Rings, Chinese Remainder Theorem

Chinese Remainder Theorem

Let H₁ H₂ … H_n be a set of coprime ideals. By coprime, we mean the sum of any two ideals spans the ring.

Let J be the product of all these ideals. We will prove R/J is isomorphic to the direct product of the quotient rings R/H_i, as i runs from 1 to n.

An element in R/J can be mapped to the i^th component in the direct product via R/H_i. This is a well defined ring homomorphism, since each H_i wholly contains J. We need to show it is 1-1 and onto.

Focus on H₁. We know x + y = 1 for some x in H₁ and y in H₂. Do the same for each H_i in the set. Multiply all these equations together, and something in H₁ + something in the product of the other ideals gives 1. Write this as x₁ + y₁ = 1. Reduce mod H₁, and y₁ = -1. If y₁ is mapped into any other quotient ring, other than R/H₁, it maps to 0,as it lives in the product of the other ideals.

With y₁ through y_n established, let's show the map is onto. Suppose z is an element in the direct product, where z_i is the i^th component. Let w be the sum of -y_iz_i, as i runs from 1 to n. Reduce w mod H_i and get c_i back again. Our ring homomorphism is onto.

If an element maps to 0 in all components, then it lies in each ideal H_i. The kernel of the map is the intersection of the ideals. If this is the same as the product J, we are done.

Whenever ideals are coprime, we will show that the intersection is the product. For two ideals, we have x+y = 1, and if w is in both ideals, w = (x+y)w = xw+yw, hence w is contained in the product H₁*H₂. The intersection and product coincide.

If H₁, H₂, and H₃ are pairwise coprime, then write x₁+y = 1 and x₂+z = 1. Let s = x₁x₂ + x₁z + x₂y. Note that s is in H₁. Let t = yz, and note that t is in H₂*H₃. Verify that s+t = 1. The ideals H₁ and H₂*H₃ are coprime. We know that the product and intersection of H₂ and H₃ coincide. Bringing in H₁, the intersection and product of all three ideals coincide. An inductive argument extends this result to finitely many ideals. The product is the intersection, the map is 1-1, and R/J is the same as the direct product of the quotient rings. The cardinality of R/J is the product of the cardinalities of R/H_i over all i.

If H₁ and H₂ are coprime then the same is true of any two powers of H₁ and H₂. Write x+y = 1 and raise to a sufficiently high power. By the binomial theorem, every term winds up in one of the two exponentiated ideals. An entire set of coprime ideals can be raised to various powers, and they remain pairwise coprime. For instance, the ideals generated by 17², 5³, and 11⁸ are coprime, simply because 17 5 and 11 are coprime.