Question

Let A, B be two subrings of a ring R such that for all a e A, b e B,

ba e A then show that

(1) A + B is a subring of R.

(2) A is an ideal of A + B.

3) An B is an ideal of B.

Answer 1

Part 1

Suppose A and B are two subrings such that

$A+B$ is a subring if $B\sube\>A$

$\therefore\exist\>a\in\>A$ such that $a\notin\>B$

$\exist\>b\in\>B$ such that $b\in\>A$

$A+B$ satisfies the axioms

$(i)A+B\ne\empty$

$0_R\in\>A$ and $0_R\in\>B$

$\implies\>A+B$ is not empty

$(ii)$

$a-b\in\>A+B$

If $a-b\in\>A$ then $a-(a-b)\in\>A\implies\>b\in\>A$

$(iii)$

$\forall\>a,b\in\>A+B,a.b\in\>A+B$

Part 2

A is a subring of R $\implies\>A\ne\empty$

If $a,b\in\>A$ then $a-b\in\>A+B$

If $r\in\R$ and $a\in\>A+B$ then $ar,ra\in\>A+B$

$\implies\>A$ is an ideal of $A+B$

Part 3

$(i)$

$A,B\ne\empty$

$\implies\>AnB\ne\empty$

$(ii)a,b\in\>AnB\implies\>a-b\in\>AnB$

$(iii)r\in\>R,b\in\>AnB\implies\>br,rb\in\>AnB$

Therefore $AnB$ is an ideal of $B$