why $A\cong A/aA\times A/bA$ when $char(A)=ab$ and $(a,b)=1$

Question

if $A$ is a ring with $char(A)=ab$ and $(a,b)=1$, can someone explain why the Chinese remainder theorem implies $A\cong A/aA\times A/bA$? i know that TCR implies directly $A/aAbA\cong A/aA\times A/bA$ but i'm confused why $aAbA$ must be $(0)$?

Answer 1

Consider the ring homomorphism $$ f\colon A\to A/aA\times A/bA,\qquad f(x)=(x+aA,x+bA) $$ Its kernel is $aA\cap bA$. If $x=ay=bz$, for $y,z\in A$, we get $$ ax=abz=0,\quad bx=bay=0 $$ If $ma+nb=1$, which is guaranteed by Bézout's identity, then $$ x=(ma+nb)x=0 $$ Therefore the homomorphism $f$ is injective.

It is also surjective, because if $x,y\in A$, we can consider $x-y=am(x-y)+bn(x-y)$, so we can set $$ z=x-am(x-y)=y+bn(x-y) $$ and $z-x\in aA$, $z-y\in bA$, so $$ (x+aA,y+bA)=(z+aA,z+bA)=f(z) $$