MAT 3343, APPLIED ALGEBRA, FALL 2003

Answers to Problem Set 5

Peter Selinger

Problem 1. In ℤ_{10}, we calculate: p(0) = 8, p(1) = 0, p(2) = 4, p(3) = 0, p(4) = 8, p(5) = 8, p(6) = 0, p(7) = 4, p(8) = 0, p(9) = 8. Thus, there are 4 roots. This does not contradict the root theorem, because ℤ_{10} is not a field.

Problem 2. (a) First note that p(x) is not a unit nor zero; thus, it is either irreducible or reducible. p(x) has a root iff it has a linear factor. Clearly, if p(x) has a linear factor, it is reducible. Conversely, if p(x) is reducible, then one of its factors must be of degree 1, since p(x), as a third-degree polynomial, cannot have two factors of degree 2.

(b) Let F = ℝ, the field of real numbers. Let q(x) = (x^{2} + 1)(x^{2} + 2). Clearly, q(x) is reducible; on the other hand, q(x) > 0 for all x ∈ ℝ, thus q has no roots in ℝ.

Problem 3. We use the Euclidean Algorithm in the Euclidean ring ℤ_{2}[x]. By repeated long division (details not shown), we find that

quotient:  remainder:
x^{8} + x^{7} + x^{5} + x^{3} + x^{2} + x + 1
= (x^{7} + x^{6} + x^{4} + x^{3} + 1) (x)  + (x^{4} + x^{3} + x^{2} + 1)
x^{7} + x^{6} + x^{4} + x^{3} + 1
= (x^{4} + x^{3} + x^{2} + 1)  (x^{3} + x) + (x^{3} + x + 1)
x^{4} + x^{3} + x^{2} + 1
= (x^{3} + x + 1)  (x + 1) + 0

Thus the gcd is x^{3} + x + 1.

Problem 4. (a) The irreducible polynomials of degree up to 4 in ℤ_{2} were given in class. They are:

linear:  x, x + 1
quadratic: x^{2} + x + 1
cubic:  x^{3} + x^{2} + 1, x^{3} + x + 1
quartic: x^{4} + x^{3} + x^{2} + x + 1, x^{4} + x^{3} + 1, x^{4} + x + 1

An irreducible polynomial of degree 5 must have highest and lowest coefficient 1, so it must be of the form x^{5} + ax^{4} + bx^{3} + cx^{2} + dx + 1. Moreover, it must have an odd number of non-zero coefficients (or else x + 1 will be a factor). This leaves 8 possibilities. Of these 8 possible choices, 2 are divisible by x^{2} + x + 1. The remaining ones are irreducible:

\begin{matrix}x^{5} + x^{4} + x^{3} + x^{2} + 1, x^{5} + x^{4} + x^{3} + x + 1, x^{5} + x^{4} + x^{2} + x + 1, \\ x^{5} + x^{3} + x^{2} + x + 1, x^{5} + x^{3} + 1, x^{5} + x^{2} + 1, \end{matrix}

(b) We check whether any of the polynomials from (a) are factors of p(x) = x^{12} + x^{10} + x^{7} + x^{6} + 1:

Degree 1: x and x + 1 are not factors, because neither 0 nor 1 is a root of p(x).

Degree 2: The only irreducible quadratic polynomial in ℤ_{2}[x], x^{2} + x + 1, is not a factor (the remainder of the division is 1).

Degree 3: Of the two irreducible polynomials of degree 3, only x^{3} + x^{2} + 1 and x^{3} + x + 1 is a factor of p(x), with quotient q(x) = x^{9} + x^{8} + x^{6} + x^{4} + x^{3} + x^{2} + 1. The quotient q(x) has no more factors of degree 3.

Degree 4: Of the three irreducible polynomials of degree 4, only x^{4} + x^{3} + 1 is a factor of q(x), with quotient q′(x) = x^{5} + x^{2} + 1. The quotient is irreducible.

