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:n"e(liction of fractional Brownian by G. Gripenberg and I. Norros J. AppJ. Probah.~ SS (1006) 4O()-410 fractional Brownian motion -t) = fJin(2r(H - ~ )) t-H+i (T - t)2r fa uH-t (0- 10 0- +T)H-t +t ,arum 1 we have II + that bj + cj = 1; thus 1= R, so there exist elements bj E I) and cj E 1such n ITCbj+c)=l. j=2 Expand the left side and observe that any product containing at least one bj belongs to I), while c 2 ... cn belongs to IT 1=2 I J , the collection of all finite sums of products x 2 ... xn 14 Ring with Xj E ~. Thus we have elements bEl) and a E TI J=2 I j (a subset of each 9 with b + a = 1. Consequently, a == 1 mod I) and a == 0 mod ~ for j > 1, as desired. 1. By the argument of part (1), for each i we can find c j with c j == 1 mod I j and c j == 0 mod ~,j 7:: i. If a = a)c) + ... + ancn, then a has the desired properties. To see this, write a - ai = a - a/c j + alc j - 1), and note that a - ajc j is the sum of the 2. j 7:: i, and is therefore congruent to 0 mod Ii" We have b == a j mod I j for all i iff b - a == 0 mod I j for all i, that is, iff b - a E 3. ajc}' n7::1 Ii' and the result follows. Define!: R -t TIJ=IR1 Ii by j{a) = (a + I), ... , a + In). If a), ... , an E R, then by part (~) there is an element a E R such that a == a j mod I j for all i. But then j{a) = (a) + I), ... , an + In)' proving that! is surjective. Since the kernel of! is the intersection of the ideals ~, 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. Problems 1. Give an example of an ideal that is not a subring,and a subring that is not an ideal. 2. If the integers m j, i = 1, ... , n, are relatively prime in pairs, and a), ... , an are arbitrary integers, show that there is an integer a such that a == a j mod mj for all i, and that any two such int.;gers are congruent modulo m) ... m n . 3. If the integers m i , i = 1, ... , n, are relatively prime in pairs and m = m) ... mn , show that there is a ring isomorphism between Z m and the direct product TI7::IZm4 . Specifically, a mod m corresponds to (a mod m), ... , a mod mn). 4. Suppose that R = R) x R2 is a direct product of rings. Let R) be the ideal R) x {O} = {(r), 0): r) E R)}, and let R '2 be the ideal {CO, r2) : r2 E R2}. Show that RlR') ~ R2 and RlR2 ~ R). 5. If I), ... , In are ideals, the product I) ... In is defined as the set of all finite sums L:: a a j li 2i ... anj' where akj Elk' k = 1, ... , n. Assume that R is a commutative ring. Under the hypothesis of the Chinese remainder theorem, show that the intersection of the ideals I j coincides with their product. Ring 15 6. Let II' ... , In be ideals in the ring R. Suppose that RI TI;R/J; via a + n, I; -? nj Ii is isomorphic to (a + II' ... , a + In). Show that the ideals I j are relatively prime in pairs. Maximal and Prime Ideals If I is an ideal of the ring R, we might ask "What is the smallest ideal containing f' and "What is the largest ideal containing f'. Neither of these questions is challenging; the smallest ideal is I itself, and the largest ideal is R. But if I is a proper ideal and we ask for a maximal proper ideal containing I, the question is much more interesting. DEFINITION A maximal ideal in the ring R is a proper ideal that is not contained in any strictly larger proper ideal. THEO