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Maximal ideals of a quotien ring |
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| Feb19-11, 05:28 PM | #1 |
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Maximal ideals of a quotien ring
1. The problem statement, all variables and given/known data
I am try to prove : Let R be a ring, I be a ideal of R. Then N is a maximal in R/I if and only if N=M/I where M is a maximal ideal in R that contains I. 2. Relevant equations 3. The attempt at a solution First I'm not 100% sure that the statement is true, but I'm trying to prove it. So N is maximal in R/I, then R/I/N is a field which means I/N is maximal in R. This is as far as I get. Thanks! |
| Feb19-11, 05:41 PM | #2 |
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This statement is 100% correct. So don't worry about it.
Have you heard of the fourth isomorphism theorem (you probably didn't call it that). It states that there is a bijective correspondance between ideals of R that contain I and ideals of R/I. In particular, if J is an ideal of R/I, then J=J'/I for some ideal J' of R. This is the thing you have to use to prove this question. If you didn't see it, then perhaps you could try to prove it... |
| Feb19-11, 06:38 PM | #3 |
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| Feb19-11, 06:45 PM | #4 |
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Maximal ideals of a quotien ring
Well, x2 isn't an irreducible polynomial of R[x], since x2=x.x
Thus R[x]/(x2) is not a field. |
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