if M is the maximal ideal of R....how can we say that R/M has only 2 ideals ....0 and R/M itself??

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- Sep 7th 2010, 06:29 AM #1

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- Sep 7th 2010, 07:42 AM #2

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- Sep 7th 2010, 11:56 AM #4

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that was what i had set out to prove....that if If R is a commutative ring with identity, and M is a maximal ideal, then R/M is a field.....they directly said that R/M has 2 ideals 0 and R/M itself....so i cld'nt get how they said this...

could you please tell me somethng abt this correspondence theorem. cz i dont seem to get what it says....

- Sep 7th 2010, 11:32 PM #5
The correspondence theorem (I believe) basically says two things,

1. Let $\displaystyle I, J \lhd R$ with $\displaystyle I \subseteq J$. Then $\displaystyle J/I \lhd R/I$.

2. If $\displaystyle J/I \lhd R/I$ then there exists an ideal $\displaystyle J^{\prime} \lhd R$ such that $\displaystyle I \subseteq J^{\prime}$ and $\displaystyle J^{\prime}/I = J/I$.

So, essentially, the ideals in $\displaystyle R/I$ correspond precisely to the ideals $\displaystyle J$ such that $\displaystyle I \lhd J \lhd R$. As there are no such proper ideals, as $\displaystyle I$ is maximal, then what can you conclude?