Modular arithmetic with polynomial Given $n=pq$ where $p\;\&\; q,p\ne q$ are primes,
P(x) is polynomial and $z\in\mathbb Z_{n}$
I need to prove that:
$p(z)\equiv 0\pmod{n}\iff p(z\pmod{p})\equiv 0\pmod{p}\;\land p(z\pmod{q})\equiv0\pmod{q}$
If i could prove the more general case: P(z) mod n ≡ P(z mod p) mod p
(q is the same) then i could also prove what i need of course. 
from my understanding so far, the above is true, but i can't figure out a way to prove this.
Back to the original question, i tried to see why the fact that p divides P(z mod p) and q divides P(z mod q) implies that pq divides P(z) and vice versa, but i didn't have much success with this.
A hint is sufficient. thank you!
 A: Hints:


*

*Chinese remainder theorem, to go from $p,q$ to $n$.

*Expand the polynomial $P$, to see that $P(z \mod m)\equiv P(z)\pmod{m}$.
A: First,  generally $\rm\,\ p,q\mid n\iff {\rm lcm}(pq)\mid n.\,$ But $\rm\ {\rm lcm}(p,q) = pq\,$ when $\rm\,p,q\,$ are coprime. 
Second  $\rm\,\ p\mid P(n)\iff p\mid P(n\ {\rm mod}\ p)\ $ holds true because $\rm\ {\rm\ mod}\ p\!:\ A\equiv a\ \Rightarrow\ P(A)\equiv P(a),\,$ for any polynomial $\rm\,P\,$ with integer coefficients. This is true because polynomials are composed of Sums and Products, and these operations respect congruences, i.e. the rules below hold true
Congruence Sum Rule $\rm\qquad\quad  A\equiv a,\quad B\equiv b\ \Rightarrow\ \color{#0a0}{A+B\,\equiv\, a+b}\ \ \ (mod\ m)$
Proof $\rm\ \ m\: |\: A\!-\!a,\ B\!-\!b\ \Rightarrow\ m\ |\ (A\!-\!a) + (B\!-\!b)\ =\ \color{#0a0}{A+B - (a+b)} $
Congruence Product Rule $\rm\quad\ A\equiv a,\ \ and \ \  B\equiv b\ \Rightarrow\ \color{#c00}{AB\equiv ab}\ \ \ (mod\ m)$
Proof $\rm\ \ m\: |\: A\!-\!a,\ B\!-\!b\ \Rightarrow\ m\ |\ (A\!-\!a)\ B + a\ (B\!-\!b)\ =\ \color{#c00}{AB - ab} $
Beware that such rules need not hold true for other operations, e.g.
the exponential analog of above $\rm A^B\equiv a^b$ is not generally true (unless $\rm B = b,\,$ so it follows by applying $\,\rm b\,$ times the Product Rule).
