Question

(The Third Isomorphism Theorem) Let I and K be ideals in a ring R such that$\mathit{K}\mathbf{⊑}\mathit{I}$ . Then I/K is an ideal in R/K by Exercises 19. Prove that$\left(R/K\right)\mathbf{/}\left(I/K\right)\mathbf{\cong }\mathit{R}\mathbf{/}\mathit{I}$ .[Hint: Show that$\mathit{f}\mathbf{}\mathbf{:}\mathbf{}\mathit{R}\mathbf{/}\mathit{K}\mathbf{\to }\mathit{R}\mathbf{/}\mathit{I}$ given by$\mathit{f}\left(r+K\right)\mathbf{=}\mathit{r}\mathbf{+}\mathit{I}$ is a well defined surjective homomorphism with kernel I/K .]

Short answer

It has been proved that$\left(R/K\right)/\left(I/K\right)\cong R/I$.

Step-by-step solution

Suppose a projection homomorphism

Let$\Pi :R\to R/I$ be the projection homomorphism, which is surjective with kernel I

Show that f is a homomorphism

Since $K⊑I$this implies that there is an induced homomorphism $f=\overline{\Pi }:R/K\to R/I$where $f\left(r+K\right)=r+I$.

So f is a homomorphism.

Show that f is surjective

Since $\Pi$is surjective,

Hence,fis surjective.

Prove that Ker f=I/K

Let $a\in I$

Then

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

Thus, $a+K\in$Ker f.

This implies $I\cap J⊑$Ker f

Conversely, let r+ K lies in the Ker f.

Then

$\begin{array}{rcl}r+I& =& f\left(r+K\right)\\ & =& I\end{array}$

and $r\in I$.

This implies Kernel $=\left\{a+K:a\in I\right\}=I/K$.

Conclusion

Hence, By First Theorem of Isomorphism we can conclude that$\left(R/K\right)/\left(I/K\right)\cong R/I$