Download 4.3 Existence of Roots

4.3
4.3
J.A.Beachy
1
Existence of Roots
from A Study Guide for Beginner’s by J.A.Beachy,
a supplement to Abstract Algebra by Beachy / Blair
a b 25. Show that the field F =
a, b ∈ R defined in Exercise 4.1.13 is isomor−b a phic to the field of complex numbers.
a b
= a + bi. The following calculations show that φ
Solution: Define φ
−b a
respects multiplication.
a b
c d
ac − bd ad + bc
φ
=φ
= (ac − bd) + (ad + bc)i
−b a
−d c
−ad − bc ac − bd
a b
c d
φ
φ
= (a + bi)(c + di) = (ac + bdi2 ) + (ad + bc)i
−b a
−d c
= (ac − bd) + (ad + bc)i
It is clear that φ is one-to-one and onto and respects addition, so φ is an isomorphism.
Comment: Appendix 5 of Abstract Algebra contains more details.
√
√
26. Show that Q( 3i) = {a + b 3i | a, b ∈ Q} is a field isomorphic to Q[x]/ x2 + 3 .
Comment:
In the course of defining the appropriate function, we will show that
√
Q( 3i) is a field.
Solution: By Proposition 4.3.3, each congruence class in Q[x]/ x2 + 3 contains
a
unique representative of the form a+bx, with a, b ∈ Q. Define φ : Q[x]/ x2 + 3 → C
√
by φ([a + bx]) = a + b 3i. Since [a + bx] = [c + dx] implies that a = c and b = d, it
follows that φ is a well-defined function.
√
√
√
If φ([a√+ bx]) = φ([c + dx]), then a + b 3i = c + d 3i, and so a − c = (d − b) 3i.
Since 3i 6∈ Q, the only way this can happen is if a − c = 0 and d − b = 0.√Then
a = c and b = d, so [a + bx] = [c + dx], and thus φ√is one-to-one. The set Q( 3i) is
the image of φ, so φ maps Q[x]/ x2 + 3 onto Q( 3i)
In calculating the product [a + bx][c + dx], we get [ac + (ad + bc)x + bdx2 ], and to
reduce this to standard form we can make the substitution x2 = −3, since x2 + 3 ≡
0 (mod x2 + 3). Thus [a + bx][c + dx] = [(ac − 3bd) + (ad + bc)x]. We then have the
following calculations.
√
φ([a + bx] + [c + dx]) = φ([(a + c) + (b + d)x]) = (a + c) + (b + d) 3i
√
√
= (a + b 3i) + (c + d 3i) = φ([a + bx]) + φ([c + dx])
√
φ([a + bx][c + dx]) = φ([(ac − 3bd) + (ad + bc)x]) = (ac − 3bd) + (ad + bc) 3
√
√
= (a + b 3i)(c + d 3i) = φ([a + bx])φ([c + dx])
4.3
J.A.Beachy
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√
Because φ : Q[x]/ x2 + 3 → Q( 3i) is a one-to-one correspondence that respects
√
so we
addition and multiplication, it transfers the properties of a field to Q( 3i), and √
do not need to go through the verification of all of the field axioms for the set Q( 3i).
It follows that φ is an isomorphism of fields.
27. Show that
thefollowing set of matrices, with entries from Z3 , is a field isomorphic to
Z3 [x]/ x2 + 1 .
0 0
1 0
−1
0
0 1
1 1
−1
1
0 −1
,
,
,
,
,
,
,
0 0
0 1
0 −1
−1 0
−1 1
−1 −1
1
0
1 −1
−1 −1
,
=F
1
1
1 −1
Solution: (This is just an outline of the solution.) Define φ : Z3 [x]/ x2 + 1 → F
1 0
0 1
by φ([a + bx]) = a
+b
. The mapping is well-defined since each
0 1
−1 0
congruence class contains a unique representative of the form a+bx, and simply listing
the possible values of φ shows it to be a one-to-one correspondence. It is clear that
φ will preserve addition. That φ preserves multiplication depends on the fact that
the congruence class [x], which satisfies the equation [x]2 = −[1], maps to the matrix
2
0 1
0 1
1 0
, which satisfies the corresponding equation
=−
.
