Download Homework 6 February 23, 2006 Math 522 Direction: This homework

Homework 6
February 23, 2006
Math 522
Direction: This homework is due on (or before) March 2, 2006. In order to receive
full credit, answer each problem completely and must show all work.
1. Using Eisenstein’s critera show that, for any prime p, the cyclotomic polynomial Φp (x) =
xp −1
x−1
is irreducible over the rationals Q.
Answer: If f (x + 1) is irreducible in Q[x], then f (x) is also irreducible in Q[x]. We will use
this fact to show Φp (x) is irreducible in Q[x]. Expanding Φ(x + 1), we get
p
X
(x + 1)p − 1
1 p p−k
=
x
(x + 1) − 1
x k
k=0
p p−2
p p−3
p
= xp−1 +
x
+
x
+ ··· +
.
1
2
p−1
Φp (x + 1) =
Since
0 6= 1 mod p
p
0=
mod p,
k = 1, 2, 3, ..., p − 1
k
p
0 6=
mod p2
p−1
by Eisenstein’s criteria Φp (x + 1) is irreducible in Q[x]. Therefore Φp (x) is irreducible over
the rational numbers Q.
2. Determine which of the polynomials below are irreducible over the rationals Q: (A)
f (x) = x5 + 9x4 + 12x2 + 6, (B) g(x) = x4 + x + 1, (C) h(x) = x4 + 3x2 + 3, (D) k(x) =
5 5
2x
+ 92 x4 + 15x3 + 37 x2 + 6x +
3
14 .
Answer: (A) Take p = 3 and apply Eisenstein’s criteria: 0 6= 1 mod 3, 0 = 9 mod 3,
0 = 12 mod 3, 0 = 6 mod 3, and 0 6= 6 mod 9. Hence f (x) = x5 + 9x4 + 12x2 + 6 is
irreducible over Q.
(B) Since g(x) = x4 +x+1, hence g̃(x) = x4 +x+1 in Z2 . Further, deg(g(x)) = 2 = deg(g̃(x)).
Since g̃(0) = 1 = g̃(1), the function g̃(x) has no linear factors. Hence if g̃(x) is reducible,
then
g̃(x) = (x2 + ax + b)(x2 + cx + d)
for some a, b, c, d ∈ Z2 . Multiplying out the two factors on the right side, and then equating
various powers of x, we get
a + c = 0,
b + d + ac = 0,
ad + bc = 1,
bd = 1.
Solving this system of equations, we see that
1
a
−b =1
b
for b 6= 0.
In Z2 , the above equation has no solution. Thus g̃(x) is irreducible in Z2 , and consequently
the polynomial g(x) = x4 + x + 1 is irreducible over Q.
(C) The polynomial h(x) = x4 + 3x2 + 3 is irreducible by Eisenstein’s criteria with p = 3.
(D) First show 14k(x) is irreducible over Q. Note
14k(x) = 35x5 + 63x4 + 210x3 + 6x2 + 84x + 3.
Take p = 3 and apply Eisenstein’s criteria to see 14k(x) is irreducible over Q. Hence k(x) is
also irreducible over Q.
3. Let f (x) ∈ F [x], where F is a field, and a ∈ F . Then f (a) is the remainder in the
division of f (x) by x − a.
Answer: THIS WAS DONE IN CLASS. LOOK AT YOUR NOTES.
4. Let f (x) ∈ F [x], where F is a field, and a ∈ F . Then a is a zero of f (x) if and only if
x − a is a factor of f (x).
Answer: By Remainder Theorem, we have
f (x) = (x − a)g(x) + f (a),
(1)
where g(x) ∈ F [x]. Suppose a is a zero of f (x). Then f (a) = 0. Hence from (1), we get
f (x) = (x − a)g(x). This proves that (x − a) is factor of f (x).
Conversely, if (x − a) is factor of f (x), then the remainder f (a) must be zero. That is
f (a) = 0 and thus a is a zero of f (x).
5. Is the polynomial f (x) = x4 + x3 + 2x2 + 3x + 2 irreducible over Z5 ?
Answer: Consider f (x) = x4 + x3 + 2x2 + 3x + 2 in Z5 [x]. Since f (2) = 0, f (x) is not
irreducible in Z5 [x].
6. Construct a field of order 4. What are the elements of this field? Show that the sum of
the elements of this field is equal to zero (that is, the additive identity of this field).
