Multiplicative map giving the roots of $f (x_1 , . . . , x_n ) ∈ \Bbb Z[x_1 , . . . , x_n ]$

Question

Let $f (x_1 , . . . , x_n ) ∈ \Bbb Z[x_1 , . . . , x_n ]$

Denote by Card{} the cardinality of a set.

For an integer $a$, Consider the map: $C_f(a)=Card\{(m_1 , . . . , m_n ) ∈ (\Bbb Z/a\Bbb Z)^n : f (m_1 , . . . , m_n ) ≡ 0 (mod\ a)\}$

Prove that $C_f$ is multiplicative in $a$.

I see that I can use Chinese remainder theorem for $(a,b)=1$ to show that $C_f(ab)=C_f(a)C_f(b)$. But I am having trouble for the case $(a,b)\ne1$.

Thank you for your help.

Answer 1

Multiplicative only means $C_f(ab) = C_f(a)C_f(b)$ for relatively prime $a$ and $b$. So you are already done. A function which satisfies such a property for all pairs (not just relatively prime pairs) is called totally multiplicative. Your $C_f$ function in fact need not be totally multiplicative. Consider the case $n = 1, f(x_1) = x_1^2$. Then $C_f(4) = 2$ but $C_f(2)C_f(2) = 1$.