We therefore obtain p(x) = (x^{5} + x^{2} + 1)(x^{4} + x^{3} + 1)(x^{3} + x^{2} + 1).

Problem 5. (a) In ℚ[x], the polynomial p(x) = x^{5} − 1 has a root x = 1, so x − 1 is a factor. We have p(x) = (x − 1)(x^{4} + x^{3} + x^{2} + x + 1). Let q(x) = x^{4} + x^{3} + x^{2} + x + 1. Then q(x) is a cyclotonic polynomial; by a theorem proved in class, q(x) is irreducible in ℚ[x]. [Recall the proof: Let y + 1 = x, then

q(y + 1) = ((y + 1)^{5} − 1)/(y + 1 − 1)
= (y^{5} + 5y^{4} + 10y^{3} + 10y^{2} + 5y + 1 − 1)/y
= y^{4} + 5y^{3} + 10y^{2} + 10y + 5.

Then q(y + 1) is irreducible in ℚ[y] by Eisenstein’s criterion (with p = 5). It follows that q(x) is irreducible in ℚ[x].]

(b) In ℤ_{2}[x], the polynomial p(x) = x^{5} + 1 has a root x = 1, and thus p(x) = (x+1)(x^{4} +x^{3} +x^{2} +x+1). The polynomial q(x) = x^{4} +x^{3} +x^{2} +x+1 has no roots, and therefore it has no linear factors. So if q(x) was reducible, it would have to have a quadratic factor; the only irreducible quadratic is x^{2} + x + 1, which is not a factor of q(x), thus q(x) is irreducible.

(c) if ℤ_{5}[x], the polynomial p(x) = x^{4} + 1 has no roots (p(0) = 1, p(1) = 2, p(2) = 2, p(3) = 2, p(4) = 2). Thus it has no linear factor. However, y^{2} +1 has roots y = ±2, thus y^{2} +1 = (y +2)(y −2). It follows, by letting y = x^{2}, that x^{4} + 1 = (x^{2} + 2)(x^{2} − 2).

(d) The polynomial p(x) = 2x^{3} + x^{2} + 4x + 2 in ℚ[x], if reducible, must have a linear factor (since deg p = 3). By the rational roots theorem, the only possible roots are of the form r/s, where r|2 and s|2. This leaves as possible roots x = ±1/2, ±1, ±2. Of these, we find that only x = −1/2 is a root. We have p(x) = (x + 1/2)(2x^{2} + 4) = (2x + 1)(x^{2} + 2). Since x^{2} + 2 has no more rational roots, it is irreducible.

(e) In ℚ[x], the polynomial p(x) = x^{4} − 9x + 3 is irreducible by Eisenstein’s criterion, with p = 3.

(f) The polynomial p(x) = x^{8} − 16 has no rational roots; in fact, its only real roots are ±\sqrt{2}. The complex roots of p(x) lie on a circle of radius \sqrt{2}; they are \sqrt{2}e^{2πiθ/8}, where θ = 0, 1, 2, …, 7. Or concretely, these roots are ±\sqrt{2}, ±i\sqrt{2}, 1 ± i, −1 ± i. We know that each conjugate pair of complex linear factors determines a real quadratic factor, so p(x) factors into irreducible factors over ℝ[x] as p(x) = (x+\sqrt{2})(x−\sqrt{2})(x^{2}+2)(x^{2}+ 2x+2)(x^{2}−2x+2). Only the first two factors are not rational; they combine to a rational factor (x^{2} − 2). So we have p(x) = (x^{2} − 2)(x^{2} + 2)(x^{2} + 2x+2)(x^{2} −2x+2). These four factors are irreducible over ℚ[x] (because their roots, as we saw, are irrational, or also by Eisenstein’s criterion).

