English | Español

# Try our Free Online Math Solver! Online Math Solver

 Depdendent Variable

 Number of equations to solve: 23456789
 Equ. #1:
 Equ. #2:

 Equ. #3:

 Equ. #4:

 Equ. #5:

 Equ. #6:

 Equ. #7:

 Equ. #8:

 Equ. #9:

 Solve for:

 Dependent Variable

 Number of inequalities to solve: 23456789
 Ineq. #1:
 Ineq. #2:

 Ineq. #3:

 Ineq. #4:

 Ineq. #5:

 Ineq. #6:

 Ineq. #7:

 Ineq. #8:

 Ineq. #9:

 Solve for:

 Please use this form if you would like to have this math solver on your website, free of charge. Name: Email: Your Website: Msg:

# Math Homework #6 Solutions

p 315, #4 Let and suppose that x−r divides f (x) for
some r ∈Q. Then we must have f(r) = 0. We will use this fact to prove that in fact r ∈Z.
If r = 0 then there is nothing to prove. So assume r ≠ 0 and write r = a/b with a, b ∈Z,
b ≠ 0 and (a, b) = 1. Then Multiplying both sides by bn then yields which is equivalent to Since this implies that b divides a n. Since
(a, b) = 1, this can only occur if b = ±1. But then r = a/b = ±a ∈ Z, as claimed.

p 315, #6 If p is prime and f(x) ∈ Zp[x] is irreducible then is a field by
Corollary 1 to Theorem 17.5, since Zp is a field. Moreover, we proved in class that since
deg f(x) = n the each element of can be expressed uniquely as  for some ai ∈ F. Since there are p choices for each coefficient
ai and n coefficients, there are exactly pn such cosets. That is, is a field with
pn elements.

p 316, #8 Let f(x) = x3 + 2x + 1 ∈ Z3[x]. It is easy to show that f(x) has no roots in Z3
and as deg f(x) = 3, this implies f(x) is irreducible in Z3[x]. So according to Exercise #6, is a field with 33 = 27 elements.

p 316, #10

a. x5 + 9x4 + 12x2 + 6 is irreducible according Eisenstein’s criterion with p = 3.

b. Consider x4 + x + 1 mod 2. It is easy to see that this polynomial has no roots in Z2,
and so to prove irreducibility in Z2 it suffices to show it has no quadratic factors. The
only quadratic polynomial in Z2[x] that does not have a root in Z2 is x2 + x + 1 which
does not divide x4 + x + 1 in Z2[x], as is also easily checked. It follows that x4 + x + 1
is irreducible in Z2[x] and so by the mod p test with p = 2 we conclude that x4 + x + 1
is irreducible in Q[x].

c. x4 + 3x2 + 3 is irreducible according to Eisenstein’s criterion with p = 3.

d. Consider x5 +5x2 +1 mod 2, which is x5 +x2 +1. It is easy to see that this polynomial
has no roots in Z 2, and so to prove irreducibility in Z2 it again suffices to show it has
no quadratic factors . The only quadratic polynomial in Z 2[x] that does not have a root
in Z2 is x2 + x + 1 which does not divide x5 + x2 + 1 in Z2[x], as is also easily checked.
It follows that x5 + x2 + 1 is irreducible in Z2[x] and so by the mod p test with p = 2
we conclude that x5 + 5x2 + 1 is irreducible in Q[x].

e. Let f(x) = (5/2)x5 + (9/2)x4 + 15x3 + (3/7)x2 + 6x + 3/14 and g(x) = 35x5 + 63x4 +
210x3 +6x2 +84x+3 = 14f(x). Since 14 is a unit in Q[x], f(x) is irreducible in Q[x] if
and only if g(x) is, and the latter statement is true by Eisenstein’s criterion with p = 3.

