Split a quotient using comaximal ideals

Try

Find an integer that is one modulo three and two modulo four. Testing the residues modulo twelve gives ten. The Chinese remainder theorem explains both existence and uniqueness modulo twelve, and generalizes the problem from integers to ideals in any commutative unital ring.

The required condition is comaximality: I+J=R. This means some i∈I and j∈J satisfy i+j=1. Distinct ideals need not be comaximal.

Predict

If i+j=1, then j is one modulo I and zero modulo J, while i has the opposite behavior. Predict a combination of a,b,i,j that has residue a modulo I and b modulo J. These two elements act as selectors for the desired coordinates.

Experiment

For I=3Z and J=4Z, use −3+4=1. The selector four preserves the modulo-three coordinate, while negative three preserves the modulo-four coordinate.

Desired residues	Constructed integer	Check
1 modulo3, 2 modulo4	1·4+2·(−3)=−2	Equivalent to10 modulo12
2 modulo3, 1 modulo4	2·4+1·(−3)=5	Both residues correct

The construction replaces a search with a formula. Changing representatives of a or b changes the constructed answer by an element of the kernel, so the resulting quotient class remains the same.

Understand

Consider the unital map R→R/I×R/J sending r to its two residue classes. Its kernel is I∩J. Comaximality makes it surjective: aj+bi maps to the prescribed pair (a+I,b+J). The first isomorphism theorem gives R/(I∩J)≅R/I×R/J.

Under the same hypothesis, I∩J=IJ. The inclusion IJ⊆I∩J always holds. For a in the intersection, write a=a(i+j)=ai+aj. In ai, the first factor lies in J and the second in I; in aj, the first lies in I and the second in J. Thus both terms belong to the product ideal. This proves the reverse inclusion and yields R/IJ≅R/I×R/J.

The theorem iterates for finitely many pairwise comaximal ideals, with the quotient by their product corresponding to the product of quotient rings. Pairwise comaximality must be justified; arbitrary distinct ideals are insufficient. For 4Z and 6Z, their sum is 2Z. Simultaneous residues must have the same parity, so the pair zero modulo four and one modulo six cannot occur.

For distinct a,b in a field F, the ideals (x−a) and (x−b) are comaximal because subtracting their generators gives the nonzero constant b−a, a unit. Hence F[x]/((x−a)(x−b))≅F×F, through evaluation at a and b. This also explains zero divisors: the nonzero coordinate pairs (1,0) and (0,1) multiply to zero.

Repeated factors behave differently. The ideals (x−a) and themselves are not comaximal unless they equal the whole ring, which they do not. Therefore a quotient by (x−a)² is not split by this CRT argument; it contains a nonzero nilpotent class represented by x−a.

Apply

Construct an explicit inverse to the residue-pair map in a coprime integer example. Then show that two solutions differ by a multiple of both moduli and hence their product. Keep existence and uniqueness as separate claims.

In polynomial examples, identify the map and its kernel before claiming a product decomposition. A matching cardinality is not enough to prove a ring isomorphism, especially when zero divisors or nilpotents distinguish possible structures.

Continue

Fraction fields next enlarge a domain by adding inverses instead of collapsing an ideal. Domain cancellation will be the essential ingredient that makes fraction equality and arithmetic consistent.