Mathematics 351 – Abstract Algebra 1 Solutions for Problem Set 2 September 15, 2007

Section 3.3

4. The tables for addition mod 10 in S and addition mod 5 in ℤ_{5} do correspond under the given mapping f : ℤ_{5} → S. However, the tables for multiplication mod 10 in S and multiplication in ℤ_{5} do not. In other words for the given mapping f : ℤ_{5} → S it is not the case that f(a ·_{5} b) = f(a) ·_{10} f(b) for all a, b ∈ ℤ_{5}. The tables are:

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

and

\begin{matrix}·_{10} & 0 & 2 & 4 & 6 & 8 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 2 & 0 & 4 & 8 & 2 & 6 \\ 4 & 0 & 8 & 6 & 4 & 2 \\ 6 & 0 & 2 & 4 & 6 & 8 \\ 8 & 0 & 6 & 2 & 8 & 4 \end{matrix}

For example, using Hungerford’s notation, note that f(\overline{2} ·_{5} \overline{3}) = f(\overline{1}) = 2, but f(\overline{2}) = 4, and f(\overline{3}) = 6, so f(\overline{2}) ·_{10} f(\overline{3}) = 4.

5. The mapping f is injective since f(a) = f(a′) implies (a, 0_{S}) = (a′, 0_{S}), so a = a′. It is also surjective since every (a, 0_{S}) in \overline{R} is f(a) for the a ∈ R appearing in the first component. Then

f(a + b) = (a + b, 0_{S}) = (a, 0_{S}) + (b, 0_{S}) = f(a) + f(b)

and

f(a · b) = (a · b, 0_{S}) = (a, 0_{S}) · (b, 0_{S}) = f(a) · f(b)

by the definition of the operations in R × S. Therefore f is an isomorphism.

10.

a) This is not an homomorphism because f(a · b) = −(a · b) ≠ (−a) · (−b) = f(a) · f(b).

b) This is an isomorphism, hence a homomorphism. The reason is that in fact −x = x for all x ∈ ℤ_{2}. This is the same as the identity mapping which is clearly an isomorphism.

c) This is not a homomorphism since, for instance g(0) = 1 ≠ 0 in ℚ shows that the property of Theorem 3.12, part (1) fails for this f. d) This is not a homomorphism since

h(ab) = ( \begin{matrix}−(ab) & 0 \\ ab & 0 \end{matrix} ) ≠ ( \begin{matrix}−a & 0 \\ a & 0 \end{matrix} ) ( \begin{matrix}−b & 0 \\ b & 0 \end{matrix} ) = h(a)h(b).

e) This is a homomorphism since by the definitions of the ring operations in ℤ_{12} and ℤ_{4},

f([x]_{12} + [y]_{12}) = f([x + y]_{12}) = [x + y]_{4} = [x]_{4} + [y]_{4} = f([x]_{12}) + f([y]_{12}),

and

f([x]_{12} · [y]_{12}) = f([x · y]_{12}) = [x · y]_{4} = [x]_{4} · [y]_{4} = f([x]_{12}) · f([y]_{12}).

17. Note that S consists of the classes [x] ∈ ℤ_{28} for x divisible by 4. This set is closed under differences and multiplication in ℤ_{28}, so by the criterion of Theorem 3.6, it is a subring. The mapping f([x]_{7}) = [8x]_{28} is seen to be injective and surjective by computing the values: f([0]_{7}) = [0]_{28}, and

f([1]_{7}) = [8]_{28}, f([2]_{7}) = [16]_{28}, f([3]_{7}) = [24]_{28}, f([4]_{7}) = [4]_{28}, f([5]_{7}) = [12]_{28}, f([6]_{7}) = [20]_{28}.

The mapping f is a homomorphism since by the definitions of the ring operations in ℤ_{7} and ℤ_{28},

f([x]_{7} + [y]_{7}) = f([x + y]_{7}) = [8(x + y)]_{28} = [8x]_{28} + [8y]_{28} = f([x]_{7}) + f([y]_{7}),

and since [8 · 8]_{28} = [64]_{28} = [8]_{28}:

f([x]_{7} · [y]_{7}) = f([x · y]_{7}) = [8(x · y)]_{28} = [8x]_{28} · [8y]_{28} = f([x]_{7}) · f([y]_{7}).

