider non-empty ideal $I$. In a field, every element is an unit. It implies $1 \in I$. It quickly implies $I=R$. Is that correct? – Nighty Dec 17 '14 at 07:05
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    @LeeKM: Yes, that’s exactly right (except that you mean that every non-zero element is a unit!). – Brian M. Scott Dec 17 '14 at 07:07
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by using answer @Brian M.scott

  1. Suppose that $F$ is a field,and $I$ is a non-zero ideal in $F$, if $x \in I$ then since $x^{-1}$ is in $F$ and since $I$ is ideal then,$$xx^{-1} \in I \rightarrow 1_{F} \in I$$ so, $I=R$

  2. if $J$ be ideal containig $I$ and $J \neq R$ then $\frac{J}{I}$ is an ideal of $\frac{R}{I}$, since $\frac{R}{I}$ is field then by 1 $\frac{J}{I}$ is $\frac{R}{I}$ or ${0}=I$ then $J =I$ or $J=R$

erfan soheil
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