Proving M is a Maximal Ideal of R: Commutative Rings and Prime Ideals

[FanofAFan]

Homework Statement

R is a commutative ring, and normal to I, let M/I be a maximal ideal of R/I. Prove that M is a maximal ideal of R?

Homework Equations

The Attempt at a Solution

Not sure where to begin, but I think since we know R is commutative then we can say R/I is commutative and since M/I is an ideal of R/I we just need to show that M/I is also prime? But I could be completely off

 

[micromass]

Have you seen the following isomorphism theorem:

There exists a bijection between ideals of R which contain I and ideals of R/I.

I usually call that the fourth isomorphism theorem, but other names or also often used. Now, I suggest you use that bijection...

 

[Hurkyl]

I'm fond of the various relationships between properties of an ideal I in a ring R, and properties of the quotient ring R/I, myself.

 

[FanofAFan]

So Let M/I be a maximal ideal of R/I and R is commutative ring, So we need to show that M is maximal ideal of R, let H be an ideal of R such that M $$\subseteq$$ H $$\subseteq$$ R and every ideal is a sub-ring, then H is a sub ring of R. Therefore M is a sub-ring of H $$\subseteq$$ R, H is normal to I. Then we have M/I is subset of H/I is an ideal of R/I. And we have M/I $$\subseteq$$ H/I $$\subseteq$$ R/I and M/I is a maximal. By definition of maximal ideal M/I = H/I or R/I = H/I if M/I = H/I then M = H if H/I = R/I then H = R thus M is a maximal ideal of R