Mapping from $\mathbb Z/2\mathbb Z \times \mathbb Z/3\mathbb Z \times \dots\times \mathbb Z/p_n\mathbb Z$ to $\mathbb Z/p_n\#\mathbb Z$.

Question

I know $\mathbb Z/2\mathbb Z \times \mathbb Z/3\mathbb Z \times \dots\times \mathbb Z/p_n\mathbb Z$ is isomorphic $\mathbb Z/p_n\#\mathbb Z$ (where $p_n\#$ is the primorial of primes up to $p_n$) by the Fundamental Theorem of Finite Abelian Groups, and I know the mapping from $\mathbb Z/p_n\#\mathbb Z$ to the product is $f(x) = (x\bmod2, x\bmod 3, x\bmod5, \dots,x\bmod{p_n})$. (Is there a good reference for this?) Since the product and $\mathbb Z/p_n\#\mathbb Z$ are isomorphic, the mapping in the reverse direction (from the product to $\mathbb Z/p_n\#\mathbb Z$) certainly exists. Is it known? I have figured out a formula - which I believe I can prove is correct - and am wondering if it might be worth writing this up as a paper.

Thanks.

Answer 1

For $$N = \prod_{i=1}^j p_i^{e_i}$$ let $$c_i \frac{N}{p_i^{e_i}} \equiv 1 \bmod p_i^{e_i}$$ then

$$(a_1,\ldots,a_j) \mapsto \sum_{i=1}^j a_i \frac{N}{p_i^{e_i}} c_i$$ is the unique isomorphism $$\prod_{i=1}^j \Bbb{Z}/p_i^{e_i}\Bbb{Z} \to \Bbb{Z}/N\Bbb{Z}$$ whose inverse is $$m \mapsto (m \bmod p_1^{e_1}, \ldots,m \bmod p_j^{e_j})$$

Answer 2

That is exactly the statement of the Chinese remainder theorem.

Of course, if you found a specialized formula that is easier to calculate, or has some peculiarity, you can write that up. We will be looking forward to that.