**17.11** Show that x^{4} + 1 is irreducible over Q but
reducible over R.

Note that f(x) = x^{4} + 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) = x^{4} + 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, x^{4} + 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 p^{2}.

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 < p^{2}, where p^{2}
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 p^{2}.

**17.24** Find all zeros of f(x) = 3x^{2} +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) = 2x^{2} + 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 ax^{2} + 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 s^{n}, 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 x^{4} + 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.