Modern Algebra I Section 1 Assignment 6

Exercise 1. (pg. 64 Warm Up a) Why is a linear polynomial in Q[x] always irreducible?

Solution:
If the linear polynomial p was not irreducible, then there would be two polynomials a, b ∈
Q[x] such that ab = p and deg a, deg b ≥1. But then 1 = deg p = deg a+deg b≥ 2 by Theorem
4.1, which is a contradiction.

Exercise 2. (pg. 64 Warm Up b) Why is a polynomial of the form x² + a ∈ Q[x], where a > 0,
always irreducible?

Solution:
If the polynomial were not irreducible, then there would be two polynomials a, b ∈Q[x]
such that x² +a = ab and deg a, deg b ≥1. By Theorem 4.1, deg a = deg b = 1. So a and b are
linear polynomials, say p x − q and r x − s say, for p, q, r, s ∈ Q. By the Root Theorem (4.3),
that means (q/ p)² + a = 0 and (s/r )² + a = 0. But it is easy to see that b² + a≥a > 0 and
c² +a ≥a > 0, a contradiction.

Exercise 3. (pg. 65Warm Up c) Determine a factorization of x ⁴−5x²+4 into ir reducibles in Q [x].

Solution:

Using the Rational Root Theorem (5.6), we know that all rational roots of the polynomial are
of the form

If we apply the Root Theorem (4.3) and substitute these into the polynomial, we find that 1, -1,
2, and -2 are roots of the polynomial. Thus it factors as

(x +1) (x −1) (x +2) (x −2) .

Exercise 4. (pg. 65 Warm Up e) We know that 7 is an irreducible integer, but is 7 an irreducible
polynomial?

Solution:
The definition of an irreducible polynomial p includes the requirement that deg p > 0. Since
deg 7 = 0, 7 is not an irreducible polynomial.

Exercise 5. (pg. 65 Warm Up g) Factor 2x³ +7x² −2x −1 completely into irreducibles in Q[x],
using Gauss’ Lemma and the Root Theorem. Adjust your factorization (if necessary) so that all factors
belong to Z[x].

Solution:
We will also use the Rational Root Theorem (5.6)—I have no idea why the book didn’t suggest
that.

By the Rational Root Theorem, we know that all rational roots of the polynomial are of the
form

If we apply the Root Theorem (4.3) and substitute these into the polynomial, we find that 1/2 is
the only rational root. We can use long division or synthetic division to show that

We can adjust the factorization using associates: factor the scalar 2 from the quadratic polynomial
and distribute it into the linear polynomial. Thus

It is easy to use the Rational Root Theorem (5.6) and the Root Theorem (4.3) to show that
this polynomial has no roots in Z. By Gauss’ Lemma (5.5) the quadratic polynomial is thus
irreducible over Q.

Exercise 6. (pg. 65 Exercise 1) Prove Theorem 5.1: A polynomial in Q[x] of degree greater than
zero is either irreducible or the product of irreducibles.

Solution:
Let f ∈ Q[x] be arbitrary, but fixed. Assume deg f > 0. We proceed by induction on n =
deg f .
For the inductive base, let n = 1. Then f is irreducible by Exercise 1 (Warm Up a).
We have shown that the assertion is true for n = 1.
Assume that n > 1. For the inductive hypothesis, assume that every polynomial of degree i is
either irreducible or the product of irreducibles, for all i : 1≤ i < n.
If f is irreducible, then we are done.
Otherwise, we can write f = pq for some p, q ∈ Q[x] where deg p, deg q > 0. Recall from
Theorem 4.1 that deg f = deg p +deg q. Thus deg p = deg f −deg q < deg f . Similarly deg q <
deg f . By the inductive hypothesis, p is either irreducible or the product of irreducibles. So we
can write for some r ∈ N and for some irreducible , for all i : 1 ≤i ≤r .
By similar reasoning, we can write for some s ∈ N and for some irreducible
, for all j : 1≤ j ≤s. Hence where r, s ∈ N and
are irreducible for all i : 1 ≤i ≤r and for all j : 1 ≤j ≤s. So f is the product of irreducibles.
We have shown that f is either irreducible or the product of irreducibles. Since n is arbitrary,
this is true regardless of the degree of f . Since f is arbitrary, it is true for all f ∈Q[x].

Exercise 7. (pg. 65 Exercise 4) Use Gauss’s Lemma to determine which of the fol lowing are irreducible
in Q[x]:

Solution:
It is fairly easy to use factoring by grouping and see that the second polynomial is not irreducible
in Z[x], since

By Gauss’s Lemma (5.5), the polynomial is also not irreducible in Q[x].

