Polynomials
17.11 Show that x4 + 1 is irreducible over Q but
reducible over R.
Note that f(x) = x4 + 1 is irreducible over Q if
is
irreducible over Q. By Eisenstein’s Criterion, f(x + 1) is irreducible over Q
(consider the prime
p = 2), and so f(x) = x4 + 1 is irreducible over Q.
Observe that over R, 
. Note that each of these degree
two factors is irreducible, for if one of them were reducible, x4 + 1 would have
a linear factor (i.e.
a root) over R, which is clearly not the case.
17.17 Show that for every prime p there exists a field of order p2.
We will show that an irreducible polynomial of degree two always exists over
.
A reducible
monic quadratic polynomial has the form (x −α )(x − β). There are p monic
polynomials of
degree one over , and so there are p(p − 1)/2 + p = p(p + 1)/2 (consider the
case α= β)
reducible monic polynomials of degree two over . But p(p + 1)/2 < p2, where p2
is the number
of monic quadratic polynomials over . Therefore, there always exists a (monic)
irreducible
polynomial f(x) of degree two over , and so
is a finite field of
order p2.
17.24 Find all zeros of f(x) = 3x2 +x+4 over
by substitution. Find all zeros
of f(x) by using
the Quadratic Formula 
. Do your answers agree? Should they? Find all
zeros of g(x) = 2x2 + x + 3 over 
by substitution. Try the Quadratic Formula
on g(x). Do your
answers agree? State necessary and sufficient conditions for the Quadratic
Formula to yield the
zeros of a quadratic from 
, where p is a prime greater than 2.
For f(x), both methods give the zeros 
Note that the quadratic formula
involves the
term
in . We do not find any zeros for g(x). In this case, the quadratic
formula produces the term 
and
is not a square in
, i.e. there is no element
such that 
(mod 5).
A quadratic polynomial ax2 + bx + c (a ≠ 0) has a zero in
if and only if
its discriminant
is a square in , i.e.
such that 
(mod p).
17.25 (Rational Root Theorem) Let 
and
. Prove
that if r and s are relatively prime integers and
, then
and 
.
If , then 
Multiplying both sides by sn, we have
Since r divides every term except
and (r,
s) = 1, we must
have . Since s divides every term except for
and (r, s) = 1, we must
have .
17.29 Show that x4 + 1 is reducible over for every prime p.
We will show that for every prime p, at least one of the elements
or 
is square in .
This will give us the desired result since:
If 
such that
then 
If such that
then 
If such that
then 
To see that one of these cases must hold, consider the homomorphism
Since 
(this argument does not hold for p = 2, but this case can
be handled separately). Suppose that 
Since
only has two cosets in
, we
have 
Additionally, is abelian,
and so 
is a group. Note that
has order 2, so the square of any element in this factor group is the
identity coset
Now we have
so
that 
Therefore,
is a square in
, so one of the three desired cases holds in
for all primes p.