Ash R.'s Abstract algebra, 1st graduate year course PDF

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An ∈ R, then by part (2) there is an element a ∈ R such that a ≡ ai mod Ii for all i. But then f (a) = (a1 + I1 , . . , an + In ), proving that f is surjective. 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.

This follows because a|b means that b = ac for some c ∈ R. For short, divides means contains. It would be useful to be able to recognize when an integral domain is a UFD. The following criterion is quite abstract, but it will help us to generate some explicit examples. 22 CHAPTER 2. 6 Theorem Let R be an integral domain. (1) If R is a UFD then R satisfies the ascending chain condition (acc) on principal ideals: If a1 , a2 , . . belong to R and a1 ⊆ a2 ⊆ . . , then the sequence eventually stabilizes, that is, for some n we have an = an+1 = an+2 = .

Pn = vq1 q2 . . qm where the pi and qi are irreducible and u and v are units. Then p1 is a prime divisor of vq1 . . qm , so p1 divides one of the qi , say q1 . But q1 is irreducible, and therefore p1 and q1 are associates. Thus we have, up to multiplication by units, p2 . . pn = q2 . . qm . By an inductive argument, we must have m = n, and after reordering, pi and qi are associates for each i. ♣ We now give a basic sufficient condition for an integral domain to be a UFD. 7 Definition A principal ideal domain (PID) is an integral domain in which every ideal is principal, that is, generated by a single element.

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


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