Question

Let $\mathit{f}\mathbf{:}\mathit{R}\mathbf{\to }\mathit{S}$be a homomorphism of rings and let $\mathit{K}\mathbf{=}\left\{r\in R/f\left(r\right)=0_s\right\}$.

Prove that K is an ideal in R.

Short answer

It is proved that K is an ideal in R.

Step-by-step solution

Theorem statement

Let R be a ring and Iand S are non-empty subsets of the ring, then I is said to be the ideal if and only if it has the following two properties:

1. If $a,b\in I$, then $a-b\in I$.
2. If $r\in R$and $a\in I$, then $ra\in I$and$ar\in I.$

Satisfied property (i) of theorem

It is given that$f\left(0_R\right)=0_R\in I$ , so K is non-empty.

Now, assume that $a,b\in K$and $s\in R$, then we have:

$$\begin{gathered} f\left(a-b\right)=f\left(a\right)-f\left(b\right) \\ =0_s-0_s \\ =0_s \end{gathered}$$

So,$\left(a-b\in K\right)$ .

Hence, property (i) of the theorem is stated in step 1.

Satisfied property (ii) of theorem

Let $r\in R$and by using the definition of ring homeomorphism in step 2,

$$\begin{gathered} f\left(ra\right)=f\left(r\right)f\left(a\right) \\ =f\left(r\right)0_s \\ =0_s \end{gathered}$$

And,

$$\begin{gathered} f\left(ar\right)=f\left(a\right)f\left(r\right) \\ =0_{s}f\left(r\right) \\ =0_s \end{gathered}$$

So, $ra\in K$and $ar\in K$.

Thus, property (ii) is also satisfied for the theorem stated in step 1.

Hence, it is proved that K is an ideal in R.