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.

  • 1
    $\begingroup$ Hint: Chinese remainder theorem. $\endgroup$ Jan 30, 2021 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, 2021 at 16:24
  • $\begingroup$ Yes, precisely. $\endgroup$ Jan 30, 2021 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, 2021 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, 2021 at 20:38

1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.