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$\mathit{K}$ is a subring of $\mathit{R}$.

Short answer

It is proved that $K$is a subring of$R$ .

Step-by-step solution

DetermineR→S is a homomorphism and K={r∈R|f(r)=0S}

Consider the given equation,

$K=\left\{\mathrm{r}\in \mathrm{R}|\mathrm{f}\left(\mathrm{r}\right)=0_{\mathrm{S}}\right\}$

Here,$f:R\to S$ is a homomorphism of ring and it is a subring of $R$.

It is easy to show that is closed under multiplication and subtraction under Theorem 3.1.

K={r∈R|f(r)=0S}

Let’s consider that $x$and $y$are arbitrary elements in $K$.

Then,

$$\begin{gathered} x,y\in K\Rightarrow f\left(x\right)=0_{S}andf\left(y\right)=0_S \\ \Rightarrow f\left(x-y\right)=f\left(x\right)-f\left(y\right)=0_S-0_S \\ \Rightarrow x-y\in K \end{gathered}$$

Similarly,

$f\left(x\right)=0_{S}andf\left(y\right)=0_S$assumed

$$\begin{gathered} \Rightarrow f\left(xy\right)=f\left(x\right)f\left(y\right)=0_S{0}_S=0_S \\ \Rightarrow xy\in K \end{gathered}$$

Therefore, $K$ is a subring of R.