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Course

MAT 246

Subject

Mathematics

Date

Nov 13, 2023

Type

Other

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4

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Homework Problems 10
Elizabeth Luong
August 6, 2023
Problem 1.
Suppose that
p
is a prime such that
p
−
1
is not a power of two and let
α
=
e
2
πi/p
.
(a) Show that if
f
is a polynomial, then
f
(
x
)
is irreducible if and only if
f
(
x
+ 1)
is irreducible.
(b) Let
f
(
x
) =
x
p
−
1
+
. . .
+
x
+ 1
and note that
f
(
x
) =
x
p
−
1
x
−
1
. Show that
f
is irreducible over
Q
.
Hint. Apply (a) and use the binomial theorem to expand it into a new polynomial.
(c) Conclude that a regular polygon with
p
sides is not constructible.
(a) Let's assume
f
(
x
) is reducible. This means that there exist polynomials
g
(
x
) and
h
(
x
) of degree
greater than 0 such that
f
(
x
) =
g
(
x
)
×
h
(
x
). Now let's consider
f
(
x
+1) =
g
(
x
+1)
×
h
(
x
+1).
Using the binomial expansion we get:
g
(
x
+ 1) =
g
(
x
) +
g
′
(
x
)
h
(
x
+ 1) =
h
(
x
) +
h
′
(
x
)
Where
g
′
(
x
) and
h
′
(
x
) are the derivatives of
g
(
x
) and
h
(
x
), respectively.
Substituting the
expressions for
g
(
x
+1) and
h
(
x
+1) into
f
(
x
+1), we get
f
(
x
+1) = (
g
(
x
)+
g
′
(
x
))
×
(
h
(
x
)+
h
′
(
x
)).
Notice that the constant term in
f
(
x
) is the product of the constant terms in
g
(
x
) and
h
(
x
).
Since both
g
(
x
) and
h
(
x
) have non-zero constant terms, the constant term in
f
(
x
+ 1) will be
non-zero. If
f
(
x
+1) is reducible, it would imply that
f
(
x
+1) can be expressed as the product
of two polynomials of degree greater than 0. However, since the constant term of
f
(
x
+ 1) is
non-zero, this is not possible. Therefore,
f
(
x
+ 1) must be irreducible. Similarly, if we assume
f
(
x
+ 1) is reducible, we can show that it leads to the conclusion that
f
(
x
) is also reducible,
which is a contradiction. Hence,
f
(
x
+ 1) is irreducible if and only if
f
(
x
) is irreducible.
1

(b) Let
f
(
x
) =
x
p
1
+
...
+
x
+ 1 and note that
f
(
x
) =
x
p
1
/x
1. Show that f is irreducible over Q.
Using the geometric series formula, we can express
f
(
x
) as
f
(
x
) =
x
p
1
+
x
p
2
+
...
+
x
+1 = (
x
p
−
1)
/
(
x
−
1). Now, assume
f
(
x
) is reducible over Q. Then, we can write
f
(
x
) as the product of two
polynomials of degree greater than 0
f
(
x
) =
g
(
x
)
×
h
(
x
) where
g
(
x
) and
h
(
x
) are polynomials
with rational coefficients and degree greater than 0.Now, consider
f
(
x
+1) =
g
(
x
+1)
×
h
(
x
+1).
We know from part (a) that
f
(
x
+1)is irreducible if and only if
f
(
x
)is irreducible. Since
f
(
x
)is
irreducible, it follows that
f
(
x
+ 1) is also irreducible. However, we can compute
f
(
x
+ 1) as
follows:
f
(
x
+1) = ((
x
+1)
p
−
1)
/
((
x
+1)
−
1) =
x
p
1
+
...
+
x
+1 =
f
(
x
). Since
f
(
x
+1) =
f
(
x
),
this implies that
g
(
x
+ 1)
×
h
(
x
+ 1) =
g
(
x
)
×
h
(
x
).Now, we can use the fact that the ring of
polynomials over Q is an integral domain, meaning that if
g
(
x
)
×
h
(
x
) = 0, then
g
(
x
) = 0 or
h
(
x
) = 0. Since both
g
(
x
) and
h
(
x
) have degrees greater than 0, this implies that at least one
of them has a degree greater than 0. However, this leads to a contradiction because
f
(
x
) is
irreducible, which means it cannot be factored into two non-constant polynomials. Therefore,
our assumption that
f
(
x
) is reducible is incorrect, and
f
(
x
) must be irreducible over Q.
(c) From part (b), we have shown that the polynomial
f
(
x
) =
x
p
1
+
...
+
x
+ 1 is irreducible over
Q. This means that the field extension
Q
(
α
)
/
Q
, where
α
=
e
(
2
πi/p
), has degree [
Q
(
α
)
/
:
Q
] =
p
−
1, which is not a power of two. Now, it is known that a regular polygon with p sides is
constructible if and only if the field extension
Q
(
α
)
/
Q
has degree [
Q
(
α
)
/
:
Q
] that is a power
of two. Since [
Q
(
α
)
/
:
Q
] =
p
−
1 is not a power of two, a regular polygon with p sides is not
constructible. This result aligns with the fact that a regular p-gon is constructible if and only
if p is a product of distinct Fermat primes and a power of 2, which is not the case for the given
prime p.
Solution.
2

Problem 2.
Determine which are constructible with compass and straightedge:
(a) An equilateral triangle with area 2.
(b) An isosceles triangle with perimeter 8 and area 2.
Solution.
1. Let's assume we can construct an equilateral triangle with area 2.
Let s be the length of
each side of the equilateral triangle.The area of an equilateral triangle with side length s is
given by
p
3
/
4
×
s
2
. Setting this equal to 2, we get:
p
3
/
4
×
s
2
= 2
Solving for s:
s
2
= (2
×
4)
/
√
3 = 8
/
√
3
We need to check if
s
2
is constructible.
If
s
2
is constructible, then
s
=
p
8
/ sqrt
3 must
also be constructible.
So, the question boils down to whether 8
/
√
3 is constructible.
To
check whether 8
/
√
3 is constructible, we need to verify if
q
8
/
√
3 is constructible.
But
(
q
8
/
√
3)
2
= 8
/
√
3 , which means that 8
/
√
3 is constructible if and only if
q
8
/
√
3 is con-
structible. However, it is known that
q
8
/
√
3 is not constructible, which means that 8
/
√
3 is
also not constructible. Therefore, it is not possible to construct an equilateral triangle with
area 2 using a compass and straightedge.
2. Let's assume we can construct an isosceles triangle with perimeter 8 and area 2. Let the two
congruent sides of the isosceles triangle have length a, and let the base have length b. The
perimeter of the triangle is 8, so we have: 2
a
+
b
= 8. The area of the triangle is 2, so we
have: (1
/
2)
×
b
×
h
= 2, where h is the height of the triangle. We can express h in terms
of a and b using the Pythagorean theorem for the right triangle formed by the height, half
the base, and one of the congruent sides:
h
2
+ (
b/
2)
2
=
a
2
.
Now, we can eliminate h by
substituting its value from the second equation:
b
2
/
4 + (
b
2
−
16)
/
4 =
a
2
b
2
+ (
b
2
−
16) = 4
a
2
2
b
2
−
16 = 4
a
2
b
2
= 2
a
2
+ 8
3