5.3 Ideals and Factor Rings

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1 5.3 J.A.Beachy Ideals and Factor Rings from A Study Guide for Beginner s by J.A.Beachy, a supplement to Abstract Algebra by Beachy / Blair 27. Give an example to show that the set of all zero divisors of a commutative ring need not be an ideal of the ring. Solution: Theelements (1,0) and (0,1) of Z Z are zero divisors, but if theset of zero divisors were closed under addition it would include (1, 1), an obvious contradiction. 28. Show that in R[x] the set of polynomials whose graph passes through the origin and is tangent to the x-axis at the origin is an ideal of R[x]. Solution: We can characterize the given set as I = {f(x)r[x] f(0) = 0 and f (x) = 0}. Using this characterization of I, it is easy to check that if f(x),g(x) I, then f(x) ± g(x) I. If g(x) R[x] and f(x) I, then g(0)f(0) = g(0) 0 = 0, and D x (g(x)f(x)) = g (x)f(x) +g(x)f (x), and so evaluating at x = 0 gives g (0)f(0) + g(0)f (0) = g (0) 0+g(0) 0 = 0. This shows that I is an ideal of R[x]. 29. ToillustrateProposition5.3.7(b), givethelatticediagramofidealsofz 100 = Z/ 100, and the lattice diagram of ideals of Z that contain 100. Solution: Since multiplication in Z 100 corresponds to repeated addition, each subgroup of Z 100 is an ideal, and the lattice diagram is the same as that of the subgroups of Z 100, given in the following diagram. Z 100 2Z 100 5Z 100 4Z Z Z Z Z The ideals of Z that contain 100 correspond to the positive divisors of 100. Z Let R be the ring Z 2 [x]/ x 3 +1.

2 5.3 J.A.Beachy 2 Comment: Table 5.1 gives the multiplication table, though it is not necessary to compute it in order to solve the problem. Table 5.1: Multiplication in Z 2 [x]/ x 3 +1 x 2 +x x+1 x 2 +1 x 2 +x+1 1 x x 2 x 2 +x x 2 +x x+1 x x 2 +x x 2 +1 x+1 x+1 x+1 x 2 +1 x 2 +x 0 x+1 x 2 +x x 2 +1 x 2 +1 x 2 +1 x 2 +x x+1 0 x 2 +1 x+1 x 2 +x x 2 +x x 2 +x+1 x 2 +x+1 x 2 +x+1 x 2 +x+1 1 x 2 +x x+1 x 2 +1 x 2 +x+1 1 x x 2 x x 2 +1 x 2 +x x+1 x 2 +x+1 x x 2 1 x 2 x+1 x 2 +1 x 2 +x x 2 +x+1 x 2 1 x (a) Find all ideals of R. Solution: By Proposition (b), the ideals of R correspond to the ideals of Z 2 [x] that contain x We have the factorization x 3 +1 = x 3 1 = (x 1)(x 2 +x+1), so the only proper, nonzero ideals are the principal ideals given below. A = [x+1] = {[0],[x 2 +x],[x+1],[x 2 +1]} B = [x 2 +x+1] = {[0],[x 2 +x+1]} Comment: We note that R = A B, accounting for its 8 elements. (b) Find the units of R. Solution: We have [x] 3 = [1], so [x] and [x 2 ] are units. To show that the only units are 1, [x], and [x 2 ], note that [x+1][x 2 +x+1] = [x 3 +1] = [0], so [x+1] and [x 2 +x+1] cannot be units. This also excludes [x 2 +x] = [x][x+1] and [x 2 +1] = [x 2 ][1+x]. (c) Find the idempotent elements of R. Solution: Using the general fact that (a +b) 2 = a 2 +2ab+b 2 = a 2 +b 2 (since Z 2 [x] has characteristic 2) and the identities [x 3 ] = [1] and [x 4 ] = [x], it is easy to see that the idempotent elements of R are [0], [1], [x 2 +x+1], and [x 2 +x]. 31. Let S be the ring Z 2 [x]/ x 3 +x. Comment: It isn t necessary to construct Table 5.2 in order to answer (a) and (b). (a) Find all ideals of S. Solution: Over Z 2 we have the factorization x 3 +x = x(x 2 +1) = x(x+1) 2, so by Proposition (b) the proper nonzero ideals of S are the principal ideals generated by [x], [x+1], [x 2 +1] = [x+1] 2, and [x 2 +x] = [x][x+1]. A = [x] = {[0],[x 2 ],[x],[x 2 +x]} C = [x 2 +x] = {[0],[x 2 +x]} B = [x 2 +1] = {[0],[x 2 +1]} Comment: We note that S = A B. B+C = [x+1] = {[0],[x+1],[x 2 +1],[x 2 +x]}

