Question

If$\mathit{f}\mathbf{}\mathbf{:}\mathbf{}\mathit{R}\mathbf{\to }\mathit{S}$ is a surjective homomorphism of rings with kernel K , prove that there is a bijective function from the set of all ideals S of to the set of all ideals of R that contain K [Hint: Part(a) and Exercise 10.]

Short answer

It has been proved that there is a bijective function from the set of all ideals of S to the set of all ideals of R that contain K.

Step-by-step solution

Define two maps

Let I(S) be the set of all ideals in the ring S.

Let $I_K\left(R\right)$ be the set of all ideals in R that contains K.

Now, define a map $f:I\left(S\right)\to I_K\left(R\right)$such that$f\left(M\right)={\alpha }^{-1}\left(M\right)=\left\{r\in R|\alpha \left(r\right)\in M\right\}$

It is a well defined map.

Now, define a map $g:I_K\left(R\right)\to I\left(S\right)$such that $g\left(I\right)=\alpha \left(I\right)$.

It is also a well defined map.

Show that the two maps are inverses of each other

It is clear that $g\left(f\left(M\right)\right)=M$.

Now claim that $f\left(g\left(I\right)\right)=I$

Now we shall prove our claim

Since $f\left(g\left(I\right)\right)=\left\{r\in R|\alpha \left(r\right)\in \alpha \left(I\right)\right\}$

This set contains I.

Conversely, if $\alpha \left(r\right)\in \alpha \left(I\right)$

Then, $\alpha \left(r\right)=\alpha \left(a\right)$for some $a\in I$.

This implies $\alpha \left(r-a\right)=0_s$so that $r-a\in K⊑I$

This implies $r\in a+I=I$

Conclusion

Hence we see that f and g are inverses of each other. So they are bijections.