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

UkusakazaL 2021-09-19 Answered

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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Expert Answer

StrycharzT
Answered 2021-09-20 Author has 12279 answers

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 \(i \times e\), \(R/ker\varnothing\cong\varnothing R)\).

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