3 5.3 J.A.Beachy 3 Table 5.2: Multiplication in Z 2 [x]/ x 3 +x x 2 x x 2 +x x 2 +1 x+1 1 x 2 +x+1 x 2 x 2 x x 2 +x 0 x 2 +x x 2 x 2 x x x 2 x 2 +x 0 x 2 +x x x x 2 +x x 2 +x x 2 +x x 2 +x x 2 +x x x 2 +1 x 2 +1 x 2 +1 x 2 +1 x+1 x 2 +x x 2 +x 0 x 2 +1 x 2 +1 x+1 x+1 1 x 2 x x 2 +x x 2 +1 x+1 1 x 2 +x+1 x 2 +x+1 x 2 x x 2 +x x 2 +1 x+1 x 2 +x+1 1 (b) Find the units of S. Solution: Since no unit can belong to a proper ideal, it follows from part (a) that we only need to check [x 2 +x+1]. This is a unit since [x 2 +x+1] 2 = [1]. (c) Find the idempotent elements of S. Solution: Since [x 3 ] = [x], we have [x 2 ] 2 = [x 2 ], and then [x 2 +1] 2 = [x 2 +1]. These, together with [0] and [1], are the only idempotents. 32. Show that the rings R and S in Problems 30 and 31 are isomorphic as abelian groups, but not as rings. Solution: Both R and S are isomorphic to Z 2 Z 2 Z 2, as abelian groups. They cannot be isomorphic as rings since S has a nonzero nilpotent element, [x 2 +x], but R does not. We can also prove this by noting that the given multiplication tables show that R has 3 units, while S has only Let I, J be ideals of the commutative ring R, and for r R, define the function φ : R R/I R/J by φ(r) = (r +I,r +J). (a) Show that φ is a ring homomorphism, with ker(φ) = I J. Solution: The fact that φ is a ring homomorphism follows immediately from the definitions of the operations in a direct sum and in a factor ring. Since the zero element of R/I R/J is (0 +I,0+J), we have r ker(φ) if and only if r I and r J, so ker(φ) = I J. (b) Show that if I +J = R, then φ is onto, and thus R/(I J) = R/I R/J. Solution: If I + J = R, then we can write 1 = x + y, for some x I and y J. Given any element (a + I,b + J) R/I R/J, consider r = bx + ay, noting that a = ax + ay and b = bx + by. We have a r = a bx ay = ax bx I, and b r = b bx ay = by ay J. Thus φ(r) = (a +I,b +J), and φ is onto. The isomorphism follows from the fundamental homomorphism theorem. Comment: This result is sometimes called the Chinese remainder theorem for commutative rings. It is interesting to compare this proof with the one given for Theorem

4 5.3 J.A.Beachy LetI,J beideals ofthecommutative ringr. Showthat ifi+j = R, theni 2 +J 2 = R. Solution: If I+J = R, then there exist a I and b J with a+b = 1. Cubing both sides gives us a 3 +3a 2 b+3ab 2 +b 3 = 1. Then we only need to note that a 3 +3a 2 b I 2 and 3ab 2 +b 3 J Show that x 2 +1 is a maximal ideal of R[x]. Solution: Since x 2 +1 is irreducible over R, it follows from results in Chapter 4 that R[x]/ x 2 +1 is a field. Therefore x 2 +1 is a maximal ideal. 36. Is x 2 +1 a maximal ideal of Z 2 [x]? Solution: Since x is not irreducible over Z 2, the ideal it generates is not a maximal ideal. In fact, x+1 generates a larger ideal since it is a factor of x Let R and S be commutative rings, and let φ : R S be a ring homomorphism. (a) Show that if I is an ideal of S, then φ 1 (I) = {a R φ(a) I} is an ideal of R. Solution: If a,b φ 1 (I), then φ(a) I and φ(b) I, so φ(a±b) = φ(a)±φ(b) I, and therefore a ± b φ 1 (I). If r R, then φ(ra) = φ(r)φ(a) I, and therefore ra φ 1 (I). By Definition 5.3.1, this shows that φ 1 (I) is an ideal of R. (b) Show that if P is a prime ideal of S, then φ 1 (P) is a prime ideal of R. Solution: Suppose that P is a prime ideal of R and ab φ 1 (P) for a,b R. Then φ(a)φ(b) = φ(ab) P, so φ(a) P or φ(b) P since P is prime. Thus a φ 1 (P) or b φ 1 (P), showing that φ 1 (P) is prime in R. 38. Prove that in a Boolean ring every prime ideal is maximal. Solution: We will prove a stronger result: if R is a Boolean ring and P is a prime ideal of R, then R/P = Z 2. Since Z 2 is a field, Proposition (a) implies that P is a maximal ideal. If a R, then a 2 = a, and so a(a 1) = 0. Since P is an ideal, it contains 0, and then a(a 1) P implies a P or a 1 P, since P is prime. Thus each element of R is in either P or 1+P, so there are only two cosets of P in R/P, and therefore R/P must be the ring Z In R = Z[i], let I = {m+ni m n (mod 2)}. (a) Show that I is an ideal of R. Solution: Let a+bi and c+di belong to I. Then a b (mod 2) and c d (mod 2), so a+c b+d (mod 2) and a c b d (mod 2), and therefore (a+bi)±(c+di) I. For m+ni R, we have (m+ni)(a+bi) = (ma nb)+(na+mb)i. Then ma nb na +mb (mod 2) since n m (mod 2) and nb nb (mod 2). This completes the proof that I is an ideal of R. (b) Find a familiar commutative ring isomorphic to R/I. Solution: Note that m + ni I if and only if m n 0 (mod 2). Furthermore, m + ni 1 + I if and only if 1 (m + ni) I, and this occurs if and only if

5 5.3 J.A.Beachy 5 1 m n (mod 2) if and only if m n 1 (mod 2). Since any element of m+ni R satisfies either m n 0 (mod 2) or m n 1 (mod 2), the only cosets in R/I are 0+I and 1+I. Therefore R/I = Z 2. ANSWERS AND HINTS 45. Let P and Q be maximal ideals of the commutative ring R. Show that R/(P Q) = R/P R/Q. Hint: If you can show that P +Q = R, then you can use Problem 33 (b). 48. Show that in Z[ 2] the principal ideal generated by 2 is a maximal ideal. Hint: Define a ring homomorphism from Z[ 2] onto Z 2.