Chinese Remainder Theorem for Commutative Rings If $R$ is a commutative ring with $1$ and $I, J$ are ideals of $R$ that are pairwise coprime or comaximal (meaning $I + J = R$), then $IJ = I \cap J$, and the quotient ring $R/I \cap J$ is isomorphic to $R/I \times R/J$.

I will first prove that $IJ = I \cap J$.

If $x \in IJ$, then $x = \sum a_i b_i$ where $a_i \in I$ and $b_j \in J$. Thus for any fixed $i$, we have that $a_i b_i \in I$ since $a_i \in I$ and $a_i b_i \in J$ since $b_i \in J$. Thus, $\sum a_i b_i \in I$ and $\sum a_i b_i \in J$, which means that x = $\sum a_i b_i \in I \cap J$. Therefore, $IJ \in I \cap J$.

Note that since $R$ contains 1, we can write $I \cap J = R(I \cap J)$. Therefore, $I \cap J = R(I \cap J) = (I+J)(I \cap J) = I(I \cap J) + J(I \cap J) \subseteq IJ + JI = IJ$.

Now, I will prove that $R/IJ \cong R/I \times R/J$.

Consider the ring homomorphism $\phi: R \rightarrow R/I \times R/J$ defined by $\phi(r) = (r+ I, r+J)$ with kernel $I \cap J$. If we can prove that $\phi$ is surjective, then we can apply the first ring isomorphism theorem to show that $R/(I \cap J) \cong R/I \times R/J$.

Since $I + J = R$, for any $r_1, r_2 \in R$, we can write $r_1 = i_1 + j_1$ and $r_2 = i_2 + j_2$, where $i_1, i_2 \in I$ and $j_1, j_2 \in J$. Let $x = i_1 + j_2$. This gives us $x - r_1 = j_2 - j_1 \in J$ and $x - r_2 = i_1 - i_2 \in I$. In particular, we have $x + I = r_2 + I$ and $x + J = r_1 + J$. Hence, for all $(r_1 + I, r_2 + J)$, we can find an element $x$ such that $\phi(x) = ( x+ I, x+ J) = (r_2 +I, r_1 + J)$. This shows that $\phi$ is surjective.

Hence, we can conclude that $\phi: R/(I\cap J) \cong R/I \times R/J$.

(Can someone verify if the proof is correct?)

Using module homomorphism, the result also hold (I think.)

So, now my concern is proving the converse.

Converse to Chinese Remainder Theorem

According to this post the converse is false if the isomorphism is of rings. I don't quite understand how the link given shows this. I will put the link here: Can $R \times R$ be isomorphic to $R$ as rings? Can someone explain to me how the result in this link shows that it is only true when the isomorphism is of $R$-modules?

If so, am I correct to say that chinese remainder theorem shows that $R/I \cap J \cong R/I \times R/J$ both as rings and $R$-modules, but the converse is only true when the isomorphism is of $R$-modules?

What about this, Show that $R/(I \cap J) \cong (R/I) \times (R/J) $

Can it be used to show that the converse is true by saying that since $\phi: R \rightarrow R/I \times R/J$ is surjective if and only if $I+J = R$, so $R/I \cap J \cong R/I \times R/J$ if and only if $I+J = R$ I think we cannot because in the post the $\phi$ is predefined, but here we are talking about the general case, $R/I \cap J \cong R/I \times R/J$ and the homomorphism might not be the same. Am I right?

I am very confused about this. Note that this is only my first algebra course, so please don't use stuff that is too advanced to explain to me. Thanks.

  • $\begingroup$ Your proof is correct and it works as a ring isomorphism and as a module isomorphism as well. $\endgroup$ – user26857 Apr 21 '15 at 7:22
  • $\begingroup$ The converse holds for modules but not for rings. (For instance, note that $R\times R\not\simeq R$ as $R$-modules for any commutative unitary ring, so one can't produce similar conterexamples.) $\endgroup$ – user26857 Apr 21 '15 at 7:23
  • $\begingroup$ @user26857 isn't the (solution of the) second link saying that it is possible to have $R \times R \cong R$ as rings? $\endgroup$ – user10024395 Apr 21 '15 at 7:34
  • $\begingroup$ @user26857 no, i meant this one math.stackexchange.com/questions/895428/…. Cos you say "The converse holds for modules but not for rings. (For instance, note that R×R≄R as R-modules for any commutative unitary ring, so one can't produce similar conterexamples.)" I don't quite understand this. Can you explain more? $\endgroup$ – user10024395 Apr 21 '15 at 7:42
  • $\begingroup$ In the link you just gave it $R\times R\simeq R$ as $R$-modules holds for some non-commutative ring (check it!). For a commutative unitary ring this is impossible! (Btw, I've corrected your question. Take a look at the last edit. If your concern was about a possible example of a ring $R$ such that $R\times R\simeq R$ as $R$-modules, then this comment made it clear that this is not possible for commutative unitary rings, but it's possible for non-commutative rings. I've also added a comment in this vein under the question you linked.) $\endgroup$ – user26857 Apr 21 '15 at 7:55

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