If $a \equiv b \mod (nm)$, then is it always true that $$a \equiv b \mod n$$ $$a \equiv b \mod m$$
I feel like it's kind of obvious because if $a-b = k_1nm$, then $a-b = k_2n$, ($k_2 = k_1n$) but I can't find any resources on the topic, so maybe I'm wrong
Well, if $m,n\geq 2$ are relatively prime, then the mapping $${\Bbb Z}_{mn}\rightarrow{\Bbb Z}_m\times {\Bbb Z}_n:a\mapsto (a\mod m, a\mod n)$$ is a ring isomorphism. In this way, $a\equiv b\mod mn$ is equivalent to $a\equiv b\mod m$ and $a\equiv b\mod n$.