Let $e=(1,0)$. Since $e$ becomes invertible in $S^{-1}(A\times B)$ and $e(0,b)=0$, every $(0,b)$ becomes zero. The map
$$
S^{-1}(A\times B)\longrightarrow A,\qquad
\frac{(a,b)}{(1,1)}\mapsto a,\quad
\frac{(a,b)}{(1,0)}\mapsto a
$$
is therefore a well-defined ring isomorphism, with inverse $a\mapsto(a,0)/1$.


Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 101 / 2 / a / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-101/2/a/ii/solution