Answer: Since 4 = 22 , we start with a field Z2 of characteristic 2 and look for an irreducible
polynomial of degree 2 in Z2 [x]. It is easy to note that p(x) = x2 + x + 1 has no zeros in Z2
since p(0) = 1 = p(1). Hence, the polynomial p(x) = x2 + x + 1 is irreducible in Z2 [x]. Thus
Z2 [x]/ < x2 + x + 1 > is a field with 4 elements. The elements of this field are given by
Z2 [x]/ < x2 + x + 1 > = {ax + b+ < x2 + x + 1 > | a, b ∈ Z2 }
= { 0+ < x2 + x + 1 >, 1+ < x2 + x + 1 >,
x+ < x2 + x + 1 >, x + 1+ < x2 + x + 1 > }
= { 0, 1, α, α + 1 },
where we have denoted 0+ < x2 +x+1 > as 0, 1+ < x2 +x+1 > as 1, and x+ < x2 +x+1 >
as α. The sum of the elements of this field is given by
0 + 1 + α + α + 1 = (0+ < x2 + x + 1 >) + (1+ < x2 + x + 1 >)
(x+ < x2 + x + 1 >) + (x + 1+ < x2 + x + 1 >)
= 2x + 2+ < x2 + x + 1 >
= 0+ < x2 + x + 1 >
= 0.
7. Construct a field of order 4. Provide the addition and multiplication tables for this field.
Answer: Note that α is a zero of the polynomial p(x) = x2 + x + 1 since
p(α) = α2 + α + 1
= (x+ < x2 + x + 1 >)2 + (x+ < x2 + x + 1 >) + (1+ < x2 + x + 1 >)
= (x2 + x + 1+ < x2 + x + 1 >)
= 0+ < x2 + x + 1 >
= 0.
Hence α2 + α + 1 = 0 or α2 = −α + 1 which is α2 = α + 1.
The addition table for Z2 [x]/ < p(x) > is the following:
+
0
1
α
α+1
0
0
1
α
α+1
1
1
0
α+1
α
α
α
α+1
0
1
α+1
α+1
α
1
0
The multiplication table for Z2 [x]/ < p(x) > is the following:
·
0
1
α
α+1
0
0
0
0
0
1
0
1
α
α+1
α
0
α
α+1
1
α+1
0
α+1
1
α
8. Construct a field of order 25. What are the elements of this field?
Answer: Since 25 = 52 , we start with a field Z5 of characteristic 5 and look for an irreducible
polynomial of degree 2 in Z5 [x]. Such a polynomial is p(x) = x2 + x + 1.
Hence Z5 [x]/ < x2 + x + 1 > is a field of order 25. The elements of this field are:
Z5 [x]/ < x2 + x + 1 > = { ax + b+ < x2 + x + 1 > | a, b ∈ Z5 }
= { 0, 1, 2, 3, 4,
α, α + 1, α + 2, α + 3, α + 4,
2α, 2α + 1, 2α + 2, 2α + 3, 2α + 4,
3α, 3α + 1, 3α + 2, 3α + 3, 3α + 4,
4α, 4α + 1, 4α + 2, 4α + 3, 4α + 4, },
where α = x+ < p(x) >.
9. Construct a field of order 49. What are the elements of this field?
Answer: Since 49 = 72 , we start with a field Z7 of characteristic 7 and look for an irreducible
polynomial of degree 2 in Z7 [x]. Such a polynomial is p(x) = x2 +1. Hence Z7 [x]/ < x2 +1 >
is a field of order 49. The elements of this field are:
Z5 [x]/ < x2 + 1 > = { ax + b+ < x2 + x + 1 > | a, b ∈ Z5 }
= { aα + b | a, b ∈ Z7 },
where α = x+ < x2 + 1 >.
10. Use Eisenstein’s criteria to show that the polynomial f (x) = x4 + 4x3 + 6x2 + 4x + 2 is
irreducible in Q[x].
Anwer: Since f (x) = x4 + 4x3 + 6x2 + 4x + 2, we have
f˜(x) = x4 + x3 + x + 2
in Z3 [x]. Since f˜(0) = 2 = f˜(1), and f˜(2) = 1, the polynomial f˜(x) has no linear factors in
Z3 [x] and therefore f (x) can be written as
x4 + x3 + x + 2 = (x2 + ax + b)(x2 + cx + d)
for some a, b, c, d ∈ Z3 . Multiplying out the two factors on the right side, and then equating
various powers of x, we get
a + c = 1,
b + d + ac = 0,
ad + bc = 1,
bd = 2.
It can be easily shown that the above system of equations has no solution in Z3 . Thus f˜(x)
is irreducible in Z3 and hence f (x) is irreducible in Q[x].
The same conclusion can be easily derived using Eisenstein’s criteria. Take p = 2 and
apply Eisenstein’s criteria to conclude f (x) is irreducible in Q[x].