# ANSWERS FOR MATH PROBLEM SET VII

Section 14B, page 301.

**15. **Following the Euclidean algorithm, we find the following instances of the
division theorem in

Thus one gcd is 2x + 2. The monic gcd is x + 1.

Now that we know the answer, let's look at the problem again. We are working
over . The two

given polynomials can be factored :

and

and x^{2} + x + 2 is irreducible in . From this
factorization, it is clear that the gcd is x + 1.

**19.** First, following the Euclidean algorithm, the have the following (obvious)
instances of the

division theorem :

Thus we see that the gcd is indeed 1, so solutions exist .

Secondly, we can unwind this to obtain the Bezout polynomials.

and

which together give

which is an equation in the desired form.

Section 14C, page 305.

**38**. We are given that q = pr for some r. Since q is irreducible and p is not a
unit, we conclude

that r is a unit. Both p and q are monic, so r = 1.

**42.** Having read the hint, we observe that
. So the two
given polynomials

are

and

The gcd is

**45.** Counterexample: In the field of polynomials over the reals, let
,
with n = 2. Then

f is reducible , but it has no irreducible factor of degree < 1/1.

The author must have meant "< n/2" to be " ≤n/2." Now the result is pretty clear .
We have

where k≥2, the g_{i}'s are irreducible,
and
So we could not have

both and
That is, either
or

**46.** In , there are of course two linear polynomials (i.e., degree 1), four
quadratic polynomials

(degree 2), eight cubic polynomials (degree 3), and sixteen quartic polynomials
(degree 4).

We know that both the linear polynomials , x and x + 1, are irreducible. For the
quadratics and the

cubics, any polynomial with a root in is reducible, because it will have a
linear factor, by the root

theorem. And if it does not have a root, then it will be irreducible, because
for 2 or 3 to be the sum of

two smaller numbers, one of those smaller numbers must be 1.

In this way, we find one quadratic , , and two cubics,
and , that are

irreducible.

For the quartics, there are four without roots. But might they be the product of
irreducible quadrat-

ics? Because there is only one irreducible quadratic, this consideration
eliminates only one quartic, its

square . The other three are irreducible: , and
.

That finishes the problem. Nonetheless, we add here a list of all 32 polynomials
in with degree

4 or less, with their factorization into irreducible polynomials.

0 | zero | 1 | unit | |

x | irreducible | x + 1 | irreducible |

irreducible

irreducible

irreducible

irreducible

irreducible

irreducible

Section 15D, page 319.

**18.** We are given the valuable clue that 3+2i is a root of f. Therefore 3-2i is
also a root, and f

is divisible by the quadratic

When we carry out the division, we obtain

Because x^{2} + 2x + 2 has no real roots, it is irreducible over the reals, and we
are done.

**21. **We seek a multiple of f that has real coefficients. Following the hint, we
calculate

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