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

UkusakazaL 2021-09-19 Answered

Let R={[ab0a];a,bQ}. One can prove that R with the usual matrix addition and multiplication is a ring. Consider J={[0b00];bQ} 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 RJ is isomorphic to Q.

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Expert Answer

StrycharzT
Answered 2021-09-20 Author has 102 answers

Claim ker()=J
If α=([abaa]) ker(),
then (α)=0 0=([abaa])=a
Then, ker()J - 1
on the other hard.
If β=[0b00] J, then
(β)=0
J ker() - 2
From 1×2 ker=J.
Since, for any α Q, we can have [a00a] R
such thet ([a00a])=a.
Then, is surgective.
By the 1st isomorphic theorem for the rings,
RJ=Q
Hence, the Quotient ring RJ is isomorphic to Q. (Proved).
1st Isomorphic Theorem for Rings:
Let be a rinh homomorphism from R to S. Then the mapping from RJ= to (R), given by r+ker(r) is an isomorphism i×e, R/kerR).

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