Question about notation in this proof of the Chinese Remainder Theorem

Question

As part of a proof in my lecture notes, it states:

Let $f$ be a function such that $f:\mathbb{Z}_{mn}\rightarrow\mathbb{Z}_m\times\mathbb{Z}_n$ such that $b\mapsto(b\mod m,b\mod n)$. Then, we now show that $f$ is injective.

Suppose $f(u)=(r,s)=f(v)$ where $u,v\in\mathbb Z_{mn}$. Then, when $u$ and $v$ by $m$, the remainder is $r$ and then $u$ and $v$ are divided by $n$, the reaminder is $s$. Hence, $u\equiv_m v$ and $u\equiv_n v$.

First, when they mention $u,v\in\mathbb{Z}_{mn}$, from my understanding, they are referring to the congruence classes of $[u]_{mn}$ and $[v]_{mn}$ (but the square brackets are just ignored).

However, then they go ahead and say that $u$ is congruent to $v\mod m$ (i.e. $u\equiv_m v$). How is congruent modulo $n$ defined for congruence classes? Do we still treat $u$ and $v$ as integers despite them being congruence classes?

$u\equiv_n v\iff [u]_n = [v]_n\iff u+n\Bbb Z = v+n\Bbb Z.,$ It is not unusual to blur the the distinction between elements in the isomorphic rings $\ \Bbb Z_n\equiv \Bbb Z/n\Bbb Z$ and/or shift on-the-fly between congruence and coset notation. — Bill Dubuque, Mar 09 '22 at 19:08

score 1 · Accepted Answer · answered Mar 09 '22 at 19:23

This is a good question and thinking about it will lead to greater understanding!

Yes, elements of $\Bbb Z_{mn}$ are congruence classes (even though we often write the integers without the congruence-class brackets), while congruences are initially defined between integers only. So if we ever write $u,v\in \Bbb Z_{mn}$ and $u\equiv v\pmod m$ ... we had better be sure that the truth or falsity of the mod-$m$ congruence doesn't depend on which representatives $u,v$ of the congruence classes we choose.

Fortunately, this turns out to be true, as we can check. Let $u_1\in[u]$ and $v_1\in[v]$ be any representatives of those congruence classes in $\Bbb Z_{mn}$. By definition, that means that $u_1 = u+jmn$ and $v_1=v+kmn$ for some integers $j,k$. But then $u_1-v_1 = (u-v) + (jn-kn)m$, which means that $u_1\equiv v_1\pmod m$ if and only if $u\equiv v\pmod m$.

The key here is that $m$ divides $mn$. It's instructive to see how the argument fails if we take congruence classes in $\Bbb Z_b$ and ask whether two such classes are congruent modulo $a$ when $a$ does not divide $b$: that isn't a sensible property. (For example, if $b$ is even then it makes sense to talk about elements of $\Bbb Z_b$ being even or odd, while if $b$ is odd then it doesn't make sense to do so.)

Thank you for the answer! Just a quick question, if there anything wrong with saying $u\equiv_m v\iff u=v$? — ianc1339, Mar 12 '22 at 01:32
That's not correct. You wrote $u=v$: if you mean equality of integers, then $\Rightarrow$ is not correct, while if you mean equality of congruence classes in $\Bbb Z_{nm}$, then $\Leftarrow$ is not correct. — Greg Martin, Mar 13 '22 at 03:07

Question about notation in this proof of the Chinese Remainder Theorem

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