19 and 3.1/18. First we show that ℤ^{∗} is a ring under the ⊕, ⊙ operations. (part of 3.1/18). Closure under ⊕, and ⊙ is clear since the elements of ℤ^{∗} are just the ordinary integers and a⊕b = a+b−1, a⊙b = a+b−ab. ⊕ is commutative since a⊕b = a+b−1 = b+a−1 = b⊕a. ⊕ is associative since (a⊕b)⊕c = (a+b−1)+c−1 = a+b+c−2, while a⊕(b⊕c) = a+(b+c−1)−1 = a+b+c−2 also. The integer 0_{ℤ^{∗}} = 1 is an identity element for the ⊕ operation since a ⊕ 1 = a + 1 − 1 = a for all a. Each integer a has an inverse −(a) = −a+2 under the ⊕ operation since a⊕(−a+2) = a−a+2−1 = 1 = 0_{ℤ^{∗}}. ⊙ is associative since (a⊙b)⊙c = (a+b−ab)+c−(a+b−ab)c = a+b+c−ab−ac−bc+abc, while a⊙(b⊙c) = a+(b+c−bc)−a(b+c−bc) = a+b+c−ab−ac−bc+abc also. Finally, we have a distributive law since a⊙(b⊕c) = a+(b+c−1)−a(b+c−1) = (a+b−ab)+(a+c−ac)−1 = (a⊙b)⊕(a⊙c). It is not necessary to prove the other distributive property because it easy to see that ⊙ is commutative as well: a ⊙ b = a + b − ab = b + a − ba = b ⊙ a for all a, b.

The ring ℤ^{∗} is an integral domain because it is a commutative ring by the previous paragraph, it has a multiplicative identity 1_{ℤ^{∗}} = 0 (because a ⊙ 0 = a + 0 − a · 0 = a for all a ∈ ℤ), and if a ⊙ b = 0_{ℤ}^{∗}, then a + b − ab = 1, so (a − 1)(b − 1) = 0, and hence by properties of the usual multiplication in ℤ, a = 1 = 0_{ℤ^{∗}}, or b = 1 = 0_{ℤ^{∗}}. This completes 3.1/18.

Now, consider the mapping f : ℤ → ℤ^{∗} defined by f(x) = 1 − x. f is an injective and surjective mapping because if y ∈ ℤ^{∗} (also an ordinary integer, then there exists exactly one x ∈ ℤ with f(x) = 1 − x = y, namely x = 1 − y. Next, we have

f(a + b) = 1 − (a + b) = (1 − a) + (1 − b) − 1 = f(a) ⊕ f(b).

Finally,

f(a · b) = 1 − ab = (1 − a) + (1 − b) − (1 − a)(1 − b) = f(a) ⊗ f(b).

Therefore the mapping f is a ring homomorphism, and ℤ and ℤ^{∗} are isomorphic as rings.

24. The mapping is g : ℝ → M_{2×2}(ℝ) with

g(r) = ( \begin{matrix}0 & 0 \\ −r & r \end{matrix} ) .

This is injective because if g(r) = g(s) for r, s ∈ ℝ, then

( \begin{matrix}0 & 0 \\ −r & r \end{matrix} ) = ( \begin{matrix}0 & 0 \\ −s & s \end{matrix} ) ,

which implies r = s (look at the lower right entries). However g is not surjective since not every 2×2 matrix is g(r) for some r (take any matrix A with nonzero entries in the first row, for instance).

27. Let f : R → S be an isomorphism. Then f is injective and surjective, so it has an inverse function g : S → R. The inverse function of an injective and surjective mapping is itself injective and surjective by “general nonsense” (i.e. this is a direct consequence of the definitions – f is an inverse function for g, so g must also be injective and surjective.) To show that g satisfies the other properties of an isomorphism of rings, let x, y ∈ S. Since f is surjective, there are a, b ∈ R with x = f(a) and y = f(b), so g(a) = x and g(b) = y. Then using the homomorphism properties for f:

g(a + b) = g(f(x) + f(y)) = g(f(x + y)) = x + y = g(a) + g(b),

and

g(a · b) = g(f(x) · f(y)) = g(f(x · y)) = x · y = g(a) · g(b).

Hence g is also an isomorphism of rings.

28. We will use the subring criterion of Theorem 3.6. First note that K is always nonempty by part 1 of Theorem 3.12: f(0_{R}) = 0_{S}, so O_{R} ∈ K. Next, let a, b ∈ K. Then since f is a homomorphism,

f(a − b) = f(a) − f(b) = 0_{S} − 0_{S} = 0_{S}.

This shows a − b ∈ K, so K is closed under differences. Similarly,

f(a · b) = f(a) · f(b) = 0_{S} · 0_{S} = 0_{S}.

This shows a · b ∈ K so K is closed under products. Therefore K is a subring of R.

29. This very similar to 28. We again use the subring criterion of Theorem 3.6. First note that P is always nonempty by part 1 of Theorem 3.12: f(0_{R}) = 0_{S} ∈ T, so O_{R} ∈ P. Next, let a, b ∈ P, so f(a), f(b) ∈ T. Then since f is a homomorphism,

f(a − b) = f(a) − f(b) ∈ T,

since f(a), f(b) ∈ T and T is a subring of S. This shows a − b ∈ P, so P is closed under differences. Similarly,

f(a · b) = f(a) · f(b) ∈ T,

since T is a subring of S. This shows a · b ∈ P so P is closed under products. Therefore P is a subring of R.