p 316, #12 Since it has degree 2, to show that x2 +x+4 is irreducible in Z11[x] it suffices
to show it has no roots in Z11, as Z11 is a field. This is straightforward and is left to the

p 316, #16
a. Since Zp is a field, a polynomial of the form x2 + ax + b ∈ Zp[x] is reducible if and
only if there exist c, d ∈ Z11 so that x2 + ax + b = (x + c)(x + d). There are such
polynomials for which c ≠ d and p for which c = d. Therefore, there are exactly reducible monic quadratic polynomials in Zp[x]. Since there are p2 polynomials of the
form x2 + ax + b and each one is either reducible or irreducible, we conclude there are irreducible monic degree 2 polynomials in Zp[x].

b. If f(x) ∈ Zp[x] is irreducible of degree 2, then f(x) = ag(x) for some a ∈ F, a ≠ 0
and g(x) ∈ F[x] irreducible, monic and of degree 2. There are p − 1 choices for a and,
by part (a), p(p − 1)/2 choices for g(x). Therefore there are p(p − 1)2/2 irreducible

p 316, #24 Substituting all of the elements of Z7 into 3x2 + x + 4 we find that it has two
roots: 4 and 5. The quadratic formula “ predicts ” the roots since 6-1 = 6 in Z7 and −47 = 2 in Z7. Since 32 = 9 = 2 in Z7, we can take and so
the two predicted roots are

1 ± 6 · 3 = 4, 5

which agree with those found by substitution.
If we substitute all of the elements of Z5 into 2x2 +x+3 we find no roots. The quadratic
formula predicts the roots are since 4-1= 4 and −23 = 2 in Z5. p However, there is no element in Z5 whose square is 2, so is not an element of Z5. Consequently the roots predicted by the quadratic formula do
not belong to Z5, which is in agreement with the fact that there are no roots in Z5.

It turns out that the quadratic formula alwaysgives the roots of ax2 + bx + c ∈ F[x] for
any field F, as long as we agree that if b2 − 4ac is not a square in F then we interpret the
formula as yielding no roots. This is easily proven using the usual proof of the quadratic
formula (i.e. completing the square ).

p 317, #28 Let f(x) ∈ Q[x] be nonzero. Choose an n ∈ Z+ so that g(x) = nf(x) ∈ Z[x].
Since n ≠ 0 it is a unit in Q[x]. So f(x) is irreducible in Q[x] if and only if g(x) is, and the
latter’s irreducibility can be tested using the mod p test.

p 317, #30 Let If p = 2 then the polynomial in
question is x − 1 which is obviously irreducible in Q[x]. If p > 2 then it is odd and so is the pth cyclotomic polynomial, which is irreducible according to the Corollary of Theorem
17.4. It follows that f(x) is irreducible, for if f(x) factored so too would g(x).

p 317, #32 Let f(x), g(x) ∈ Z[x] and suppose that Then there is an
h(x) ∈ Z[x] so that f(x)g(x) = (x2 + 1)h(x). Since x2 + 1 is primitive and irreducible in
Q[x], it is also irreducible in Z[x]. We apply Theorem 17.6 to write where the ai, bi and ci are primes in Z and the pi(x), qi(x) and ri(x) are irreducible polynomials
of positive degree in Z[x]. Substituting these expressions into f(x)g(x) = (x2 +1)h(x)
and rearranging we obtain Theorem 17.6 and the irreducibility of x2 + 1 now imply that x2 + 1 = ±pi(x) for some i or
x2 + 1 = ±qi(x) for some i. In the first case x2 + 1 divides f(x) and in the second x2 + 1
divides g(x). That is, either Therefore is prime
in Z[x].

Let p ∈ Z+ be any prime. We will show that is properly contained in which is not equal to Z[x]. This will prove that is not maximal. Since every nonzero
element of has degree at least 2, This proves that is properly
contained in Now suppose, for the sake of contradiction, that Then there exist f(x), g(x) ∈ Z[x] so that f(x)(x2 + 1) + g(x)p = 1. If we consider this
equation mod p we obtain which is impossible since in Zp[x] is either 0 or has degree at least 2. This contradiction establishes that is
not equal to Z[x], which completes the proof that is not maximal in Z[x].

 Prev Next