Homework 9 Solutions to Selected Problems
June 11, 2012
1
Chapter 17, Problem 12
Since x2 + x + 4 has degree 2 and Z11 is a eld, we may use Theorem 17.1 and
show that f (x) is irreducible because it has no zeros:
x
0
1
2
3
4
5
6
7
8
9
10
x2 + x + 4 mod 11
4
6
10
16 = 5
24 = 2
34 = 1
46 = 2
60 = 5
76 = 10
94 = 6
114 = 4
Since the polynomial has no zeros in Z11 , it is irreducible.
1
2
Chapter 17, Problem 14
Let us start by nding zeros of f (x) in Z2 . It turns out that f (0) = 1, while
f (1) = 4 = 0 (mod 2). Thus, x − 1 divides f (x). By using long division
(remember to reduce mod 2), we see that
x−1
−
3
x
x3
0
x2
+x2
−x2
0
−
+x
x
x
+1
+1
+1
−1
0
f (x) = (x − 1)(x2 + 1).
Is x2 + 1 irreducible? Let us nd zeros again. Since 12 + 1 = 2 = 0 (mod 2),
x − 1 divides x2 + 1. We can use long division again, but since we are working
mod 2, 1 = −1, so we can write
x2 + 1 = x2 − 1 = (x − 1)(x + 1) = (x − 1)(x − 1)
and therefore
f (x) = (x − 1)(x2 + 1) = (x − 1)(x − 1)(x − 1) = (x − 1)3 .
3
Chapter 17, Problem 30
Note that
f (x) = xp−1 −xp−2 +xp−3 −. . .−x1 +1 = (−x)p−1 +(−x)p−2 +(−x)p−3 +. . .+(−x)+1
= Φp (−x).
That is, f (x) is the pth cyclotomic polynomial, but with x replaced by −x.
Therefore, if
f (x) = g(x)h(x),
then
Φp (x) = f (−x) = g(−x)h(−x).
By the Corollary on page 310, Φp (x) is irreducible, so either g(−x) or h(−x) is
a unit, so either g(−x) = a or h(−x) = b where a, b ∈ Zp and a, b 6= 0. Thus,
g(x) = a, or h(x) = b, so either g(x) is a unit or h(x) is a unit. Therefore, f (x)
is irreducible.
2
4
Chapter 17, Problem 32
4.1
hx2 + 1i
is a prime ideal
Let f (x), g(x) ∈ Z[x], and suppose f (x)g(x) ∈ x2 + 1 . Then there is a polynomial q(x) ∈ Z[x] such that
f (x)g(x) = q(x)(x2 + 1).
Since x2 + 1 is monic (in particular its leading coecient is a unit in Z), we
can use the division algorithm from page 296. Although the book proves the
division algorithm for polynomials with coecients in a eld, if you look at the
proof, the only multiplicative inverse needed is for the leading coecient of the
dividing polynomial. Thus,
f (x) = q1 (x)(x2 + 1) + r1 (x),
g(x) = q2 (x)(x2 + 1) + r2 (x),
where r1 (x) and r2 (x) have degree less than 2 (the degree of x2 + 1). Then
f (x)g(x) = q1 (x)q2 (x)(x2 + 1)2 + q1 (x)(x2 + 1)r2 (x)
+q2 (x)(x2 + 1)r1 (x) + r1 (x)r2 (x) = q(x)(x2 + 1).
We can get rid of the terms involving x2 + 1 by plugging in x = i. Since
i2 + 1 = 0, we get
f (i)g(i) = r1 (i)r2 (i) = q(i)(i2 + 1) = 0.
Since r1 (x) and r2 (x) have degree less than 2, write r1 (x) = a1 x + a0 and
r2 (x) = b1 x + b0 . Then
r1 (i)r2 (i) = (a1 i + a0 )(b1 i + b0 ) = 0.
Since C is a eld, and hence an integral domain, either a1 i+a0 = 0 or b1 i+b0 = 0.
Without loss of generality, say a1 i + a0 = 0. Then a1 = 0 and a0 = 0, so
r1 (x) = 0x + 0 = 0,
and therefore
f (x) = q1 (x)(x2 + 1) + r1 (x) = q1 (x)(x2 + 1) ∈ x2 + 1 ,
so x2 + 1 is a prime ideal.
