Mathematics 351 – Abstract Algebra 1 Solutions for Problem Set 6 October 18, 2007

Section 6.1

2. We let

I = {p(x) = a_{0} + a_{1}x + ⋯ ∈ ℤ[x] : a_{0} is even}.

I is clearly nonempty since 0, 2, 4 + x^{2} ∈ I and so forth. Moreover, if p(x) = a_{0} + a_{1}x + ⋯ and q(x) = b_{0} + b_{1}x + ⋯ are in I, then p(x) − q(x) = a_{0} − b_{0} + (a_{1} − b_{1})x + · · ·. Since a_{0}, b_{0} are divisible by 2, so is a_{0} − b_{0}. Hence I is closed under differences. If r(x) = c_{0} + c_{1}x + ⋯ is an arbitrary polynomial in ℤ[x], and p(x) as above is in I, then r(x)p(x) = a_{0}c_{0} + (a_{0}c_{1} + a_{1}c_{0})x + · · ·. Since a_{0} is divisible by 2, so is a_{0}c_{0}. Hence I is closed under multiplication by arbitrary r(x) ∈ ℤ[x]. This shows I is an ideal by the definition.

6.

a. The set of nonunits in ℤ_{8} is I = {0, 2, 4, 6}. This (nonempty!) set is closed under differences and products by arbitrary elements in ℤ_{8}. This is true because these are the classes of the even numbers mod 8. Since 8 is also even, a ≡ 2k mod 8 and b ≡ 2ℓ mod 8 implies a − b ≡ 2(k − ℓ) mod 8. If 2(k − ℓ) ≡ c mod 8, then 2(k − ℓ) − c = 8m for some integer m. But then c = 2(k − ℓ) − 8m is also even. Similarly if r is an arbitrary class mod 8, and a ≡ 2k mod 8, then if 2rk ≡ c mod 8, we have 2k − c = 8m for some integer m, so c is also divisible by 2.

b. The set of nonunits in ℤ_{9} is I = {0, 3, 6}. This is an ideal for a similar reason to what happened in part a – this is the set of classes of numbers divisible by 3 in ℤ_{9}.

7.

a. In ℤ_{5}, (0) = {0}, and (1) = (2) = (3) = (4) = ℤ_{5}.

b. In ℤ_{9}, (0) = {0}, (1) = (2) = (4) = (5) = (7) = (8) = ℤ_{9}, and (3) = (6) = {0, 3, 6}.

c In ℤ_{12}, (0) = {0}, (1) = (5) = (7) = (11) = ℤ_{12}, (2) = (10) = {0, 2, 4, 6, 8, 10}, (3) = (9) = {0, 3, 6, 9}, (4) = (8) = {0, 4, 8}, and (6) = {0, 6}.

12.

a An example is (2) = (−2). This follows since every even integer n = 2k is also n = (−2)(−k), and vice versa. (This is the only way the equality (c) = (d) can happen if c ≠ d in ℤ, since c ∈ (d) says c = dm for some integer m. Conversely, d ∈ (c) says d = cn for some integer n. Then substituting, we see c = c(nm), so nm = 1 ∈ ℤ. This is only possible if n = m = 1 or n = m = −1.) b To show the equality, we show inclusions in both directions. ⊆: Let n ∈ (4, 6). Then

n = 4a + 6b for some integers a, b. But then n = 2(2a + 3b), and 2a + 3b ∈ ℤ, so n ∈ (2). ⊇: Note that 2 = 4(−1) + 6(1) ∈ (4, 6). Then for all n ∈ ℤ, 2n = 4(−n) + 6(n) ∈ (4, 6). Hence (2) ⊆ (4, 6).

c Similar to part b. First we show ⊆: Let n ∈ (6, 9, 15). Then n = 6a+9b+12c = 3(2a+3b+4c) for some a, b, c ∈ ℤ. This shows (6, 9, 15) ⊆ (3). On the other hand, to show ⊇: since 3 = 6(−1) + 9(1), we have 3n = 6(−n) + 9(n) + 12(0) ∈ (6, 9, 15). Comment: Note that this argument actually shows somewhat more than the problem asked. In fact (6, 9) = (3), and the third generator is unnecessary.

