Question

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 mi ... mn. 4. If the integers mi, i = 1,..., n, are relatively prime in pairs and m = mi...mn, show that there is a ring isomorphism between Zm and the direct product Ili=1 Zmi. Specifically, a mod m corresponds to (a mod mi,..., a mod mn). 5. Suppose that R=R Ris a direct product of rings. Let Ribe the ideal R; x {0} = {(r1,0): rı € Ri}, and let Rs be the ideal {(0,r2): r2 € R2}. Show that R/R R2 and R/R, R. 6. If I1, ... ,In are ideals, the product ll ...In is defined as the set of all finite sums Si aliazi... ani, where aki e Ik, k = 1,...,n. (See the proof of part (1) of 2.3.7; a brief check shows that the product of ideals is an ideal.] Assume that R is a commutative ring. Under the hypothesis of the Chinese remainder theorem, show that the intersection of the ideals le coincides with their product. 7. Let I1,..., In be ideals in the ring R. Suppose that R/ ni li is isomorphic to II, R/I; via a +nili (a +11,..., a +In). Show that the ideals Ii are relatively prime in pairs.

Answer 1

37 To show, 7 an integer a such that a=ai (mod mi) vi. and any two such integers are congruent to modulo mm..... ma ie, We have to find a solution of the system of congruence a= a (mod mi) aza (med me) aan(modmn) For this we shall use a theorem. The theorem is- a, b,m be integers with my and ged(a,m) eI. Then the congruence ax=b[mod m) has a solution : Further it to is a solution, then any intoger &, is a solution of the given Linear congruence it and only it x = x (mod m). In this case, we say that that the solution is unique modulo m. ret, M= m, m... Man and W;= M, where is 1,2,...,n. since, for it), ged (mi, mi) = 1, we find that gedo - 1 Then by theorem, the linear congruence N; x = 1 (mod mi) has a unique solution, saj bi modulo mi. let, a= a, b, a, b 2 NL + . . . + a, b Nn. We show, a, is a soln. of the pirer system of congruences. We first observe that, i = 2 mm. M. Miti mno . (mal mi) for ja 1, 2, . ., tit, ...N. then, az= a; b;N; (mod mi) for 121,2,...,n. But Ni bi = 1 (mod mi) and implies that Nib. 9 = a. (mod m.).

Hence, a= a; (mod mi). This shows ao is a solution of the system of congruence. Now, we have to prove and part. let, va' be a solution of the System of congreence. Then, a'=a; (mod mi) for iz 1, 2, - -,n. Because ged (mi, m; )= 1 for it), it follows that m.m.... min divides osa-ao Thus, we find that a = a (mod m m.....my).