Let R={[ab0a];a,b∈Q}. One can prove that R with the usual matrix addition and multiplication is a ring. Consider J={[0b00];b∈Q} 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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StrycharzT · Accepted Answer

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/ker∅≅∅R).

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

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