Let $r,s \in \mathbb{Z}^+$ such that $(r,s)=1.$ Consider the ring $\mathbb{Z}_{r} \times \mathbb{Z}_{s}$ (with respect to ordinary multiplication and addition, component by component), prove that the application $f: \mathbb{Z} ⟶ \mathbb{Z}_{r} \times \mathbb{Z}_{s}$ given by $a ⟶ (ā_r, ā_s)$ is an epimorphism of rings.

I've already proved that it's a homomorphism since

$(a+b)$ maps to $([a+b]_r, [a+b]_s)$, which is $([a]_r + [b]_r) + ([a]_s + [b]_s)$.

$(ab)$ maps to $([ab]_r, [ab]_s)$, which is $([a]_r [b]_r) ([a]_s [b]_s$)$.

These proposition are true owning to the operations defined in modular arithmetic. So it's a homomorphism.

If the map is surjective then for $(x,y) \in ℤ_r \times ℤ_s$ there's an $a \in ℤ$. This means that for any $(x,y)$, the equations $[a]_r = x$, and $[a]_s = y$ are simultaneously solvable. But I'm failing at showing the latter proposition.

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    $\begingroup$ Hint: Chinese remainder theorem. $\endgroup$ Jan 30 '21 at 16:15
  • $\begingroup$ ok, so we have $a ≡ x \quad mod\, r \Longleftrightarrow [a]_r = x$, and  $a ≡ y \quad mod\, s \Longleftrightarrow [a]_s = y$, and these are simultaneously solvable by crt since $r,s$ are coprime? $\endgroup$
    – gioretikto
    Jan 30 '21 at 16:24
  • $\begingroup$ Yes, precisely. $\endgroup$ Jan 30 '21 at 16:39
  • $\begingroup$ I believe that this is part of the proof of the Chinese Remainder Theorem (for arbitrary rings), hence it feels a little bit cheap to invoke the Chinese Remainder Theorem to prove that this homomorphism is surjective. I have given a more general proof below. $\endgroup$ Jan 30 '21 at 17:37
  • $\begingroup$ Remark: A ring epimorphism is not the same as a surjective ring homomorphism. A surjective ring homomorphism is always a ring epimorphism, but the converse is not true. $\endgroup$
    – azif00
    Jan 30 '21 at 20:38

Given any relatively prime integers $r$ and $s,$ consider the ring homomorphism $\pi : \mathbb Z \to (\mathbb Z / r \mathbb Z) \times (\mathbb Z / s \mathbb Z)$ defined by $\pi(n) = (n + r \mathbb Z, n + s \mathbb Z).$ By hypothesis that $r$ and $s$ are relatively prime, Bézout's Lemma implies that there exist integers $a$ and $b$ such that $ra + sb = 1.$ Consequently, for any element $(x + r \mathbb Z, y + s \mathbb Z)$ of $(\mathbb Z / r \mathbb Z) \times (\mathbb Z / s \mathbb Z),$ we have that $$(x + r \mathbb Z, y + s \mathbb Z) = (xra + xsb + r \mathbb Z, yra + ysb + s \mathbb Z) = (xsb + r \mathbb Z, yra + s \mathbb Z).$$ Can you find an element $n$ of $\mathbb Z$ such that $\pi(n) = (xsb + r \mathbb Z, yra + s \mathbb Z)$ to finish the proof?

Like I indicated in my comment above, the original question is part of the proof of the Chinese Remainder Theorem for arbitrary rings; the only thing to do is to replace "relatively prime integers" with "comaximal ideals $I$ and $J.$" From here, it follows that there exist elements $i \in I$ and $j \in J$ such that $i + j = 1_R,$ and the rest follows similarly to the above proof.


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