By Ash R.

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Submit yr observe: First released October nineteenth 1989

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**Example text**

The integers Z form an integral domain that is not a ﬁeld. 2. Let Zn be the integers modulo n, that is, Zn = {0, 1, . . , n − 1} with addition and multiplication mod n. (If a ∈ Zn then a is identiﬁed with all integers a + kn, k = 0, ±1, ±2, . . }. Thus, for example, in Z9 the multiplication of 3 by 4 results in 3 since 12 ≡ 3 mod 9, and therefore 12 is identiﬁed with 3. Zn is a ring, which is an integral domain (and therefore a ﬁeld, since Zn is ﬁnite) if and only if n is prime. 4. Note that by deﬁnition of characteristic, any ﬁeld of prime characteristic p contains an isomorphic copy of Zp .

3 CHAPTER 2. RING FUNDAMENTALS Theorem If R is an integral domain, then a nonzero polynomial f in R[X] of degree n has at most n roots in R, counting multiplicity. Proof. 2), possibly applied several times, we have f (X) = q1 (X)(X − a1 )n1 , where q1 (a1 ) = 0 and the degree of q1 is n − n1 . If a2 is another root of f , then 0 = f (a2 ) = q1 (a2 )(a2 − a1 )n1 . e. a2 is a root of q1 (X). Repeating the argument, we have q1 (X) = q2 (X)(X −a2 )n2 , where q2 (a2 ) = 0 and deg q2 = n−n1 −n2 . 2), the quotient becomes constant, and we have f (X) = c(X − a1 )n1 .

8). If p is a nonzero prime ideal of R, then p is contained in the maximal ideal q , so that q divides the prime p. Since a maximal ideal must be proper, q cannot be a unit, so that p and q are associates. But then p = q and p is maximal. The proof of the converse is given in the exercises. 6 Problems 1-6 form a project designed to prove that if R is a UFD and every nonzero prime ideal of R is maximal, then R is a PID. Let I be an ideal of R; since {0} is principal, we can assume that I = {0}. Since R is a UFD, every nonzero element of I can be written as up1 .