So I have encountered in some notes on number theory, that :
If $n=p_1^{e1}p_2^{e2}...p_k^{ek}$ is the prime factorization of n, and integers $x_1,x_2,...,x_k$ satisfy $$f(x_i)\equiv0 \mod p_i^{ei}$$ for $i=1,2,..,k$, then we can find $x$ such that $x\equiv x_i\mod p_i^{ei}$ for all i using the Generalized CRT. But then such an x would satisfy $f(x)\equiv0\mod p_i^{ei}$ for all i, and therefore $f(x)\equiv 0\mod n$. From here it follows that if each congruence $f(x)\equiv0\mod p_i^{ei}$ has $s_i$ solutions, then the congruence $f(x)\equiv0\mod n$ has $s_1s_2...s_k$ solutions.
I understood the first part, so if you have a $x$ that is equivalent to each of the $x_i$ in their respective modulo, then $f(x)$ basically evaluates to $f(x_i)$ in each case and with this we can conclude that such a $x$ is a solution to the general congruence modulo n. Now, I think in the last statement, they are trying to say that since you have $s_i$ solutions for each, you can have $s_1s_2...s_k$ ways of finding a different set of solutions for the k congruences. With each way, you can find a $x$ with the CRT, so you would have $s_1s_2...s_k$ solutions for $x$. Now, I am wondering whether any of the $x$ would duplicate in this case. Can someone provide a more rigorous/convincing proof for "if each congruence $f(x)\equiv0\mod p_i^{ei}$ has $s_i$ solutions, then the congruence $f(x)\equiv0\mod n$ has $s_1s_2...s_k$ solutions"?
Here, $s_i$ solutions mean that there are $s_i$ different solutions $\rm{mod} \, p_i^{e_i}$. We just can take representatives $h_{p,1},\ldots,h_{p,s_i} < p^{e_i}$. Now you can just use the Chinese remainder theorem (this is a ring-homomorphis!) to get for any choice $(h_{p_1,i_1},\ldots,h_{p_k,i_k})$ of this numbers exactly one number $s < p_1^{e_1} \ldots p_k^{e_k}$ and they are all different. Thus, the number of such solutions is at least $s_1 \ldots s_k$. On the other hand, any solution $\rm{mod}\,{n}$ gives also a tuple $(x_1,\ldots,x_n)$ which solves $f(x_i) = 0 \mod{p_i^{e_i}}$, i.e. we already counted this solution. Thus, the number of solutions is in fact $s_1 \ldots s_k$.