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

Let $$R=\left\{\begin{bmatrix}a & b \\0 & a \end{bmatrix}; a,b\in Q\right\}$$. One can prove that R with the usual matrix addition and multiplication is a ring. Consider $$J =\left\{\begin{bmatrix}0 & b \\0 & 0 \end{bmatrix};b\in Q\right\}$$ 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 $$\displaystyle\frac{R}{{J}}$$ is isomorphic to Q.

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StrycharzT

Claim $$\rightarrow ker(\varnothing)=J$$
If $$\alpha=\left(\begin{bmatrix}a & b \\a & a \end{bmatrix}\right)\in$$ $$ker(\varnothing)$$,
then $$\varnothing(\alpha)=0$$ $$\rightarrow0=\varnothing\left(\begin{bmatrix}a & b \\a & a \end{bmatrix}\right)=a$$
Then, $$ker(\varnothing)\subset J$$ - 1
on the other hard.
If $$\beta=\begin{bmatrix}0 & b \\0 & 0 \end{bmatrix}\in$$ J, then
$$\varnothing(\beta)$$=0
$$\displaystyle\rightarrow$$ J $$\displaystyle\subset$$ $$ker(\varnothing)$$ - 2
From $$\displaystyle\frac{{1}}{\times}{2}$$ $$ker\varnothing=J$$.
Since, for any $$\displaystyle\alpha\in$$ Q, we can have $$\begin{bmatrix}a & 0 \\0 & a \end{bmatrix}\in$$ R
such thet $$\varnothing\left(\begin{bmatrix}a & 0 \\0 & a \end{bmatrix}\right)=a$$.
Then, $$\varnothing$$ is surgective.
By the 1st isomorphic theorem for the rings,
$$\displaystyle\frac{{R}}{{J}}\stackrel{\sim}{=}$$Q
Hence, the Quotient ring $$\displaystyle\frac{R}{{J}}$$ is isomorphic to Q. (Proved).
1st Isomorphic Theorem for Rings:
Let $$\varnothing$$ be a rinh homomorphism from R to S. Then the mapping from $$\displaystyle\frac{R}{{J}}\stackrel{\sim}{=}$$ to $$\varnothing(R)$$, given by $$r+ker\varnothing\rightarrow\varnothing$$(r) is an isomorphism $$i \times e$$, $$R/ker\varnothing\cong\varnothing R)$$.