Prove the ideal correspondence if J is an ideal in R and pi

Prove the ideal correspondence: if J is an ideal in R and pi: R rightarrow R/J is the projection, then the map taking I to I/J is a bijection between the set of ideals in R containing J and the set of all ideals in R/J. What is the inverse map? Prove that, in the above, the prime ideals correspond to the prime ideals, and the maximal ideals to the maximal ideals.

Solution

Let us first prove the the thorem:

Statement:If I is an ideal of the ring R,then the map
S S/I sets up a one-to-one correspondence between the set of all subrings of R containing
I and the set of all subrings of R/I,as well as a one-to-one correspondence between the set
of all ideals of R containing I and the set of all ideals of R/I. The inverse of the map is
Q 1(Q),where is the canonical map: R R/I.
Proof:

If S is a subring of R then S/I is closed under addition,subtraction and multiplication. For example,if s and s belong to S,we have (s + I)(s\' + I) = ss\' + I S/I. Since 1_R S we have 1_R + I S/I

thus S/I is a subring of R/I.

Conversely,if S/I is a subring of R/I,then S is closed under
addition,subtraction and multiplication,and contains the identity,hence is a subring or R.
For example,if s, s S then (s + I)(s\' + I) S/I,so that ss\' + I = t + I for some t S,
and therefore ss\' t I. But I S,so ss\' S.
Now if J is an ideal of R containing I,then J/I is an ideal of R/I by the third
isomorphism theorem for rings. Conversely,let J/I be an ideal of R/I. If r R and x J
then (r+I)(x+I) J/I,that is, rx+I J/I. Thus for some j J we have rxj I J,
so rx J. A similar argument shows that xr J,and that J is an additive subgroup of R.
It follows that J is an ideal of R.

We know that every ideal is a subring.Choosing S=I and ideal J of ring R (instead of I)in above theorem gives the proof.