Question

Let T and I be as in Exercise 44 of Section 6.1. Prove that $\mathit{T}\mathbf{/}\mathit{I}\cong \mathrm{ℝ}$ .

Short answer

It has been proved that$T/I\cong \mathrm{ℝ}$.

Step-by-step solution

Definition as per reference

T is a subring of $M\left(\mathrm{ℝ}\right)$such that it contains matrices of the form $\left(\begin{array}{cc}a& b\\ 0& a\end{array}\right)$with $a,b\in \mathrm{ℝ}$

I is an ideal in T containing matrices of the form$\left(\begin{array}{cc}0& b\\ 0& 0\end{array}\right)$ with$b\in \mathrm{ℝ}$ .

Suppose a map φ : T→ℝ

Define a map$\phi :T\to \mathrm{ℝ}$such that $\phi \left(\begin{array}{cc}a& b\\ 0& a\end{array}\right)=a$.

It is a well defined map.

Prove that φ is a homomorphism\

Let$\left(\begin{array}{cc}a& b\\ 0& a\end{array}\right),\left(\begin{array}{cc}a\text{'}& b\text{'}\\ 0& a\text{'}\end{array}\right)\in T$ then $\phi \left(\left(\begin{array}{cc}a& b\\ 0& a\end{array}\right)+\left(\begin{array}{cc}a\text{'}& b\text{'}\\ 0& a\text{'}\end{array}\right)\right)=\phi \left(\begin{array}{cc}a+a\text{'}& b+b\text{'}\\ 0& a+a\text{'}\end{array}\right)$

Now,

$\begin{array}{rcl}\phi \left(\begin{array}{cc}a+a\text{'}& b+b\text{'}\\ 0& a+a\text{'}\end{array}\right)& =& a+a\text{'}\\ & =& \phi \left(\begin{array}{cc}a& b\\ 0& a\end{array}\right)+\phi \left(\begin{array}{cc}a\text{'}& b\text{'}\\ 0& a\text{'}\end{array}\right)\end{array}$

Similarly, $\phi \left(\left(\begin{array}{cc}a& b\\ 0& a\end{array}\right)·\left(\begin{array}{cc}a\text{'}& b\text{'}\\ 0& a\text{'}\end{array}\right)\right)=\phi \left(\begin{array}{cc}a& b\\ 0& a\end{array}\right)·\phi \left(\begin{array}{cc}a\text{'}& b\text{'}\\ 0& a\text{'}\end{array}\right)$

Therefore, $\phi$is a homomorphism.

Prove that φ is surjective

It is a surjective map since $\phi \left(\begin{array}{cc}a& b\\ 0& a\end{array}\right)=a$for any $a\in \mathrm{ℝ}$

Also,$Ker\left(\phi \right)=I$.

Conclusion

Hence, By First Theorem of Isomorphism we can conclude that$T/I\cong \mathrm{ℝ}$