# 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 b^{n} 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) ∈
Z_{p}[x] is irreducible then
is a
field by

Corollary 1 to Theorem 17.5, since Z_{p} 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 a_{i}
∈ F. Since there are p choices for
each coefficient

a_{i} and n coefficients, there are exactly p^{n} such cosets. That is,
is a field with

p^{n} elements.

p 316, #8 Let f(x) = x^{3} + 2x + 1 ∈ Z_{3}[x]. It is easy to show that f(x) has no
roots in Z_{3}

and as deg f(x) = 3, this implies f(x) is irreducible in Z_{3}[x]. So according to
Exercise #6,

is a field with 3^{3} = 27 elements.

p 316, #10

a. x^{5} + 9x^{4} + 12x^{2} + 6 is irreducible according Eisenstein’s criterion with p =
3.

b. Consider x^{4} + x + 1 mod 2. It is easy to see that this polynomial has no
roots in Z_{2},

and so to prove irreducibility in Z_{2} it suffices to show it has no quadratic
factors. The

only quadratic polynomial in Z_{2}[x] that does not have a root in Z_{2} is x^{2} + x + 1
which

does not divide x^{4} + x + 1 in Z_{2}[x], as is also easily checked. It follows that
x^{4} + x + 1

is irreducible in Z_{2}[x] and so by the mod p test with p = 2 we conclude that x^{4}
+ x + 1

is irreducible in Q[x].

c. x^{4} + 3x^{2} + 3 is irreducible according to Eisenstein’s criterion with p = 3.

d. Consider x^{5} +5x^{2} +1 mod 2, which is x^{5} +x^{2} +1. It is easy to see that this
polynomial

has no roots in Z _{2}, and so to prove irreducibility in Z_{2} 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 Z_{2} is x^{2} + x + 1 which does not divide x^{5} + x^{2} + 1 in Z_{2}[x], as is also
easily checked.

It follows that x^{5} + x^{2} + 1 is irreducible in Z_{2}[x] and so by the mod p test
with p = 2

we conclude that x^{5} + 5x^{2} + 1 is irreducible in Q[x].

e. Let f(x) = (5/2)x^{5} + (9/2)x^{4} + 15x^{3} + (3/7)x^{2} + 6x + 3/14 and g(x) = 35x^{5} +
63x^{4} +

210x^{3} +6x^{2} +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 x^{2} +x+4 is irreducible in Z_{1}1[x]
it suffices

to show it has no roots in Z _{11}, as Z_{11} is a field. This is straightforward and
is left to the

reader.

p 316, #16

a. Since Z_{p} is a field, a polynomial of the form x^{2} + ax + b ∈ Z_{p}[x] is
reducible if and

only if there exist c, d ∈ Z_{11} so that x^{2} + 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 Z_{p}[x]. Since there are p^{2} polynomials
of the

form x^{2} + ax + b and each one is either reducible or irreducible, we conclude
there are

irreducible monic degree 2 polynomials in Z_{p}[x].

b. If f(x) ∈ Z_{p}[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

quadratic polynomials in Z_{p}[x].

p 316, #24 Substituting all of the elements of Z_{7} into 3x^{2} + x + 4 we find that
it has two

roots: 4 and 5. The quadratic formula “predicts” the roots

since 6^{-1} = 6 in Z_{7} and −47 = 2 in Z_{7}. Since 3^{2} = 9 = 2 in Z_{7}, 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 Z_{5} into 2x^{2} +x+3 we find no roots. The
quadratic

formula predicts the roots are

since 4^{-1}= 4 and −23 = 2 in Z_{5}. p However, there is no element in Z_{5} whose
square is 2, so

is not an element of Z_{5}. Consequently the roots predicted by the quadratic
formula do

not belong to Z_{5}, which is in agreement with the fact that there are no roots in
Z_{5}.

It turns out that the quadratic formula alwaysgives the roots of ax^{2} + bx
+ c ∈ F[x] for

any field F, as long as we agree that if b^{2} − 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) = (x^{2} + 1)h(x). Since x^{2} + 1 is primitive and
irreducible in

Q[x], it is also irreducible in Z[x]. We apply Theorem 17.6 to write

where the a_{i}, b_{i} and c_{i} are primes in Z and the p_{i}(x), q_{i}(x) and r_{i}(x) are
irreducible polynomials

of positive degree in Z[x]. Substituting these expressions into f (x)g(x) = (x^{2}
+1)h(x)

and rearranging we obtain

Theorem 17.6 and the irreducibility of x^{2} + 1 now imply that x^{2} + 1 = ±p_{i}(x) for
some i or

x^{2} + 1 = ±q_{i}(x) for some i. In the first case x^{2} + 1 divides f(x) and in the
second x^{2} + 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)(x^{2} + 1) + g(x)p = 1. If we
consider this

equation mod p we obtain
which is impossible since

in Z_{p}[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].

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