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The ring $\mathbb{Z}_5 \times \mathbb{Z}_3$ with two operations defined as follows:

$(u_1,u_2) +(v_1,v_2) = ((u_1+v_1) \mod 5, (u_2 + v_2) \mod 3)$

$(u_1,u_2) (v_1,v_2) = ((u_1v_1) \mod 5, (u_2 v_2) \mod 3)$

Is this ring isomorphic to the ring $\mathbb{Z}_{15}$ ?

The definition of an isomorphism of rings is that it is a bijection and a ring homomorphism. I often prove that it is either not bijective or not a ring homomorphism. However, the results I found suggest that these two rings are isomorphic to each other, and I am somewhat confused about how to prove this statement. Can anyone help me?

JuztHb
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  • What results did you find? – Michael Burr Jan 18 '25 at 12:25
  • @MichaelBurr Chinese Remainder Theorem, but that theorem is not covered in my curriculum. – JuztHb Jan 18 '25 at 12:27
  • It states that $\mathbb{Z}{m} \times \mathbb{Z}{n} \cong \mathbb{Z}_{m \cdot n}$ when gcd (m,n) = 1. – JuztHb Jan 18 '25 at 12:30
  • The map given by the canonical projections works: $\mathbb{Z}{mn} \to \mathbb{Z}_m\times\mathbb{Z}_n$ given by $[x]{mn} \mapsto ([x]{m}, [x]{n})$, for $\gcd(m,n)=1$. It is injective, and since the cardinalities of both rings are equal, we are done. You don't need more. – Dietrich Burde Jan 18 '25 at 12:56

1 Answers1

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Define the function $f: \mathbb{Z}_{15} \to \mathbb{Z}_5 \times \mathbb{Z}_3 $ by $ f(n) = (n \mod 5, n \mod 3) $. Using this definition, we have $ f(0) = (0, 0) $, $ f(1) = (1, 1) $, $ f(2) = (2, 2) $, $ f(3) = (3, 0) $.

Note that $ f(m + n) = (m + n \mod 5, m + n \mod 3) = (m \mod 5, m \mod 3) + (n \mod 5, n \mod 3) = f(m) + f(n) $. Similarly, $ f(mn) = f(m)f(n) $. This proves that $ f $ is a homomorphism.

If $ f(n) = 0 $, then $ n \equiv 0 \mod 5 $ and $ n \equiv 0 \mod 3 $. By the Chinese Remainder Theorem, we must have $ n \equiv 0 \mod 15 $. This shows that the homomorphism $ f $ is injective. Since the sets are finite, $ f $ is a bijective homomorphism and, consequently, an isomorphism.

Adriano GS
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  • Is there a way to prove this without using the Chinese Remainder Theorem ? Because that theorem is not covered in my curriculum. – JuztHb Jan 18 '25 at 12:39
  • Since the sets are finite, you can list all the values of $n \in \mathbb{Z}_{15}$ for which $f(n) = 0$ (or better, $f(n) = (0, 0)$). You will see that the only $n$, $0 \leq n < 15$, for which $f(n) = 0$ is $n = 0$. Using this, you conclude that the homomorphism is injective. – Adriano GS Jan 18 '25 at 12:44
  • Is it not necessary to show that f is surjective, but just to mention that the sets are finite? – JuztHb Jan 18 '25 at 12:49
  • Please don't answer duplicate questions. Please also read the meta announcement regarding quality standards. – Martin Brandenburg Jan 18 '25 at 12:55
  • @JuztHb The function is injective, and the sets are finite with the same number of elements; therefore, the function is also surjective. For this reason, I only mentioned that the sets are finite.

    Sorry MartinBrandenburg. We will be more careful next time.

     – Adriano GS Jan 18 '25 at 13:08