13.

a. If 1_{R} ∈ I and I is an ideal, then for all r ∈ R, r · 1_{R} = r ∈ I. Hence R ⊆ I. Since I ⊆ R by definition, this shows I = R.

b. If u ∈ I and u is a unit, then say a ∈ R satisfies au = 1_{R}. By definition, since I is an ideal, au ∈ I, so 1_{R} ∈ I, and I = R by part a.

14. Let I be an ideal in a field F, and let I ≠ (0_{F}) = {0_{F} }. Then there is some a ≠ 0_{F} in I. But then since F is a field, a is a unit in F, and then I = F by problem 13.

21.

a. I = {0, 3} is clearly nonempty, and closed under differences and products by arbitrary elements in ℤ_{6}. (A formal proof would proceed as in 6 a above.) The distinct cosets are 0 + I = I = 3 + I, 1 + I = {1, 4} = 4 + I, and 2 + I = {2, 5} = 5 + I.

b. I = {0, 3, 6, 9, 12} is clearly nonempty, and closed under differences and products by arbitrary elements in ℤ_{15} (since this subset consists of the classes of multiples of 3 mod 15). (A formal proof would proceed as in 6 a above.) The distinct cosets are 0+I = I, 1+I = {1, 4, 7, 10, 13}, and 2 + I = {2, 5, 8, 11, 14}.

23. I ≠ ∅ since r = 0_{R} satisfies 0_{R}t = 0_{R} for all t ∈ J. Hence 0_{R} ∈ I. Next, let a, b ∈ I. Then at = 0_{R} and bt = 0_{R} for all t ∈ J. But then for all t ∈ J, (a − b)t = at − bt = 0_{R} − 0_{R} = 0_{R}. This shows that a − b ∈ I. Finally let a ∈ I and r ∈ R. By associativity, we have (sa)t = s(at) = s · 0_{R} = 0_{R} for all t ∈ J. Hence sa ∈ I. Similarly for all t ∈ J, (as)t = a(st). Now J is an ideal in R and s ∈ R so st ∈ J. Hence a(st) = 0_{R}. This shows as ∈ I also. By definition, I is an ideal in R.

24. K ≠ ∅ since a = 0_{R} satisfies r0_{R} = 0_{R} for all r ∈ R. J is an ideal so 0_{R} ∈ J, and hence r0_{R} ∈ J for all r ∈ R. Hence 0_{R} ∈ K. Next, let a, b ∈ K. Then ra ∈ J and rb ∈ J for all r ∈ R. But then for all r ∈ R, r(a − b) = ra − rb ∈ J, since J is closed under differences. This shows that a − b ∈ K. Finally let a ∈ K and s ∈ R. By associativity, we have r(sa) = (rs)a ∈ J since rs ∈ R. Hence sa ∈ K. Similarly for all r ∈ R, r(as) = (ra)s. Now ra ∈ J, and J is an ideal in R so (ra)s ∈ J. Hence as ∈ K. By the definition, K is an ideal in R.

38. If I = {0}, then I = (0) is principal. Hence we may assume now that I ≠ {0}. We claim that I always contains positive elements in this case. This is true since I must contain some n ≠ 0. If n > 0, then we are done. If n < 0, then (−1)(n) = −n > 0 and −n ∈ I as well. Now, by the Well-Ordering Principle, there is a smallest strictly positive element in I, call it c. We will show that I = (c), so I is principal. The ⊇ inclusion is automatic since c ∈ I implies ck ∈ I for all k ∈ ℤ. For the other (⊆) inclusion, let n ∈ I, and divide c into n by integer division: n = qc + r, where 0 ≤ r < c. However this equation shows r = n − qc. Since n, c ∈ I, this implies r ∈ I as well. However, c was the smallest strictly positive element in I, so the possibility is r = 0. This shows n = qc ∈ (c) so I ⊆ (c). Therefore I = (c) is a principal ideal.

