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I've known the Chinese Remainder Theorem like

Let the Ring $R$ and its Ideals $I$ and $J$ s.t. $I+J=R$

Then $R/(I \cap J) \simeq R/I +R/J$

The question is

First) What is the condition of $IJ =I\cap J$ statement is true?

I.e.) Want to know the case that $(R/IJ) \simeq (R/(I \cap J))\simeq (R/IJ)$ is true.

Second) Let the R is the commutative ring with the unity like a domain.

Then, Does $(R/IJ) \simeq (R/(I \cap J))\simeq (R/IJ) $ always true?

My proof(It is just my thought) : $I,J \subset I+J \subset R$ and $I+J =R$

Hence $1_R \in I+J$, I and J are relatively prime. Therefore $IJ = I \cap J$

(Might be a wrong, But I could't find which point I was wrong.)

Third) Find the integers $m,n$

$Z[i]/\langle 4+8i \rangle \simeq Z_m \times Z_n$

Is it possible utilizing the chinese remainder theorem with this question?

e.g.) $(Z[i]/\langle 4+8i \rangle) \simeq (Z[i]/\langle 4 \rangle) \times (Z[i]/\langle 1+2i \rangle) $

Any help or advice would be appreciated.

Thank you.

se-hyuck yang
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1 Answers1

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I am not sure of necessary conditions for your first question, but a sufficient condition for $IJ = I \cap J$ is that $I,J$ are comaximal (ie. $I + J = R$), the same condition in the chinese remainder theorem.

If you are asking if the chinese remainder theorem is true in an integral domain (which is what it looks like you are asking in the second question), then the answer is no. Consider $\mathbb{Z}$ and the ideals $I = 4\mathbb{Z}$ and $J = 2\mathbb{Z}$. Then, $\mathbb{Z}/IJ \cong \mathbb{Z}/8\mathbb{Z}$ and $\mathbb{Z}/(I \cap J) \cong \mathbb{Z}/4\mathbb{Z}$. Hence, $\mathbb{Z}/IJ \not \cong \mathbb{Z}/(I\cap J)$.

I think the answer to the third question is Yes. This is because $4 + 8i = 4(1 +2i)$. Defining $I = \langle 4 \rangle $ and $J = \langle 1 + 2i \rangle $ we see that $\langle 4 + 8i\rangle = IJ$. Moreoever (and you should check this), $I + J = \mathbb{Z}[i]$. Thus, by the chinese remainder theorem $\mathbb{Z}[i]/ \langle 4+8i \rangle \cong \mathbb{Z}[i]/\langle 4 \rangle \times \mathbb{Z}[i]/ \langle 2 + i \rangle$ which as you elude to is $\mathbb{Z}_4[i] \times \mathbb{Z}_5$

Mike
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  • Oops. I mistyped the one that should have written the $ \langle 4+8i \rangle$ not $4+3i$. So What if the $Z[i]/\langle 4+8i \rangle$ in third question? Then Can we use the C.R.T? – se-hyuck yang Jul 05 '19 at 05:30
  • @se-hyuckyang The answer is yes, I have just edited my response. Please take a look. – Mike Jul 05 '19 at 13:55
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    @se-hyuckyang I should correct myself about the previous example you asked about $3 + 4i$. You still cannot use the Chinese remainder theorem. But, not for the reason that I stated. In fact, $3 + 4i$ is reducible over $\mathbb{Z}[i]$ with $3 + 4i = (2 + i)^2$. Clearly, the ideals $(2 + i)$ and $(2+ i)$ are not comaximal. So, I do not see how you could use the Chinese remainder Theorem here. In this case, $\mathbb{Z}[i]/(3+ 4i) \cong \mathbb{Z}_{25}$. Sorry for any confusion. – Mike Jul 05 '19 at 17:55
  • Thanks for response and edition! :) But would you please more explain Why does $R = (Z[i]/\langle 3+4i \rangle) $ is isomorphic with $Z_{25}$ ? All I know its order and $char(R)$ are 25. I meant there might be a some other rings that order and $char(R)$ are 25. So Are there any clear evidence that I can say $R \simeq Z_{25}$? – se-hyuck yang Jul 06 '19 at 01:39
  • @se-hyuckyang A nice discussion of this can be seen in this post https://math.stackexchange.com/questions/23358/quotient-ring-of-gaussian-integers – Mike Jul 06 '19 at 01:56