§ Proof of chinese remainder theorem on rings

§ General operations on ideals

We have at our hands a commutative ring $R$, and we wish to study the ideal structure on the ring. In particular, we can combine ideals in the following ways:
1. $I + J \equiv \{ i + j : \forall i \in I, j \in J \}$
2. $I \cap J \equiv \{ x : \forall x \in I \land x \in J \}$
3. $I\oplus J \equiv \{ (i, j) : \forall i \in I \land j \in J \}$
4. $IJ \equiv \{ ij : \forall i \in I \land j \in J \}$
We have the containment:
$IJ \subseteq I \cap J \subseteq I, J \subseteq I + J \subseteq R$

§ $IJ$ is a ideal, $IJ \subseteq I \cap J$

it's not immediate from the definition that $IJ$ is an ideal. The idea is that given a sum $\sum_k i_k j_k \in IJ$, we can write each $i_k j_k = i'_k$, since the ideal $I$ is closed under multiplication with $R$. This gives us $\sum i'_k = i'' \in I$. Similarly, we can interpret $\sum_k i_k j_k = \sum_k j'_k = j''k \in J$. Hence, we get the containment $IJ \subseteq I \cap J$.

§ $I \cap J subseteq I$, $I \cap J \subseteq J$

Immediate from the inclusion function.

§ $I, J \subseteq I + J$

Immediate from inclusion

§ CRT from an exact sequence

There exists an exact sequence:
\begin{aligned} 0 \rightarrow I \cap J \xrightarrow{f} I \oplus J \xrightarrow{g} I + J \rightarrow 0 \\ &f(r) = (r, r) \\ &g((i, j)) = i + j \end{aligned}
We are forced into this formula by considerations of dimension. We know:
\begin{aligned} &dim(I \oplus J) = dim(I) + dim(J) \\ &dim(I + J) = dim(I) + dim(J) - dim(I \cap J) \text{[inclusion-exclusion]} \\ &dim(I + J) = dim(I \oplus J) - dim(I \cap J) \\ &dim(I + J) - dim(I \oplus J) + dim(I \cap J) = 0\\ &V - E + F = 2 \end{aligned}
By analogy to euler characteristic which arises from homology, we need to have $I \oplus J$ in the middle of our exact sequence. So we must have:
$0 \rightarrow ? \rightarrow I \oplus J \rightarrow ?\rightarrow 0$
Now we need to decide on the relative ordering between $I \cap J$ and $I + J$.
• There is no universal way to send $I oplus J \rightarrow I \cap J$. It'san unnatural operation to restrict the direct sum into the intersection.
• There is a universal way to send $I \oplus J \rightarrow I + J$: sumthe two components. This can be seen as currying the addition operation.
Thus, the exact sequence must have $I + J$ in the image of $I \oplus J$. This forces us to arrive at:
$0 \rightarrow I \cap J \rightarrow I \oplus J \rightarrow I + J \rightarrow 0$
The product ideal $IJ$ plays no role, since it's not possible to define a product of modules in general (just as it is not possible to define a product of vector spaces). Thus, the exact sequence better involve module related operations. We can now recover CRT:
\begin{aligned} 0 \rightarrow I \cap J \xrightarrow{f} I \oplus J \xrightarrow{g} I + J \rightarrow 0 \\ 0 \rightarrow R \xrightarrow{f} R \oplus R \xrightarrow{g} R \rightarrow 0 \\ 0 \rightarrow R / (I \cap J) \rightarrow R/I \oplus R /J \rightarrow R/(I + J) \rightarrow 0 \end{aligned}