§ Proof of chinese remainder theorem on rings

§ General operations on ideals

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

§ IJIJ is a ideal, IJIJIJ \subseteq I \cap J

it's not immediate from the definition that IJIJ is an ideal. The idea is that given a sum kikjkIJ\sum_k i_k j_k \in IJ, we can write each ikjk=iki_k j_k = i'_k, since the ideal II is closed under multiplication with RR. This gives us ik=iI\sum i'_k = i'' \in I. Similarly, we can interpret kikjk=kjk=jkJ\sum_k i_k j_k = \sum_k j'_k = j''k \in J. Hence, we get the containment IJIJIJ \subseteq I \cap J.

§ IJsubseteqII \cap J subseteq I, IJJI \cap J \subseteq J

Immediate from the inclusion function.

§ I,JI+JI, J \subseteq I + J

Immediate from inclusion

§ CRT from an exact sequence

There exists an exact sequence:
0IJfIJgI+J0f(r)=(r,r)g((i,j))=i+j \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:
dim(IJ)=dim(I)+dim(J)dim(I+J)=dim(I)+dim(J)dim(IJ)[inclusion-exclusion]dim(I+J)=dim(IJ)dim(IJ)dim(I+J)dim(IJ)+dim(IJ)=0VE+F=2 \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 IJI \oplus J in the middle of our exact sequence. So we must have:
0?IJ?0 0 \rightarrow ? \rightarrow I \oplus J \rightarrow ?\rightarrow 0
Now we need to decide on the relative ordering between IJI \cap J and I+JI + J.
  • There is no universal way to send IoplusJIJI 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 IJI+JI \oplus J \rightarrow I + J: sumthe two components. This can be seen as currying the addition operation.
Thus, the exact sequence must have I+JI + J in the image of IJI \oplus J. This forces us to arrive at:
0IJIJI+J0 0 \rightarrow I \cap J \rightarrow I \oplus J \rightarrow I + J \rightarrow 0
The product ideal IJIJ 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:
0IJfIJgI+J00RfRRgR00R/(IJ)R/IR/JR/(I+J)0 \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}

§ References