Exact sequence $0\to I\cap J \to R \to (R/ I) \times (R/ J) \to R/ (I+J) \to 0$ and Chinese Remainder Theorem

Question

I need to find an exact sequence such that where $I$, $J$ are ideals of $R$. I know that I have to find homomorphisms

\begin{align*} &a\colon I\cap J \to R \\ &b\colon R \to (R/I) \times (R/J) \\ &c\colon(R/I) \times (R/J) \to R/(I+J) \end{align*}

such that $\ker a=(0), \ker b=\operatorname{im}a, \operatorname{im}b=\ker c,\operatorname{im}c= R/(I+J)$ but I fail to. This is my first touch with this kind of sequences and I do not know how to think. Can someone help? Then I need to prove that this chain is a result of the Chinese remainder theorem.

Edit: my tries so far for the functions (I do not think it is correct) \begin{align*} a(k)&=rk , r\in R \\ b(x)&=(\frac{1}{k} x-r , \frac{1}{k} x-r), \frac{1}{k}\in I\cap J \\ c(x,y)&=x+y+2r-\frac{x}{k}-\frac{y}{k}\\ \end{align*}

Answer 1

Your definitions are, well, weird. Your $a$ contains an unnecessary factor from $R$ which can be simply put as $r=1$. And once you have introduced this unneeded factor there you have to carry it around and take care of it while defining $b$ and $c$ which only complicates the matter.

On a final note: This appears as P. Aluffi's "Algebra: Chapter $0$" exercise V$\mathbf{.6.1}$. I am not sure if you are aware of this but it might be good to know for future readers.

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There are rather natural choices for $a$ and $b$ given by $I\cap J\to R$ being the inclusion and $R\to R/I\times R/J$ being componentwise reduction (i.e. $r\mapsto (r+I,r+J)$).
You can check quite easily that the $a$ is injective (which is equivalent to $\ker a=(0)$) and that $\ker b=I\cap J$ as $(r+I,r+J)=(I,J)$ iff $r\in I$ and $r\in J$ iff $r\in I\cap J$. The tricky part is $c$. One possible definition is

$$ c\colon R/I\times R/J\to R/(I+J),\,(r+I,s+J)\mapsto r-s+(I+J)\,. $$

It is easy to check that this map is well-defined. Clearly $\operatorname{im}b\subseteq\ker c$. For the reverse inclusion take $(r+I,s+J)\in\ker c$. Then, $r-s\in I+J$, that is, there are $i\in I,j\in J$ such that $r-s=i+j$ iff $r-i=s+j$. It follows that

$$ b(r-i)=(r-i+I,s+j+J)=(r+I,s+J) $$

which concludes the proof. The surjectivity of $c$, i.e. $\operatorname{im}c=R/(I+J)$, is straigtforward to check.

Regarding your second question: It is the other way around! That is, the Chinese Remainder Theorem is a direct consequences of this exact sequences. Indeed, the only assumption missing is comaximality of $I$ and $J$ which means $I+J=R$. In this case $R/(I+J)=0$ and exactness dictates $R/(I\cap J)\cong R/I\times R/J$.

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Just for reference - for future readers and myself - I will include a more advanced proof of this proposition. First, note that $$ 0\to R\xrightarrow{x\mapsto(x,x)}R^2\xrightarrow{(x,y)\mapsto x-y} R\to 0 $$ is exact and likewise is the variation $$ 0\to I\cap J\to I\oplus J\to I+J\to 0 $$ as also mentioned here. Now consider the following diagram:

$\require{AMScd}$ \begin{CD} @. 0 @. 0 @. 0 @. \\ @. @VVV @VVV @VVV \\ 0 @>>> I\cap J @>>> I\oplus J @>>> I+J @>>>0 \\ @. @VVV @VVV @VVV \\ 0 @>>> R @>>> R^2 @>>> R @>>>0 \\ @. @VVV @VVV @VVV \\ 0 @>>> R/I\cap J @>>> R/I\oplus R/J @>>> R/I+J @>>>0 \\ @. @VVV @VVV @VVV @. \\ @. 0 @. 0 @. 0 \end{CD}

All columns are trivially exact (being quotient sequences) and the first two rows are exact too as indicated above. Moreover, the upper two rows are compatible being variations of diagonal embeddings and difference projections. If we now induce the last row similarily we are in the situation of the $3\times3$ lemma (one of the basic diagram lemmata) and may conclude that $$ 0\to R/I\cap J\to R/I\oplus R/J\to R/I+J\to 0 $$ is exact. From hereon it is easy to recover the original statement.

Answer 2

Define: \begin{align*} a(k) &= k &\text{ for }k\in I\cap J.\\ b(r) &= (r+I,r+J) &\text{ for }r\in R.\\ c(r+I,s+J) &= (r-s)+(I+J) &\text{ for }(r+I,s+J)\in R/I\times R/J. \end{align*}

$a$ and $b$ are the "obvious" maps $I\cap J\to R$ and $R\to (R/I)\times (R/J)$. There is also an "obvious" map $(R/I)\times (R/J)$ given by $(r+I,s+J) \mapsto (r+s)+(I+J)$. But for exactness, we need to have $0 = c(b(r)) = c(r+I,r+J)$ for all $r\in R$, and this motivates using subtraction instead of addition in the definition of $c$ above.

I'll leave it to you to check that these are homomorphisms (this includes checking that $c$ is well-defined) and that the resulting sequence is exact.