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are ±1, ±2, ±3, ±4, ±6, ±12. We have f(1) = 21, so for any root we must
have (r - 1)|21, so this eliminates all but ±2, 4, -6 as possibilities. Then
f(2) = 32, f(-2) = -294, and finally we obtain the factorization f(x) =
(x - 4)(x4 - 6x3 + 9x + 3). The second factor is irreducible over Q since it
satisfies Eisenstein s criterion for p = 3.
7. Factor x5 - 2x4 - 2x3 + 12x2 - 15x - 2 over Q.
Solution: The possible rational roots are ±1, ±2, and since 2 is a root we have
the factorization x5 - 2x4 - 2x3 + 12x2 - 15x - 2 = (x - 2)(x4 - 2x2 + 8x + 1).
The only possible rational roots of the second factor are 1 and -1, and these
do not work. (It is important to note that since the degree of the polynomial
is greater than 3, the fact that it has not roots in Q does not mean that it
is irreducible over Q.) Since the polynomial has no linear factors, the only
possible factorization has the form x4-2x2+8x+1 = (x2+ax+b)(x2+cx+d).
This leads to the equations a+c = 0, ac+b+d = -2, ad+bc = 8, and bd = 1.
We have either b = d = 1, in which case a + c = 8, or b = d = -1, in which
CHAPTER 4 SOLUTIONS 89
case a + c = -8. Either case contradicts a + c = 0, so x4 - 2x2 + 8x + 1 is
irreducible over Q.
As an alternate solution, we could reduce x4 - 2x2 + 8x + 1 modulo 3 to
get p(x) = x4 + x2 + 2x + 1. This polynomial has no roots in Z3, so the
only possible factors are of degree 2. The monic irreducible polynomials of
degree 2 over Z3 are x2 + 1, x2 + x + 2, and x2 + 2x + 2. Since the constant
term of p(x) is 1, the only possible factorizations are p(x) = (x2 + x + 2)2,
p(x) = (x2 + 2x + 2)2, or p(x) = (x2 + x + 2)(x2 + 2x + 2). In the first the
coefficient of x is 1; the second has a nonzero cubic term; in the third the
coefficient of x is 0. Thus p(x) is irreducible over Z3, and hence over Q.
8. (a) Show that x2 + 1 is irreducible over Z3.
Solution: To show that p(x) = x2 + 1 is irreducible over Z3, we only need
to check that it has no roots in Z3, and this follows from the computations
p(0) = 1, p(1) = 2, and p(-1) = 2.
(b) List the elements of the field F = Z3[x]/ x2 + 1 .
Solution: The congruence classes are in one-to-one correspondence with the
linear polynomials, so we have the nine elements [0], [1], [2], [x], [x+1], [x+2],
[2x], [2x + 1], [2x + 2].
(c) In the multiplicative group of nonzero elements of F , show that [x + 1] is
a generator, but [x] is not.
Solution: The multiplicative group of F has 8 elements, and since [x]2 = [-1],
it follows that [x] has order 4 and is not a generator. On the other hand,
[x + 1]2 = [x2 + 2x + 1] = [-1 + 2x + 1] = [2x] = [-x], and so [x + 1]4 =
[-x]2 = [-1], which shows that [x + 1] does not have order 2 or 4. The only
remaining possibility (by Lagrange s theorem) is that [x + 1] has order 8, and
so it is a generator for the multiplicative group of F .
9. (a) Express x4 + x as a product of polynomials irreducible over Z5.
Solution: In general, we have x4 + x = x(x3 + 1) = x(x + 1)(x2 - x + 1).
The factor p(x) = x2 - x + 1 is irreducible over Z5 since it can be checked
that it has no roots in Z5. (We get p(0) = 1, p(1) = 1, p(-1) = 3, p(2) = 3,
p(-2) = 2.)
(b) Show that x3 + 2x2 + 3 is irreducible over Z5.
Solution: If p(x) = x3 + 2x2 + 3, then p(0) = 3, p(1) = 1, p(-1) = -1,
p(2) = 4, and p(-2) = 3, so p(x) is irreducible over Z5.
10. Express 2x3 + x2 + 2x + 2 as a product of polynomials irreducible over Z5.
Solution: We first factor out 2, using (2)(-2) = -4 a" 1 (mod 5). This
reduces the question to factoring p(x) = x3 - 2x2 + x + 1. (We could also
multiply each term by 3.) Checking for roots shows that p(0) = 1, p(1) = 1,
p(-1) = -3, p(2) = 3, and p(-2) a" -2, so p(x) itself is irreducible over Z5.
90 CHAPTER 4 SOLUTIONS
11. Construct an example of a field with 343 = 73 elements.
Solution: We only need to find a cubic polynomial over Z7 that has no roots.
The simplest case would be to look for a polynomial of the form x3 + a. The
cube of any element of Z7 gives either 1 or -1, so x3 = 2 has no root over Z7,
and thus p(x) = x3-2 is an irreducible cubic over Z7. Using the modulus p(x),
the elements of Z7[x]/ p(x) correspond to polynomials of degree 2 or less,
giving the required 73 elements. With this modulus, the identities necessary
to determine multiplication are [x3] = [5] and [x4] = [5x].
12. In Z2[x]/ x3 + x + 1 , find the multiplicative inverse of [x + 1].
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