Question

Question 10 (a): Let $\mathit{f}\mathbf{:}\mathit{R}\mathbf{\to }\mathit{S}$is a surjective homomorphism of rings and let $l$be an ideal in $\mathit{R}$. Prove that $\mathit{f}\mathbf{\left(}\mathit{l}\mathbf{\right)}$is an ideal in$\mathit{S}$ where for some$\mathit{f}\mathbf{\left(}\mathit{l}\mathbf{\right)}\mathbf{=}\left\{s\in S|s=f(a)\text{forsomea}\text{∈l}\right\}$ .

Short answer

Answer:

It can be proved that is $f\left(l\right)$an ideal in$S$

Step-by-step solution

Applying First condition for ideal

In order to prove the ideal,

Let us suppose that $a,b\in f\left(l\right)$, then

$a=f\left(s\right)$and $b=f\left(t\right)$for some, $s,t\in l$ .

Then,

$$\begin{gathered} a+b=f\left(s\right)+f\left(t\right) \\ =f(s+t) \\ \in f\left(l\right) \end{gathered}$$

Applying the second condition for ideal

Since$f$is a surjective homomorphism,

Therefore,

For any$t\in S$, there exists$r\in R$with $t=f\left(r\right)$.

So,

$$\begin{gathered} ta=f\left(r\right)f\left(s\right) \\ =f\left(rs\right) \\ \in f\left(l\right) \end{gathered}$$

Conclusion

$$\begin{gathered} a,b\in f\left(l\right)\Rightarrow a+b\in f\left(l\right) \\ t\in S\Rightarrow ta\in f\left(l\right) \\ \text{hence}f\left(l\right)\text{isanideal} \end{gathered}$$