Question

Let $\mathit{f}\mathbf{}\mathbf{:}\mathbf{}\mathit{R}\mathbf{\to }\mathit{Z}$ be a homomorphism of rings. If J is an ideal in S and localid="1653373309021" $\mathit{I}\mathbf{=}\left\{r\in R\overline{)f\left(r\right)\in J}\right\}$, prove that I is an ideal in R that contain kernel of f.

Short answer

It can be proved that Iis an ideal in R that contain kernel of f .

Step-by-step solution

 Proving I is an ideal first condition

$I=\left\{r\in R\overline{)f\left(r\right)}\in J\right\}$

In order to proveI is an ideal in R,

Consider$a,b\in I$

Then$f\left(a\right)\in J$ and$f\left(b\right)\in J$

Since J is also an ideal so

role="math" localid="1653299100337" $$\begin{gathered} f\left(a\right)-f\left(b\right)\in J \\ f\left(a-b\right)\in J \\ a-b\in I \end{gathered}$$

Second Condition

Further,

Let$r\in R$ and$a\in I$

We have

$f\left(r\right)f\left(a\right)=f\left(ra\right)$

Thus $ra\in I$

Similarly,$ar\in I$

I is an ideal in R

From step 1 and step 2 it is clear thatis an ideal in R

Prove ker f⊆I

Ker $f=\left\{r\in R\overline{)f\left(r\right)}=0\right\}$.

Every ideal contains $0$

In particular, $0\in J$

Thus, Ker $f\subseteq I$

Hence, I contains ker f .