Is $f([a]_{mn}) = ([a]_m,[a]_n)$ a bijection?

Question

Given, $f : Z/mnZ → Z/mZ × Z/nZ$, is $f([a]_{mn}) = ([a]_m,[a]_n)$ a bijection?

I have already done the work to prove that this function is well-defined.

Can I say that this is bijective though, without knowing the values of $m$ and $n$?

Answer 1

No, this is not in general bijective. (For instance, consider $m = n$ with $m > 1$, then $[m]_{mn}$ is a non-zero element in the kernel of $f$.) In general, it is only bijective if $m$ and $n$ are coprime. This is known as the Chinese Remainder Theorem.

Edit: the wikipedia article doesn't quite seem to cover this, but if $\gcd(m,n) > 1$ holds then it is not so hard to find a non-zero element in $\ker(f)$. (Hint: try to find some common multiple of $m$ and $n$.) Proving that it is indeed an isomorphism in the case $\gcd(m, n) = 1$ is a little harder; that statement is contained in the Chinese Remainder Theorem.