Let I and J be ideals of R with I J a Show JI is an ideal of

Let I and J be ideals of R with I J. (a) Show J/I is an ideal of R/I. (b) Define phi: R rightarrow (R/I)/(J/I) by phi(r) = (r + I) + J/I for r R. Show phi is a subjective homomorphism with ker phi = J. (c) Conclude that (R/I)/(J/I) R/J.

Solution

Let I and J be ideal of R with IJ

Certainly 0+I J/I

since J is an ideal and hence 0J.

The additive identity of R/I lies in J/I

take a+I, b+IJ/I so a,bJ

since J is an ideal, we get a+b, -aJ

Now (a+I)+(b+I)=(a+b)+IJ/I and -(a+I)=-a+IJ/I

therefore J/I is a subgroup of R/I under addition

now take x+IR/I so xR

since J is an ideal, we know xaJ where aJ

so (x+I)(a+I)=xa+IJ/I

Hence J/I is an ideal of R/I

consider the map :R/IR/J given by (x+I)=x+J

let x+I=x+I, implies x-xI, implies x-xJ since IJ

we get x+J=x+J

is well-defined.

hence is surjective.

now for kernel we have

ker={x+IR/Isuchthat (x+I)=0}

={x+IR/Isuchthat xJ}

= J/I

Im()={a+Jsuchthat aR}=R/J

Hence we conclude that (R/I)/(R/J)isomorphic to R/J.theorem is proved.