If the other two polynomials factor, then one of the factors is linear. (See the discussion in
the text on page 62.) For the third polynomial, the Root Theorem (4.3) and the Rational Root
Theorem (5.6) tell us that any linear factor has the form x ±1. Since neither 1 nor -1 are roots
of the polynomial, it must be irreducible. For the first polynomial, the same theorems tell us
that any linear factor has one of the forms x ±1, x ±2, 2x ±1, 4x ±1. Exhaustive inspection (via
synthetic division) shows that none of these factors the polynomial, so it is also irreducible.

Exercise 8. (pg. 65 Exercise 5) Show that x⁴ +2x² +4 is irreducible in Q[x].

Solution:
For convenience, we denote the polynomial as p.
For any rational number a/b , then

so p has no rational roots. By the Root Theorem (4.3), it has no linear factors in Q[x]. This
would exclude cubic factors as well, since if p = q r for some q, r ∈ Q[x], and if deg q = 3, then
by Theorem 4.1 deg r = deg p −deg q = 4−3 = 1. Since p has no linear factors, r cannot have
degree 1, so q cannot have degree 3.

It remains to show that the polynomial has no quadratic factors. Assume to the contrary that
p = x⁴+2x²+4 has quadratic factors p = q r . So q = ax²+b x +c and r = d x²+e x + f where
a, b, c, d, e, f ∈Q. By Gauss’ Lemma (5.5), we can assume q, r ∈ Z[x], so that a, b, c, d, e, f ∈Z.
If we multiply q , r , we can collect like terms to obtain

Two polynomials are equal iff their coefficients are equal , so

Since a, d are integers and ad = 1, it must be that a = d = 1. The system now becomes

Observe that b = −e, so we have

From equation (2), we know that b = 0 or f = c. We consider two cases.

Case 1: If f = c, equation (3) tells us that c = ±2. Substituting this into equation (1) we see
that b² = 2 or b² = −4, neither of which has an integer solution. Since b must be an integer,
f ≠ c.

Case 2: If b = 0, equation (1) tells us that f + c = 2, or f = 2− c. Substituting into equation
(3), we have

The quadratic formula shows that this has no integer solution for c. Since c must be an integer,
b ≠ 0.

Neither case gives a solution for the coefficients. Hence f cannot factor as the product of two
quadratic polynomials. Thus f is irreducible in Z[x]. By Gauss’ Lemma (5.5), f is irreducible
in Q[x].

Exercise 9. (pg. 65 Exercise 7) Use the Rational Root Theorem 5.6 to factor

Solution:

By the rational root theorem, the only roots of the polynomial possible are

Using substitution (or division), we find that the roots are 9, -1, and 1/2. The polynomial factors
as

(You may obtain a different ex pression ; I’m trying to emphasize that 1/2, −1, and 9 are all roots.

Exercise 10. (pg. 66 Exercise 8) Use the Rational Root Theorem 5.6 to argue that

is irreducible over Q[x]. Use elementary calculus to argue that this polynomial does have exactly
one real root.

Solution:
If the polynomial factors as pq, then by Theorem 4.1, 3 = deg p +deg q. Since this is a sum of
integers, either deg p = 1 or deg q = 1. By the Rational Root Theorem (5.6), all roots would be
of the form

However, none of those are roots. By the Root Theorem (4.3), the polynomial has no linear
factors. Since the polynomial has no linear factors, deg p ≠ 1 and deg q ≠ 1. So the polynomial
cannot factor.

Exercise 11. (pg. 66 Exercise 13) (a) Prove that the equation a²  = 2 has no rational solution; that is,
prove that is irrational. (This part is a repeat of Exercise 2.14.)

(b) Generalize part a, by proving that aⁿ = 2 has no rational solutions, for all positive integers
n≥ 2.

Solution:
(a) If a² = 2 has a rational solution, then a² − 2 has a rational root. By the Rational Root
Theorem (5.6), any such root has the form

Since neither of those is a root of a² −2, it follows that a² = 2 has no rational solution.

(b) If aⁿ = 2 has a rational solution, then aⁿ −2 = 0 has a rational solution, so the polynomial
aⁿ−2 has a rational root. By the Rational Root Theorem (5.6), any such root is of the form

Is either a root of aⁿ −2? Substitute and see:

Since none of the possibilities is a root of aⁿ −2, it follows that aⁿ = 2 has no rational solution.