Problem 6. Over ℤ_{5}, a quadradic polynomial x^{2} + ax + b is reducible iff it is of the form (x + c)(x + d), for some c, d ∈ ℤ_{5}. Here, the order of c, d does not matter, so there are 15 possibilities for c, d:

\begin{matrix}c & d & (x + c)(x + d) \\ 0 & 0 & x^{2} + 0x + 0 \\ 0 & 1 & x^{2} + 1x + 0 \\ 0 & 2 & x^{2} + 2x + 0 \\ 0 & 3 & x^{2} + 3x + 0 \\ 0 & 4 & x^{2} + 4x + 0 \end{matrix} \begin{matrix}c & d & (x + c)(x + d) \\ 1 & 1 & x^{2} + 2x + 1 \\ 1 & 2 & x^{2} + 3x + 2 \\ 1 & 3 & x^{2} + 4x + 3 \\ 1 & 4 & x^{2} + 0x + 4 \\ 2 & 2 & x^{2} + 4x + 4 \end{matrix} \begin{matrix}c & d & (x + c)(x + d) \\ 2 & 3 & x^{2} + 0x + 1 \\ 2 & 4 & x^{2} + 1x + 3 \\ 3 & 3 & x^{2} + 1x + 4 \\ 3 & 4 & x^{2} + 2x + 2 \\ 4 & 4 & x^{2} + 3x + 1 \end{matrix}

Thus these 15 polynomials are reducible. The 10 remaining ones are irreducible:

\begin{matrix}x^{2} + 2 & x^{2} + x + 1 & x^{2} + 2x + 3 & x^{2} + 3x + 3 & x^{2} + 4x + 1 \\ x^{2} + 3 & x^{2} + x + 2 & x^{2} + 2x + 4 & x^{2} + 3x + 4 & x^{2} + 4x + 2 \end{matrix}

Problem 7. (a) x^{3} + x^{2} + x + 1 = (x^{2} + 1)(x + 1): not irreducible in ℚ[x].

(b) 3x^{8} − 4x^{6} + 8x^{5} − 10x + 6 is irreducible in ℚ[x] by Eisenstein’s criterion, with p = 2.

(c) x^{4} + x^{2} − 6. Note that y^{2} + y − 6 has roots y = 2 and y = −3, thus y^{2} + y − 6 = (y + 3)(y − 2), thus x^{4} + x^{2} − 6 = (x^{2} + 3)(x^{2} − 2). Thus not irreducible in ℚ[x].

(d) p(x) = 4x^{3}+3x^{2}+x+1 has no roots in ℤ_{5}[x], because p(0) = 1, p(1) = 4, p(2) = 2, p(3) = 4, p(4) = 4. Therefore it has no linear factors. Since p(x) is of degree 3, it must be irreducible.

Problem 8. (a) a = \sqrt{2}/\sqrt[3]{5}. We find that a^{6} = 8/25, hence a is a root of x^{6} − 8/25, or of 25x^{6} − 8. By Eisenstein’s criterion with p = 2, this is irreducible in ℚ[x], hence it has no rational root. Therefore a is irrational.

(b) a = \sqrt{2} + \sqrt{3}. We find that a^{2} = 2 + 2\sqrt{6} + 3, therefore a^{2} − 5 = 2\sqrt{6}. Squaring again, we get (a^{2} − 5)^{2} = 24, or a^{4} − 10a^{2} + 25 = 24. Therefore, a is a root of x^{4} − 10x^{2} + 1. By the rational roots theorem, the only possible rational roots are ±1; however, these are not actually roots. Thus, x^{4} − 10x^{2} + 1 has no rational roots. This proves that a is irrational.

Problem 9. Let p(x) = 3x^{3} + 4x^{2} − x − 2. By the rational roots theorem, all possible rational roots of p(x) are of the form r/s, where r|2 and s|3. Thus, r = ±1, ±2 and s = ±1, ±3. This leaves eight potential rational roots: ±1, ±2, ±1/3, ±2/3. Of these, we find that only 2/3 and −1 are actual roots. [In fact, p(x) = (3x − 2)(x + 1)(x + 1).]