Section 4.1

5.

a) q(x) = 3x^{2} − 5x + 8, and r(x) = −4x − 6.

c) q(x) = x^{3} + 3x^{2} + 2x + 3 and r(x) = 4.

14.

a) Consider what happens at each step of the division process. If the leading term of the intermediate dividend polynomial is ax^{s} and the leading term of g is ux^{r}, where u is a unit in R, then the term that goes into the quotient is u^{−1}ax^{s−r}, and the process proceeds exactly as in the case of polynomials with coefficients in a field F.

b) Consider f = x + 1 and g = 2x + 1. There are no integer polynomials q, r as in the statement of the division algorithm because if q is any nonzero polynomial with integer coefficients, the leading term of qg will have a coefficient that is divisible by 2. Hence f cannot be written as f = qg + r with deg(r) < deg(g) or r = 0. Comment: When q ≠ 0, the leading term in qg + r will come from qg in the situation of the division algorithm because of the condition deg(r) < deg(g), or r = 0. This follows because deg(qg) ≥ deg(g).

17. φ is surjective because every [a_{m}]x^{m}+⋯+[a_{0}] ∈ ℤ_{n}[x] is φ(a_{m}x^{m}+⋯+a_{0}), for the polynomial a_{m}x^{m} + ⋯ + a_{0} ∈ ℤ[x]. We next show φ is a homomorphism. First, if f = a_{m}x^{m} + ⋯ + a_{0} and g = b_{k}x^{k} + ⋯ + b_{0}, then f + g is the polynomial where for each ℓ ≥ 0, the coefficient of x^{ℓ} is a_{ℓ} + b_{ℓ}. Hence the coefficient of x^{ℓ} in φ(f +g) is [a_{ℓ}+b_{ℓ}] = [a_{ℓ}]+[b_{ℓ}] ∈ ℤ_{n}. This shows φ(f +g) = φ(f)+φ(g). Similarly, in the polynomial f · g, for each ℓ ≥ 0, the coefficient of x^{ℓ} is ∑_{i=0}^{ℓ} a_{i}b_{ℓ−i}. Hence using the definition of the sum and product in ℤ_{n}, the coefficient of x^{ℓ} in φ(f · g) is

[∑_{i=0}^{ℓ} a_{i}b_{ℓ−i}] = ∑_{i=0}^{ℓ}[a_{i}][b_{ℓ−i}],

which is the coefficient of x^{ℓ} in φ(f) · φ(g). This shows φ(f · g) = φ(f) · φ(g).

18. D is neither a homomorphism of rings, nor an isomorphism. This can be seen because D does not satisfy the multiplicative property of homomorphisms. For example D(x · x) = D(x^{2}) = 2x ≠ D(x) · D(x) = 1. Comment: Of course, D also fails to be injective, since D(1) = 0 = D(0), but 1 ≠ 0.

Section 4.2

5 b) Here is a Maple session carrying out the computations of the remainders in the Euclidean Algorithm:

> f := x^5+x^4+2*x^3-x^2-x-2;
                   5    4      3    2
                  x  + x  + 2 x  - x  - x - 2
> g := x^4+2*x^3+5*x^2+4*x+4;
                    4      3      2
                   x  + 2 x  + 5 x  + 4 x + 4
> r1 := rem(f, g, x);
                                3
                           2 - x  - x
> r2 := rem(g, r1, x);
                                2
                         8 + 4 x  + 4 x
> r3 := rem(r1, r2, x);
                               0

The last nonzero remainder, made monic, is the gcd, so

gcd(f, g) = x^{2} + x + 2.

14. Since f, g are relatively prime, gcd(f, g) = 1. Hence there are polynomials u, v ∈ F [x] such that 1 = uf + vg. Multiplying by h on both sides, we get

h = ufh + vgh.  (1)

Now if f|h, then h = kf for some polynomial k ∈ F [x]. Similarly, if g|h then h = ℓg for some polynomial ℓ ∈ F [x]. Substitute h = ℓg for the h in the first h in (1) and h = kf for the second. We get h = ufℓg + vgkf = h. Hence h = (uℓ + vk)(fg). Since uℓ + vk ∈ F [x], this shows (fg)|h.

15. We assume gcd(f, g) = 1. Hence there are u, v ∈ F [x] with uf +vg = 1. Now if h|f, then f = hk for some k ∈ F [x]. But then (uk)h + vg = 1. This shows that 1 ∈ S = {Uh + V g : U, V ∈ F [x]}. By the proof of Theorem 4.5, it follows that 1 = gcd(h, g), so h and g are relatively prime.