−1 0
−1 0
0 1
28. Construct a field with 25 elements.
Solution: We will construct a field of the form F [x]/ hp(x)i. The fact that 25 = 52
shows that we need to take F = Z5 , and that p(x) must be an irreducible polynomial
of degree 2. A bit of trial and error (checking for roots in Z5 ) identifies x2 + x + 1 as
one of the possible choices for a polynomial of degree 2 irreducible in Z5 [x].
The field Z5 [x]/ x2 + x + 1 has 25 elements, as required. We should also give the
identity necessary to reduce products [a + bx][c + dx] to standard form:
x2 ≡ −x − 1 (mod x2 + x + 1).
29. Find all powers of [x] in Z3 [x]/ x2 + x − 1 , and then find [x]−1 .
Solution: Since x2 ≡ 1 − x (mod x2 + x − 1), we have the following list:
[x]0 = [1], [x]1 = [x], [x]2 = [1−x], [x]3 = [x][1−x] = [x−x2 ] = [x−(1−x)] = [−1−x],
[x]4 = [x][−1 − x] = [−x − x2 ] = [−x − (1 − x)] = [−1], [x]5 = [−x], [x]6 = [−x2 ] =
[−1 + x], [x]7 = [x]4 [x]3 = [1 + x], [x]8 = ([x]4 )2 = [1].
Since [x] has order 8 in the multiplicative group of the field, its inverse is [x]7 = [1+x].
Comment: Each nonzero element of Z3 / x2 + x − 1 is a power of [x], so we have
shown that the multiplicative group of this finite field is cyclic, with generator [x].
30. In Z2 [x]/ x3 + x + 1 , find the multiplicative inverse of [x + 1].
Solution: We first give a solution using the Euclidean algorithm. For p(x) = x3 +x+1
and f (x) = x+1, the first step of the Euclidean algorithm gives p(x) = (x2 +x)f (x)+1.
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Thus p(x)− (x2 + x)f (x) = 1, and so reducing modulo p(x) gives [−x2 − x][f (x)] = [1],
and therefore [x + 1]−1 = [−x2 − x] = [x2 + x].
Alternate Solution: We next give a solution that uses the identity [x3 ] = [x + 1] to
solve a system of equations. We need to solve [1] = [x + 1][ax2 + bx + c] or
[1] = [ax3 + bx2 + cx + ax2 + bx + c]
= [ax3 + (a + b)x2 + (b + c)x + c]
= [a(x + 1) + (a + b)x2 + (b + c)x + c]
= [(a + b)x2 + (a + b + c)x + (a + c)] ,
so we need a + b ≡ 0 (mod 2), a + b + c ≡ 0 (mod 2), and a + c ≡ 1 (mod 2). This
gives c ≡ 0 (mod 2), and therefore a ≡ 1 (mod 2), and then b ≡ 1 (mod 2). Again, we
see that [x + 1]−1 = [x2 + x].
31. Is [x] a generator of the multiplicative group of the field Z5 [x]/ x2 + x + 1 ? Is [1+ x]
a generator?
Comment: This is the field defined in the solution to Problem 28.
Solution: The field has 25 elements, so its multiplicative group has 24 elements, and
[x] is a generator if and only if it has order 24. We have x2 ≡ −1 − x (mod x2 + x + 1),
so [x]3 = [x][−1 − x] = [−x − x2 ] = [−x − (−x − 1)] = [−x + x + 1] = [1]. Thus [x]
has order 3, so it is not a generator.
Since 1 + x ≡ −x2 (mod x2 + x + 1), we have [1 + x]2 = [x] and [1 + x]3 = −[x]6 =
−([x]3 )2 = −[1]. Thus [1 + x]6 = [1], but no smaller power is the identity, showing
that [1 + x] has order 6, so it is not a generator.
32. Show that x4 + x + 1 is irreducible over Z2 . Factor x4 + x + 1 over Z2 [x]/ x4 + x + 1 .
Comment: This question could be quite confusing. The expression x4 +
x + 1 is first
of all an element of Z2 [x]. It is used to construct a field F = Z2 [x]/ x4 + x + 1 .
