Maximal ideals of a quotien ring

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Homework Statement



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.

Homework Equations





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!
 
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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...
 
micromass said:
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...

Yay I figured it out using your hint! Thanks a lot!
 
Well, x2 isn't an irreducible polynomial of R[x], since x2=x.x
Thus R[x]/(x2) is not a field.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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