Will $\mathbb Z/{mn}\simeq \mathbb Z/m\times\mathbb Z/n$ as rings if $(m, n)=1$

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We know that $\mathbb Z/{mn}\simeq \mathbb Z/m\times\mathbb Z/n$ as groups iff $(m, n)=1$.

But what about as a ring?

Will $\mathbb Z/{mn}\simeq \mathbb Z/m\times\mathbb Z/n$ hold good if $(m, n)=1$ and $\mathbb Z/{mn}, \mathbb Z/m, \mathbb Z/n$ are rings ?

I am unable to prove it but I assume it should be true. But no idea how to establish. Please help me

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Yes, the canonical map $\mathbb{Z}/mn \to \mathbb{Z}/m \times \mathbb{Z}/n$ is obviously also a ring homomorphisms, and as you know bijective, hence a ring isomorphism.

More generally, if $I,J$ are coprime ideals of a ring $R$, then $R/(I \cap J) \cong R/I \times R/J$. This is the Chinese Remainder Theorem for rings. (If $I,J$ are not coprime, we still have $R/(I \cap J) \cong R/I \times_{R/(I+J)} R/J$ by the way.)