How to understand RSA Proofs of correctness at Wikipedia From Wikipedia(https://en.wikipedia.org/wiki/RSA_%28cryptosystem%29#Proofs_of_correctness):

$m^{ed}$ ≡ $m$ $\pmod{q}$
$m^{ed}$ ≡ $m$ $\pmod{p}$
 then $m^{ed}$ ≡ $m$ $\pmod{pq}$

Question:
if $m^{ed}$ ≡ $m$ $\pmod{q}$,then $q$|$m^{ed}$- $m$ 
if $m^{ed}$ ≡ $m$ $\pmod{p}$,then $p$|$m^{ed}$- $m$ 
that is to say, $m^{ed}$- $m$ is the common multiple of both $p$ and $q$

At this time, if there is $lcm(p,q)=pq$, then $m^{ed}$ ≡ $m$ $\pmod{pq}$.
However,  I am not sure $lcm(p,q)=pq$ holds true or not, so how can I prove $m^{ed}$ ≡ $m$ $\pmod{pq}$?
If $m^{ed} \equiv m \pmod{q}$, then $q\;|\;(m^{ed}-m)$;
if $m^{ed} \equiv m \pmod{p}$, then $p\;|\;(m^{ed}-m)$;
thus $m^{ed}-m$ is a multiple of both $p$ and $q$;
thus $m^{ed}-m$ is a multiple of $\operatorname{lcm}(p,q)$.

Because both $p$ and $q$ are distinct primes, $\operatorname{lcm}(p,q)=pq$ holds;
Thus, $m^{ed}-m$ is a multiple of $pq$.

So $(pq)\;|\;(m^{ed} - m)$ clearly holds.
Thus finally, $m^{ed} \equiv m \pmod{pq}$.