image of injective function from $Z_{pqr} to Z_{pq}×Z_{qr}×Z_{rp}$

Question

Fix three distinct primes $p, q, r$, prove that the map

$Z_{pqr}→Z_{pq}×Z_{qr}×Z_{pr}$ by $[x]_{pqr} → ([x]_{pq}, [x]_{qr}, [x]_{pr})$

is injective and determine its image.

My attempt:

To prove it is injective, I said let there be $x$ and $y$ such that $[x]_{pq}=[y]_{pq}$. This means $pq|(x−y)$ So, one of $p$ or $q$ must divide $(x−y)$. Similarly, either $p$ or $r$ and either $q$ or $r$ must divide $x−y$ as well. Therefore, one of $p,q,r$ must divide $x−y$.

This implies $pqr|(x−y)$, So, $[x]_{pqr}=[y]_{pqr}$ This proves that the function is injective (if I'm correct in my implications).

Now, how do I determine its image? Maybe I can use the chinese remainder theorem and establish a bijection from $Z_p×Z_q×Z_r$ to $Z_{pqr}$ and use that somehow, but can anyone help me on this?

Answer 1

As the other answer indicates, your proof is not quite right. You are not trying to show that a single prime divides $x-y$, but that all three do.

As for characterizing the image, suppose we have some triple $(a,b,c) \in Z_{pq}\times Z_{qr} \times Z_{pr}$. How can we check that it actually came from an element $x\in Z_{pqr}$? One approach is to compare, for example, the image of $a$ in $Z_p$ with the image of $c$ in $Z_p$. If there is such an $x$, these two things have to be equal.

Do the same thing for $Z_q$ and $Z_r$ you'll have three conditions on $(a,b,c)$. To actually construct $x$ using these conditions, you can use the Chinese Remainder Theorem.

More explicitly, consider the map $\psi(a,b,c) = (a-b, b-c, c-a) \in Z_q \times Z_r \times Z_p$. Prove that everything in the image of your map $\phi$ is in the kernel of $\psi$ (manipulate the definitions), and prove that everything in the kernel of $\psi$ is in the image of $\phi$ (requires the Chinese Remainder Theorem).

Answer 2

Your proof of the first part is completely wrong. "One of $p,q,r$ divides ..." does not imply "All of $p,q,r$ divide ...". I suggest you fix that problem first before trying the second part. Also, it is easy and has nothing to do with homomorphisms and kernels. Hint: if $x \mid n$ then $n = ax$ for some integer $a$, and if $y \mid n$ such that $\gcd(x,y) = 1$, then $n \equiv ax \pmod{y}$ and $x$ is invertible mod $y$, so ...

I'm not sure what the question means by saying "determine its image", since it seems the simplest description of its image is literally $\{ ([x]_{pq},[x]_{qr},[x]_{rp}) : x \in \mathbb{Z} \}$.

If you wish to invert the mapping, note that $x \bmod p = ( x \bmod pq ) \bmod p = ( x \bmod rp ) \bmod p$, so each triple $(a,b,c)$ in the image must satisfy $p \mid c-a$ and $q \mid a-b$ and $r \mid b-c$. What else can we say?