Showing Ring Isomorphism of direct product of $n$ rings

Question

For context I will include the original problem and then what I'm actually asking about.

Context: I'm trying to show that the ring $$\mathbb{Z}_{r_1} \times \mathbb{Z}_{r_2} \times \ldots \times \mathbb{Z}_{r_n} \simeq \mathbb{Z}_{r_1 \cdot r_2 \cdots r_n}$$ if $\forall i\neq j, \gcd(r_i, r_j) = 1$. Starting with the base case of $n=2$, I'm trying to use strong induction to show the result.

Now, with the help of the $n=2$ case, I've shown that $$\mathbb{Z}_{r_1} \times \mathbb{Z}_{r_2} \times \ldots \times \mathbb{Z}_{r_n\cdot r_{n+1}} \simeq \mathbb{Z}_{r_1\cdots r_n \cdot r_{n+1}} $$ with $\phi(n) = n\langle1, \ldots, 1\rangle$.

Question: At this point, I can find a $\phi: \mathbb{Z}_{r_1 \cdots r_n r_{n+1}} \rightarrow \mathbb{Z}_{r_1} \times \mathbb{Z}_{r_2} \times \ldots \times [\mathbb{Z}_{r_n} \times \mathbb{Z}_{r_{n+1}}]$ that is an isomorphism, namely $$\phi(n)= n\langle 1, \ldots, \langle 1, 1\rangle \rangle$$ but what is a possible way to extend this to the expected $$\phi(n) = \langle 1,\ldots,1\rangle$$ without using any linear algebra arguments? I feel like just stating that those two functions are effectively the "same" would not be rigorous enough for a proof.

Answer 1

You can do without building the isomorphism you have doubts about. Consider the map $$ \varphi\colon\mathbb{Z}\to \mathbb{Z}_{r_1} \times \mathbb{Z}_{r_2} \times \dots \times \mathbb{Z}_{r_n} $$ defined by $\varphi(x)=([x]_{r_1},[x]_{r_2},\dots,[x]_{r_n})$. It should be clear this is a ring homomorphism. Also, its kernel is $$ r_1\mathbb{Z}\cap r_2\mathbb{Z}\cap\dots\cap r_n\mathbb{Z} $$ Now it should be easy to prove by induction that $$ r_1\mathbb{Z}\cap r_2\mathbb{Z}\cap\dots\cap r_n\mathbb{Z}= r_1r_2\dots r_n\mathbb{Z} $$ The facts you need are

1. if $\gcd(r_i,r)=1$ for $i=1,2,\dots,n$, then $\gcd(r_1r_2\dots r_n,r)=1$;
2. if $\gcd(r,s)=1$, then $r\mathbb{Z}\cap s\mathbb{Z}=rs\mathbb{Z}$.

Once you have proved that $\ker\varphi=r_1r_2\dots r_n\mathbb{Z}$, you have that $\varphi$ induces an injective homomorphism $$ \bar\varphi\colon \mathbb{Z}/r_1r_2\dots r_n\mathbb{Z}\to \mathbb{Z}_{r_1} \times \mathbb{Z}_{r_2} \times \dots \times \mathbb{Z}_{r_n} $$ Counting elements yields that $\bar\varphi$ is also surjective.

Answer 2

The usual induction argument goes as follows: $$ \mathbb{Z}_{r_1} \times \mathbb{Z}_{r_2} \times \cdots \times \mathbb{Z}_{r_n} \times \mathbb{Z}_{r_{n+1}} \cong \mathbb{Z}_{r_1 r_2 \cdots r_n} \times \mathbb{Z}_{r_{n+1}} \cong \mathbb{Z}_{r_1 r_2 \cdots r_n r_{n+1}} $$ where the induction hypothesis is used in the first part and the base case in the second part.