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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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