Question

Question 9: $\mathit{R}\mathbf{=}\left\{\left(\begin{array}{cc}a& 0\\ b& c\end{array}\right)|a,b,c\in \mathbb{Z}\right\}$ is a ring with identity by Example 19 in Section 3.1.

a. Show that the map $\mathit{f}\mathbf{:}\mathit{R}\mathbf{\to }\mathbf{\mathbb{Z}}$given by $\mathit{f}\left(\begin{array}{cc}a& 0\\ b& c\end{array}\right)\mathbf{=}\mathit{a}$ is a surjective homomorphism.

Short answer

Answer:

It has been proved that $f$is a surjective homomorphism.

Step-by-step solution

Prove homomorphism

Consider$f:R\to \mathrm{\mathbb{Z}}$given bylocalid="1649755322671" $f\left(\begin{array}{cc}\mathrm{a}& 0\\ \mathrm{b}& \mathrm{c}\end{array}\right)=a$

In order to prove homomorphism:

Consider that:

$$\begin{gathered} f\left(\left(\begin{array}{cc}a& 0\\ b& c\end{array}\right)+\left(\begin{array}{cc}a\text{'}& 0\\ b\text{'}& c\text{'}\end{array}\right)\right)=f\left(\left(\begin{array}{cc}a+a\text{'}& 0\\ b+b\text{'}& c+c\text{'}\end{array}\right)\right) \\ =a+a\text{'} \\ =f\left(\left(\begin{array}{cc}a& 0\\ b& c\end{array}\right)+f\left(\begin{array}{cc}a\text{'}& 0\\ b\text{'}& c\text{'}\end{array}\right)\right) \\ Also, \\ f\left(\left(\begin{array}{cc}a& 0\\ b& c\end{array}\right)+\left(\begin{array}{cc}a\text{'}& 0\\ b\text{'}& c\text{'}\end{array}\right)\right)=f\left(\left(\begin{array}{cc}a+a\text{'}& 0\\ ba\text{'}+cb\text{'}& c.c\text{'}\end{array}\right)\right) \\ =a.a\text{'} \\ =f\left(\left(\begin{array}{cc}a& 0\\ b& c\end{array}\right).f\left(\begin{array}{cc}a\text{'}& 0\\ b\text{'}& c\text{'}\end{array}\right)\right) \\ Thusfisahomomorphism \end{gathered}$$

Prove that is f surjective

Since for every $a\in \mathrm{\mathbb{Z}},f\left(\begin{array}{cc}a& 0\\ 0& 0\end{array}\right)=a$, thus is$f$ surjective.

Hence,$f$ is a surjective homomorphism.