39.

a. S is nonempty since 0 = \frac{0}{1} ∈ S. If \frac{a}{b}, \frac{c}{d} ∈ S, then the difference \frac{a}{b} − \frac{c}{d} = \frac{da−bc}{bd}. To put this in lowest terms, we would compute the gcd of the top and bottom, say n = (da − bc, bd). Then we have \frac{da−bc}{bd} = \frac{nA}{nB} = \frac{A}{B} where (A, B) = 1. Note that since B|(bd) and b, d are odd, then B must also be odd. Hence \frac{a}{b} − \frac{c}{d} ∈ S. Similarly, \frac{a}{b} · \frac{c}{d} = \frac{ac}{bd}. To put this in lowest terms, we would compute the gcd of the top and bottom, say n = (ac, bd). Then we have \frac{ac}{bd} = \frac{nA}{nB} = \frac{A}{B} where (A, B) = 1. Note that B|(bd) must be odd since b, d are odd. Hence \frac{a}{b} · \frac{c}{d} ∈ S.

b. Let I be the set of elements of S with even numerators. I is nonempty since 0 = \frac{0}{1} ∈ I. I is closed under differences, since if \frac{2k}{b}, \frac{2ℓ}{d} ∈ I, then \frac{2k}{b} − \frac{2ℓ}{d} = \frac{2(dk−bℓ)}{bd}. If this is not already in lowest terms, then we might have to cancel some common factors. However note that bd is odd so 2 |̸ (bd). Hence, because of unique factorization in ℤ, the factor of 2 on the top will remain after any common factors are cancelled. This shows that I is closed under differences. Similarly, if \frac{2k}{b} ∈ I, and \frac{c}{d} ∈ S, then \frac{2k}{b} · \frac{c}{d} = \frac{2kc}{bd}. If this is not already in lowest terms, then we might have to cancel some common factors. However note that bd is odd so 2 |̸ (bd). Hence the factor of 2 on the top will remain after any common factors are cancelled. Thus I is an ideal in S.

c. We claim that 1 + I is the only coset of I different from I itself, and we will show that by showing that 1 + I coincides with the set of all elements of S where the numerator is odd. (Since every element of S either has an even numerator or an odd numerator, there cannot be any additional cosets.) Note first that 1 + \frac{2k}{b} = \frac{b+2k}{b} has an odd numerator so 1 + I is contained in the set of elements in S with odd numerators. Conversely, if \frac{2k+1}{b} has an odd numerator, then we can rewrite this as \frac{2k+1}{b} = 1 + \frac{2k+1−b}{b} Since b is odd, 2k + 1 − b is even so \frac{2k+1−b}{b} is in I. Hence \frac{2k+1}{b} ∈ 1 + I.

40. Notice that 39 is the special case of 40 where p = 2. The proofs of the more general statements here are similar. To prepare, note that the fact used in the proof of 39 that b, d odd implies bd odd carries over to this setting because of the contrapositive form of Euclid’s Lemma in ℤ: If p is prime and p |̸ b and p |̸ d, then p |̸ (bd). a. T is nonempty since 0 = \frac{0}{1} ∈ S. If \frac{a}{b}, \frac{c}{d} ∈ T, then the difference \frac{a}{b} − \frac{c}{d} = \frac{da−bc}{bd}. To put this

in lowest terms, we would compute the gcd of the top and bottom, say n = (da − bc, bd). Then we have \frac{da−bc}{bd} = \frac{nA}{nB} = \frac{A}{B} where (A, B) = 1. Note that since B|(bd) and b and d are not divisible by p, then bd and hence B must also be not divisible by p. Hence \frac{a}{b} − \frac{c}{d} ∈ T. Similarly, \frac{a}{b} · \frac{c}{d} = \frac{ac}{bd}. To put this in lowest terms, we would compute the gcd of the top and bottom, say n = (ac, bd). Then we have \frac{ac}{bd} = \frac{nA}{nB} = \frac{A}{B} where (A, B) = 1. Note that B|(bd) must be not divisible by p since b, d are not divisible by p. Hence \frac{a}{b} · \frac{c}{d} ∈ T.

b. Let I be the set of elements of T with numerators divisible by p. I is nonempty since 0 = \frac{0}{1} ∈ I. I is closed under differences, since if \frac{pk}{b}, \frac{pℓ}{d} ∈ I, then \frac{pk}{b} − \frac{pℓ}{d} = \frac{p(dk−bℓ)}{bd}. If this is not already in lowest terms, then we might have to cancel some common factors. However note that b and d are not divisible by p, so p |̸ (bd). Hence, because of unique factorization in ℤ, the factor of p on the top will remain after any common factors are cancelled. This shows that I is closed under differences. Similarly, if \frac{pk}{b} ∈ I, and \frac{c}{d} ∈ T, then \frac{pk}{b} · \frac{c}{d} = \frac{pkc}{bd}. If this is not already in lowest terms, then we might have to cancel some common factors. However note that bd is divisible by p so p |̸ (bd). Hence the factor of p on the top will remain after any common factors are cancelled. Thus I is an ideal in T.

