By Ash R.
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Extra info for Abstract algebra, 1st graduate year course
Cn belongs to j=2 Ij , the collection of all ﬁnite sums of products x2 . . xn n with xj ∈ Ij . Thus we have elements b ∈ I1 and a ∈ j=2 Ij (a subset of each Ij ) with b + a = 1. Consequently, a ≡ 1 mod I1 and a ≡ 0 mod Ij for j > 1, as desired. (2) By the argument of part (1), for each i we can ﬁnd ci with ci ≡ 1 mod Ii and ci ≡ 0 mod Ij , j = i. If a = a1 c1 + · · · + an cn , then a has the desired properties. To see this, write a − ai = a − ai ci + ai (ci − 1), and note that a − ai ci is the sum of the aj cj , j = i, and is therefore congruent to 0 mod Ii .
Since the kernel of f is the intersection of the ideals Ij , the result follows from the ﬁrst isomorphism theorem for rings. ♣ The concrete version of the Chinese remainder theorem can be recovered from the abstract result; see Problems 3 and 4. 4. 3 1. 4, Problems 1 and 2, are ring isomorphisms as well. 2. Give an example of an ideal that is not a subring, and a subring that is not an ideal. 3. If the integers mi , i = 1, . . , n, are relatively prime in pairs, and a1 , . . , an are arbitrary integers, show that there is an integer a such that a ≡ ai mod mi for all i, and that any two such integers are congruent modulo m1 .
The representation is unique because the only way to produce the i-th component hi of g is for hi to be the ith component of the factor from H i . If a group G contains normal subgroups H1 , . . , Hn such that G = H1 · · · Hn , and each g ∈ G can be uniquely represented as h1 · · · hn with hi ∈ Hi , i = 1, 2, . . , n, we say that G is the internal direct product of the Hi . As in the case of two factors, if G is the internal direct product of the Hi , then G is isomorphic to the external direct product H1 × · · · × Hn ; the isomorphism f : H1 × · · · × Hn → G is given by f (h1 , .
Abstract algebra, 1st graduate year course by Ash R.