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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.