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## Abstract algebra, 1st graduate year course by Ash R. By Ash R.

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K), then the intersection ∩ki=1 Ini is Ir where r is the least common multiple of the ni . If the ni are relatively prime in pairs, then r is the product of the ni . (iv) If R1 , . . , Rn are rings, the direct product of the Ri is deﬁned as the ring of ntuples (a1 , . . , an ), ai ∈ Ri , with componentwise addition and multiplication, that is, 12 CHAPTER 2. RING FUNDAMENTALS with (a1 , . . , an ) + (b1 , . . , bn ) = (a1 + b1 , . . , an + bn ), (a1 , . . , an )(b1 , . . , bn ) = (a1 b1 , .

9. Show that a prime ideal P cannot be the intersection of two strictly larger ideals I and J. 5 Polynomial Rings In this section, all rings are assumed commutative. To see a good reason for this restriction, consider the evaluation map (also called the substitution map) Ex , where x is a ﬁxed element of the ring R. This map assigns to the polynomial a0 + a1 X + · · · + an X n in R[X] the value a0 + a1 x + · · · + an xn in R. It is tempting to say that “obviously”, Ex is a ring homomorphism, but we must be careful.

An ), ai ∈ Ri , with componentwise addition and multiplication, that is, 12 CHAPTER 2. RING FUNDAMENTALS with (a1 , . . , an ) + (b1 , . . , bn ) = (a1 + b1 , . . , an + bn ), (a1 , . . , an )(b1 , . . , bn ) = (a1 b1 , . . , an bn ). The zero element is (0, . . , 0) and the multiplicative identity is (1, . . , 1). 7 Chinese Remainder Theorem Let R be an arbitrary ring, and let I1 , . . , In be ideals in R that are relatively prime in pairs, that is, Ii + Ij = R for all i = j. (1) If a1 = 1 (the multiplicative identity of R) and aj = 0 (the zero element of R) for j = 2, . 