Question

Question 8: Let and be rings. Show that $\mathit{\pi }\mathbf{:}\mathit{R}\mathbf{\times }\mathit{S}\mathbf{\to }\mathit{R}$given by $\mathit{\pi }\mathbf{(}\mathit{r}\mathbf{,}\mathit{s}\mathbf{)}\mathbf{=}\mathit{r}$ is a surjective homomorphism whose kernel is isomorphic to S .

Short answer

It is proved that $\pi$ is a surjective homomorphism whose kernel is isomorphic to S.

Step-by-step solution

Prove homomorphism

In order to prove $\mathrm{\pi }$is a homomorphism, let us consider $(r,s),(r\text{'}s\text{'})\in R\times S$.

$$\begin{gathered} Then,\pi \left(\right(r,s)+(r\text{'}s\text{'}\left)\right)=\mathrm{\pi }(\mathrm{r}+\mathrm{r}\text{'},\mathrm{s}+\mathrm{s}\text{'})\mathrm{Thus} \\ \mathrm{\pi }\left(\right(\mathrm{r},\mathrm{s})+(\mathrm{r}\text{'}\mathrm{s}\text{'}\left)\right)=\mathrm{\pi }(\mathrm{r}+\mathrm{r}\text{'},\mathrm{s}+\mathrm{s}\text{'}) \\ =\mathrm{r}+\mathrm{r}\text{'} \\ =\mathrm{\pi }(\mathrm{r},\mathrm{s})+\mathrm{\pi }(\mathrm{r}\text{'}\mathrm{s}\text{'}) \\ \mathrm{Also},\mathrm{\pi }\left(\right(\mathrm{r},\mathrm{s}).(\mathrm{r}\text{'}\mathrm{s}\text{'}\left)\right)\mathrm{can}\mathrm{be}\mathrm{written}\mathrm{as}: \\ \mathrm{\pi }\left(\right(\mathrm{r},\mathrm{s}).(\mathrm{r}\text{'}\mathrm{s}\text{'}\left)\right)=\mathrm{\pi }(\mathrm{r}.\mathrm{r}\text{'},\mathrm{s}.\mathrm{s}\text{'}) \\ =\mathrm{r}.\mathrm{r}\text{'} \\ =\mathrm{\pi }(\mathrm{r},\mathrm{s}).\mathrm{\pi }(\mathrm{r}\text{'}\mathrm{s}\text{'}) \\ \mathrm{Thus},\mathrm{\pi }\mathrm{is}\mathrm{a}\mathrm{homomorphism} \end{gathered}$$

Prove is π surjective

Here, $\mathrm{\pi }$is surjective because for every$r\in R$ , there exists $(r,0_s)$, such that $r=\pi (r,0_s)$.

Thus,$\mathrm{\pi }$ is surjective.

Prove that kernel is isomorphic to  S

The kernel is

$K=\left\{\right(0_R,s|s\in S\}$

It defines a map $\theta :K\to S$, such that$\theta (0_R,s)=s$.

Thus, by above map $K$it is proved that is isomorphic to S

Hence, is a$\pi$surjective homomorphism whose kernel is isomorphic to S.