c. We claim that 0 + I, 1 + I, 2 + I, …, (p − 1) + I are the only cosets of I. We will show that by showing that every element of T belongs to one of these cosets. This is significantly trickier than the corresponding statement in 39, though(!) To see it, we need to recall from Algebraic Structures that an integer b has a multiplicative inverse mod p exactly when (b, p) = 1 or p |̸ b. So let \frac{a}{b} ∈ T. We claim that this element belongs to the coset s + I where s ∈ {0, 1, …, p − 1} ≡ ua mod p, where u is the multiplicative inverse of b mod p: ub ≡ 1 mod p. To see this, write a = qp + i with i ∈ {0, 1, …, p − 1}, and ub = 1 + kp. Then

\frac{a}{b} = \frac{i + qp}{b} = \frac{i(ub − kp) + qp}{b} = ui + \frac{p(q − ki)}{b} ∈ ui + I.

But now ui ≡ ua ≡ s mod p, so ui = s + ℓp for some ℓ ∈ ℤ and hence

\frac{a}{b} = s + (\frac{pℓ}{1} + \frac{p(q − ki)}{b}) ∈ s + I.

This establishes our claim.

Section 6.2

2. By the First Isomorphism Theorem, if φ : F → S is any ring homomorphism, then im(φ) ≃ F/ ker(φ). Moreover ker(φ) is an ideal in F. By 6.1/14, this says ker(φ) = (0_{F}) or ker(φ) = F. In the first case im(φ) ≃ F/(0_{F}) ≃ F (each element of F defines a distinct coset). In the second case, im(φ) ≃ F/F = {0}, the zero ring (all elements of F belong to the same coset).

3. Continuing the reasoning from 2, if f : F → R is a surjective homomorphism and R is not the zero ring, there must be at least two elements in R (0_{R} and something else). Hence we cannot be in the second case above since f surjective says im(f) = R. It follows that im(f) ≃ F/(0_{F}) ≃ F. Since ker(f) = (0_{F}), f is injective as well as surjective, hence is an isomorphism.

6.

a. See Section 6.1/7c above. This is the same question(!)

b. We will show the case of I = (6) = {0, 6}. The others are similar, and simpler, since there are fewer distinct cosets. Here the distinct cosets are 0 + I, 1 + I, 2 + I, 3 + I, 4 + I, 5 + I. The tables for ℤ_{12}/I are as follows (writing i for i + I in all cases to save typing!)

+ 0 1 2 3 4 5
0  0 1 2 3 4 5
1  1 2 3 4 5 0
2  2 3 4 5 0 1
3  3 4 5 0 1 2
4  4 5 0 1 2 3
5  5 0 1 2 3 4

and

\begin{matrix}· & 0 & 1 & 2 & 3 & 4 & 5 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 2 & 3 & 4 & 5 \\ 2 & 0 & 2 & 4 & 0 & 2 & 4 \\ 3 & 0 & 3 & 0 & 3 & 0 & 3 \\ 4 & 0 & 4 & 2 & 0 & 4 & 2 \\ 5 & 0 & 5 & 4 & 3 & 2 & 1 \end{matrix}

For instance, (2+I)+(5+I) = 7+I = 1+I and (2+I)(5+I) = 10+I = 4+I. By inspection, we see that this quotient ring is isomorphic to ℤ_{6}. Similarly, ℤ_{12}/(0) ≃ ℤ_{12}, ℤ_{12}/(1) ≃ {0} (the zero ring), ℤ_{12}/(2) ≃ ℤ_{2}, ℤ_{12}/(3) ≃ ℤ_{3}, ℤ_{12}/(4) ≃ ℤ_{4}.

c. This follows from the First Isomorphism Theorem. Every ring homomorphism φ : ℤ_{12} → S has im(φ) ≃ ℤ_{12}/I for I = ker(φ). Every ideal I in ℤ_{12} is one of the six principal ideals found in 6.1/7c. Hence by part b, every homomorphic image of ℤ_{12} is isomorphic to one of the rings listed.

