Number of zeros of a polynomial modulo n is a multiplicative function

Question

Let $f$ be a polynomial with integer coeffcients. For $n\geq1$ let $N_f(n)$ denote the number of pairwise incongruent solutions of $f(x)=0$ mod n. I need help proving that $f$ is a multiplicative function. I tried building an isomorphism between the zeros of f mod m cross zeros of f mod n to zeros of f mod mn, and I send these to their multiplication, but I don't know how to prove that if x is a zero of $f$ mod n, and y a zero of $f$ mod m where nm are co-prime that xy is a zero of $f$ mod mn.

Answer 1

For positive integers $ m,n$ with $GCD(m.n)=1$ : If $1\le a\le m$ and $1\le b\le n $, there is a unique $S(a,b)$ such that $1\le S(a,b)\le m n$ and $ S(a,b)\equiv a\pmod m$ and $S(a,b)\equiv b\pmod n $ .PROOF: (1)Existence.The set $ T= \{a+zm : z \in Z \text{ and }1\leq a+zm\le mn\}$ is a complete residue class modulo $ n$ because $GCD(m,n)=1 $ .So $t\equiv b \pmod n$ for some $t\in T $ . (2)Uniqueness.If $1\le t\le m n$ and $1\le t^8\le m n$ with $t\equiv t^*\equiv a\pmod m$ and $t\equiv t^*\equiv b\pmod n$ then $|t-t^*|$ is divisible by both $ m$ and $n$, which implies that $m n$ divides $|t-t^*|$ (Again, because $GCD(m,n)=1 $ .But $0\le |t-t^*|<m n $, so $t-t^*=0.$ $$NOW$$ to show that your function is multiplicative, let $A=\{a :1\le a\le m \text { and } f(a)\equiv 0\pmod m\}$ and $B=\{b : 1\le b\le n \text{ and }f(b)\equiv 0\pmod n\}$. Then $ \forall (a,b)\in A\times B (f(S(a,b)) \equiv 0 \pmod {m n}.$ (Is this obvious?) But if $1\le x \le m$ and $1\le y\le n$ and $(x,y)\not \in A\times B$ then $f((x,y))\not \equiv 0\pmod m$ OR $f(S(x,y))\not \equiv 0 \pmod n$....... Since $S$ is a 1-to-1 function, we are done.