Question

(The Second Isomorphism Theorem) Let I and J be ideals in a ring R. Then$\mathit{I}\mathbf{}\mathbf{\cap }\mathbf{}\mathbf{J}$ is an ideal in I , and Jis an ideal in I + J by Exercises 19 and 20 of Section 6.1. Prove that$\frac{\mathbf{I}}{\mathbf{I}\mathbf{\cap }\mathbf{J}}\mathbf{\cong }\frac{\mathbf{I}\mathbf{+}\mathbf{J}}{\mathbf{J}}$ .[Hint: Show that$\mathit{f}\mathbf{}\mathbf{:}\mathbf{}\mathit{I}\mathbf{\to }\frac{\mathbf{I}\mathbf{+}\mathbf{J}}{\mathbf{J}}$ given by$\mathit{f}\left(a\right)\mathbf{=}\mathit{a}\mathbf{+}\mathit{j}$ is a surjective homomorphism with kernel$\mathit{I}\mathbf{\cap }\mathit{J}$ .]

Short answer

It has been proved that$\frac{I}{I\cap J}\cong \frac{I+J}{J}$.

Step-by-step solution

Suppose a map f : I→I+JJ

Define a map$f:I\to \frac{I+J}{J}$such that $f\left(a\right)=a+J$.

It is a well defined map.

Show that is a homomorphism

It is a homomorphism since it is the restriction of projection homomorphism $\Pi :R\to R/J$.

So f is also a homomorphism.

Prove that f is surjective

Every element of $\left(I+J\right)/J$equals some coset $i+j+J$for $i\in I$and $j\in J$.

But this coset equals $i+J=f\left(i\right)$.

Hence,f is surjective.

Prove that Ker f=I∩J

Let $a\in I\cap J$

Then

$\begin{array}{rcl}f\left(a\right)& =& a+J\\ & =& J\end{array}$

Thus, $a\in$Ker f.

This implies $I\cap J⊑$Ker f

Conversely, let $a\in I$ lies in the Ker f.

Then

$\begin{array}{rcl}f\left(a\right)& =& a+J\\ & =& J\end{array}$

Thus, $a\in J$

This implies $a\in I\cap J$

This implies Ker $f⊑I\cap J$.

Hence, Ker$f=I\cap J$

Conclusion

Hence, By First Theorem of Isomorphism we can conclude that $\frac{I}{I\cap J}\cong \frac{I+J}{J}$.