Then it is regarded as an element of F [x], and has to be factored in F [x]. To try to
clarify the situation, we will use α to denote the congruence class of x in F .
Some of the calculations may also look confusing, because we will use the following
shortcut. Since we are dealing with cosets of polynomials whose coefficients come
from Z2 , for a, b ∈ F we have 2ab = 0, and therefore (a + b)2 = a2 + b2 .
Solution: The answer to Exercise 4.2.12 (given in the text) states that p(x) = x4 +x+1
is irreducible over Z2 . For the sake of completeness, we should still write out a proof
here. It is easy to see that p(x) has no roots in Z2 , so it has no linear factors in Z2 [x].
Can it be the product of two quadratic factors? The only irreducible quadratic in
Z2 [x] is x2 + x + 1, since the other three quadratics all have roots. (Here is the list:
4
2
2
x2 , x2 + 1, x2 + x.) Since (x2 + x + 1)2 = x4 + x2 +
1,4 we havex + x + 1 6= (x + x + 1) ,
4
so x + x + 1 is irreducible over Z2 , and Z2 [x]/ x + x + 1 is a field.
In F = Z2 [x]/ x4 + x + 1 , let [x] = α. Then F has 24 = 16 elements, which have
the general form a + bα + cα2 + dα3 , where a, b, c, d ∈ Z2 . We will show that there
4.3
J.A.Beachy
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are 4 roots of x4 + x + 1 in F , given by the cosets corresponding to α, α2 , α + 1, and
α2 + 1. Remember that x4 + x + 1 ≡ 0, so x4 ≡ x + 1.
To check that the above cosets are roots, We have
α4 + α + 1 = [x4 + x + 1] ≡ 0,
(α2 )4 + (α2 ) + 1 = [x4 ]2 + [x2 ] + [1] ≡ [x + 1]2 + [x2 + 1] ≡ [x2 + 1 + x2 + 1] ≡ 0,
(α + 1)4 + (α + 1) + 1 ≡ [x + 1]4 + α ≡ [x4 + 1 + x] ≡ 0, and
(α2 + 1)4 + (α2 + 1) + 1 ≡ (α4 )2 + 1 + α2 ≡ [x + 1]2 + [1 + x2 ] ≡ [x2 + 1 + 1 + x2 ] ≡ 0.
Thus x4 + x + 1 factors as a product of 4 linear terms. Remembering that over Z2 we
have −1 ≡ 1, we get x4 + x + 1 = (x + α)(x + α2 )(x + α + 1)(x + α2 + 1).
ANSWERS AND HINTS
33. Factor f (x) = x5 − x4 + x3 − x2 and g(x) = x3 − x2 + x − 1 over Z3 and use the
factorizations to find their greatest common divisor.
Hint: See Problem 4.2.33. The gcd is (x − 1)(x2 + 1).
34. Factor f (x) = x5 + x4 − 2x3 − x2 + 2x − 2 and g(x) = x3 − x2 + x − 1 over Z5 and use
the factorizations to find their greatest common divisor.
Hint: See Problem 4.2.34. The gcd is (x + 2)(x − 2).
35. Factor f (x) = x5 + 3x4 − 2x3 − 3x − 3 and g(x) = x3 − x2 + x − 1 over Z7 and use the
factorizations to find their greatest common divisor.
Hint: See Problem 4.2.35. The gcd is x2 + 1.
37. Find a generator for the multiplicative group of the field Z2 [x]/ x4 + x + 1 . As in
the solution to Problem 32, let [x] = α.
Hint: A nontrivial element that does not have order 3 or 5 must have order 15.
38. Find the multiplicative inverses of [1 + x] and [1 − x] in Z3 [x]/ x2 + x + 2 .
Answer: From the multiplication table in Exercise 4.3.18, we have [1 + x]−1 = [x] and
[1 − x]−1 = −[1 − x].
1
√
√ .
40. Use a calculation in Q[x]/ x3 − 2 to rationalize the denominator of
3
1− 2+ 3 4
Hint: Find the multiplicative inverse of 1 − x + x2 in Q[x]/ x3 − 2 .
√
−1 − 3 2
Answer:
3