# Let R=\left\{\begin{bmatrix}a & b \\0 & a \end{bmatrix}; a,b\in Q\right\}ZS

Let $R=\left\{\left[\begin{array}{cc}a& b\\ 0& a\end{array}\right];a,b\in Q\right\}$. One can prove that R with the usual matrix addition and multiplication is a ring. Consider $J=\left\{\left[\begin{array}{cc}0& b\\ 0& 0\end{array}\right];b\in Q\right\}$ be a subset of the ring R.
(a) Prove that the subset J is an ideal of the ring R; (b) Prove that the quotient ring $\frac{R}{J}$ is isomorphic to Q.

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StrycharzT

Claim $\to ker\left(\varnothing \right)=J$
If $\alpha =\left(\left[\begin{array}{cc}a& b\\ a& a\end{array}\right]\right)\in$ $ker\left(\varnothing \right)$,
then $\varnothing \left(\alpha \right)=0$ $\to 0=\varnothing \left(\left[\begin{array}{cc}a& b\\ a& a\end{array}\right]\right)=a$
Then, $ker\left(\varnothing \right)\subset J$ - 1
on the other hard.
If $\beta =\left[\begin{array}{cc}0& b\\ 0& 0\end{array}\right]\in$ J, then
$\varnothing \left(\beta \right)$=0
$\to$ J $\subset$ $ker\left(\varnothing \right)$ - 2
From $\frac{1}{×}2$ $ker\varnothing =J$.
Since, for any $\alpha \in$ Q, we can have $\left[\begin{array}{cc}a& 0\\ 0& a\end{array}\right]\in$ R
such thet $\varnothing \left(\left[\begin{array}{cc}a& 0\\ 0& a\end{array}\right]\right)=a$.
Then, $\varnothing$ is surgective.
By the 1st isomorphic theorem for the rings,
$\frac{R}{J}\stackrel{\sim }{=}$Q
Hence, the Quotient ring $\frac{R}{J}$ is isomorphic to Q. (Proved).
1st Isomorphic Theorem for Rings:
Let $\varnothing$ be a rinh homomorphism from R to S. Then the mapping from $\frac{R}{J}\stackrel{\sim }{=}$ to $\varnothing \left(R\right)$, given by $r+ker\varnothing \to \varnothing$(r) is an isomorphism $i×e$, $R/ker\varnothing \cong \varnothing R\right)$.