4.2
hx2 + 1i
is not a maximal ideal
Consider x2 + 1, 2 = p1 (x)(x2 + 1) + p2 (x) · 2: p1 (x), p2 (x) ∈ Z[x] . It is the
ideal generated by x2 + 1 and 2. We need to show that
2
x + 1 ( x2 + 1, 2 ( Z[x].
3
4.2.1
x2 + 1 ( x2 + 1, 2
Note that every nonzero element of x2 + 1 has the form q(x)(x2 + 1) where
q(x) 6= 0, so the degree of a nonzero element is the sum of the degree
of q(x)
and the degree of x2 + 1. Hence the degree of any nonzero element of x2 + 1
is
to 2, so there
are no nonzero
2greater
than or equal
constant
polynomials
in
x + 1 . Thus 2 ∈
/ x2 + 1 , but 2 ∈ x2 + 1, 2 , so x2 + 1 ( x2 + 1, 2 .
4.2.2
x2 + 1, 2 ( Z[x]
We need to show
that 1 ∈
/ x2 + 1, 2 . We will proceed by contradiction and
2
assume that 1 ∈ x + 1, 2 , so there exist polynomials p1 (x) and p2 (x) in Z[x]
such that
1 = p1 (x)(x2 + 1) + p2 (x) · 2.
Warning: We cannot conclude that f (x) = 0 because the left hand side is a
constant polynomial. I apologize for this mistake from Problem 24, Chapter 16
of HW6 (a corrected proof has been posted).
We can get rid of p1 (x)(x2 + 1) by plugging in x = i:
1 = p1 (i)(i2 + 1)p2 (i) · 2 = p2 (i) · 2,
so
p2 (i) = 2−1 =
1
.
2
However, the coecients of p2 (x) are integers, and since i2 = −1 ∈ Z, p2 (i)
must be a complex number of the form a + bi where a and b are integers (that
is,
p2 (i) cannot equal 12 , a contradiction. Therefore, 1 ∈
/
2p2 (i) ∈ Z[i]
. Thus,
2
x + 1, 2 , so x + 1, 2 ( Z[x].
Alternate Method:
Take
1 = p1 (x)(x2 + 1) + p2 (x) · 2
and reduce both sides mod 2 (so we will be working in Z2 [x]). This gets rid of
the p2 (x) · 2 term and leaves us with
1 = p̄1 (x)(x2 + 1)
where the coecients of p̄1 (x) are those of p1 (x) reduced mod 2. If p̄1 (x) = 0,
then we get a contradiction (1 = 0). If p̄1 (x) 6= 0, then by Problem 17 from
Chapter 16, the degree of p̄1 (x)(x2 + 1) equals the sum of the degrees of p̄1 (x)
and x2 + 1, so the degree of p̄1 (x)(x2 + 1) is greater than or equal to the degree
of x2 + 1, which is 2. However, the constant polynomial 1 has degree zero, a
contradiction.
4
4.2.3
Conclusion
Since x2 + 1 ( x2 + 1, 2 (
Z[x], x2 + 1 is strictly contained in an ideal
which is not equal to Z[x], so x2 + 1 is not a maximal ideal.
5
Chapter 17, Problem 34
Suppose r is a real number and
r + r−1 = 2m + 1
where m is an integer (so 2m + 1 is an odd integer). Let us multiply both sides
by r:
r2 + 1 = (2m + 1)r.
Move everything to the left:
r2 − (2m + 1)r + 1 = 0.
Thus, r is a zero of the polynomial
f (x) = x2 − (2m + 1)x + 1.
Does f (x) have any zeros in Q? Since it has degree two, this is equivalent
(by Theorem 17.1) to asking whether f (x) is irreducible in Q[x]. Since the
coecients are integers, we can use Theorem 17.3 and reduce the polynomial
mod 2 (I chose 2 because the coecient of x is only known to be odd, so it will
denitely reduce to 1 mod 2):
f¯(x) = x2 − (1)x + 1 = x2 + x + 1 mod 2.
Note that f¯(x) has no zeros in Z2 :
f¯(0) = 0 + 0 + 1 = 1 6= 0,
f¯(1) = 12 + 1 + 1 = 3 = 1 6= 0.
Since f¯(x) has degree 2, by Theorem 17.1, f¯(x) is irreducible over Z2 , and since
it has the same degree as f (x), by Theorem 17.3, f (x) is irreducible over Q, so
by Theorem 17.1, f (x) has no zeros in Q. Since r is a zero of f (x), r cannot be
in Q, so it must be irrational.
5