8. These can be done by the same method illustrated in the solution for 6b above. To see a less computational method for part a, notice first that I is an ideal by Section 6.1/21a. We have ℤ_{6}/([3]) = {[0] + I, [1] + I, [2] + I}. Then we define a mapping φ : ℤ_{6} → ℤ_{3} by φ([i]_{6}) = [i]_{3} (the added subscripts indicate which ring the congruence class lives in, for clarity). This is well-defined since if [i]_{6} = [j]_{6}, then 6|(i − j) which implies 3|(i − j), so [i]_{3} = [j]_{3}. We have that φ is a ring homomorphism since

φ([i]_{6} + [j]_{6}) = φ([i + j]_{6}) = [i + j]_{3} = [i]_{3} + [j]_{3} = φ([i]_{6}) + φ([j]_{6}),

and

φ([i]_{6} · [j]_{6}) = φ([ij]_{6}) = [ij]_{3} = [i]_{3} · [j]_{3} = φ([i]_{6}) · φ([j]_{6}).

We see that φ is surjective since every element in ℤ_{3} is in the image. Finally, ker(φ) = {[0]_{6}, [3]_{6}}. Hence by the First Isomorphism Theorem,

ℤ_{3} = im(φ) ≃ ℤ_{6}/ ker(φ) = ℤ_{6}/([3]_{6}).

The proof of part b is exactly analogous and will be omitted.

28. Define φ : S → ℤ_{2} by φ (\frac{a}{b}) = [a] ∈ ℤ_{2}. Since the denominator b is odd, we have [b] = [1] in ℤ_{2}. Hence

φ (\frac{a}{b} + \frac{c}{d}) = φ (\frac{ad + bc}{bd}) = [ad + bc] = [a][d] + [b][c] = [a] + [c] ∈ ℤ_{2}.

Since this equals φ (\frac{a}{b}) + φ (\frac{c}{d}), the first homomorphism property holds. Similarly,

φ (\frac{a}{b} · \frac{c}{d}) = φ (\frac{ac}{bd}) = [ac] = [a] · [c] ∈ ℤ_{2}.

Since this equals φ (\frac{a}{b}) · φ (\frac{c}{d}), the second homomorphism property holds. Now, ker(φ) = I since it is exactly the elements in S with even numerators that map to [0] in ℤ_{2}. φ is also surjective since the elements in S with odd numerators map to [1]. So by the First Isomorphism Theorem,

S/ ker(φ) = S/I ≃ im(φ) = ℤ_{2}.

29. The idea here is similar to what we did in 28 above. However, as noted in the solution to 6.1/40, this is somewhat trickier. A mapping φ : T → ℤ_{p} that works here is to define φ (\frac{a}{b}) = [a][b]^{−1} ∈ ℤ_{p}. (Recall that [b] has a multiplicative inverse in ℤ_{p} exactly when (p, b) = 1 and that is the condition for \frac{a}{b} ∈ T.) This is a ring homomorphism since

φ (\frac{a}{b} + \frac{c}{d}) = φ (\frac{ad + bc}{bd})
= [ad + bc][bd]^{−1}
= ([a][d] + [b][c])[b]^{−1}[d]^{−1}
= [a][b]^{−1} + [c][d]^{−1}
= φ (\frac{a}{b}) + φ (\frac{c}{d})

and

φ (\frac{a}{b} · \frac{c}{d}) = φ (\frac{ac}{bd})
= [ac][bd]^{−1}
= [a][c][b]^{−1}[d]^{−1}
= ([a][b]^{−1})([c][d]^{−1})
= φ (\frac{a}{b}) · φ (\frac{c}{d}) .

The kernel of φ is exactly the set of \frac{a}{b} ∈ T with [a] = [0] ∈ ℤ_{p}, which is I. The image of φ is all of ℤ_{p} since φ (\frac{i}{1}) = [i] for i = 0, 1, …, p − 1. Hence by the First Isomorphism Theorem, ℤ_{p} = im(φ) ≃ T/ ker(φ) = T/I.