Download Homework Chapter 20 Proof Solutions 1. Describe the elements in Q

MATH 403: Homework Chapter 20 Proof Solutions
1. Describe the elements in Q
√
3
−2 . Justify carefully using an appropriate theorem.
Solution: Since x3 + 2 is irreducible over Q our Theorem states that
√ Q 3 −2 ≈ Q[x]/ x3 + 2
and the elements are
2. Prove that Q
√
√
a + b 3 −2 + c( 3 −2)2 | a, b, c ∈ Q
√ √ √ √
2 + 3 = Q 2, 3 .
√
√
Proof: Since
√ √of
√ our
√ theorem tell us that elements in Q( 2 + 3) look like linear combinations
powers of 2+ 3 (up to one less than√
the √
degree of whichever√
irreducible
polynomial
2+ 3
√
√
2
2,
3)
look
like
a
+
b
2
+
c
3
+
d
6
(using
x
−2
is a root of) and that elements in Q(
√
√
√ √
√ √
√
√
2
and x − 3) it suffices to show that 2 + 3 ∈ Q( 2, 3) and 2, 3 ∈ Q( 2 + 3) because
each of the right sides is by definition a field and will then contain all linear combinations of
all powers and all products.
√
√
√ √
• Clearly 2 + 3 ∈ Q( 2, 3).
√
√
√
√
1√
• Since Q( 2 + 3). is a field it contains √2+
= 3 − 2 and hence it contains
3
√
√
√
√
√
√
√
√
√
√
1
2) + 12 ( 2 + 3) = 3 and − 21 ( 3 − 2) + 21 ( 2 + 3) = 2.
2( 3 −
3. Consider the polynomial f (x) = x2 + x + 1 ∈ Z5 [x].
(a) Prove that f (x) is irreducible over Z5 .
Proof: Since there are no roots (try x = 0, 1, 2, 3, 4) we know f (x) is irreducible over Z5
by the degree 2/3 test.
(b) It has a root in the field extension Z5 [x]/ x2 + x + 1 .
(c) In which field extension does it have a zero?
Answer: It has a root in the field extension Z5 [x]/ x2 + x + 1 .
(d) Factor it in this field extension.
Solution: For convenience write <> in place of x2 + x + 1 and write the polynomial
as z 2 + z + (1+ <>) because we’re thinking of it as a polynomial with coefficients in the
field extension; really it’s a polynomial in (Z3 [x]/ <>) [x]. Then since x+ <> is a root
we know that z − (x+ <>) is a factor. Long division yields the other:
z +(x + 1+ <>)
z − (x+ <>) z 2
+z
+(1+ <>)
z2
−(x+ <>)z
(x + 1+ <>)z
+(1+ <>)
(x + 1+ <>)z −(x2 + x+ <>)
0+ <>
Thus z 2 + z + (1+ <>) = [z − (x+ <>)] · [z + (x + 1+ <>)].
6. Let a, b ∈ R with b 6= 0. Prove that R (a + bi) = C.
Proof: First note that since a + bi is a root of x2 − 2ax + a2 + b2 which is irreducible (over R,
since there are no real roots) we know the elements in R(a + bi) have the form A + B(a + bi)
with A, B ∈ R.
Clearly R(a + bi) ⊆ C since everything of the form A + B(a + bi) with A, B ∈ R is a complex
number.
Now let α + βi ∈ C. We wish to show that α + βi = A + B(a + bi) for some A, B ∈ R. If
we equate real and imaginary parts and solve we get B = β/b and A = α − (β/b)a. Thus
C ⊆ R(a + bi).
√ 7. Define an isomorphism φ : Q 3 7 → Q [x] / x3 − 7 and prove it is an isomorphism.
Proof: Define
√
3
φ : Q( 7) → Q[x]/ x3 − 7
by
φ a + b(7)1/3 + c(7)2/3 = a + bx + cx2 + x3 − 7
Clearly φ is 1-1 and onto by its very construction.
For addition
φ a + b(7)1/3 + c(7)2/3 + d + e(7)1/3 + f (7)2/3
= φ (a + d) + (b + e)(7)1/3 + (c + f )(7)2/3
= (a + d) + (b + e)x + (c + f )x2 + x3 − 7
and
φ a + b(7)1/3 + c(7)2/3 + d + e(7)1/3 + f (7)2/3
= a + bx + cx2 + x3 − 7 + d + ex + f x2 + x3 − 7
= (a + d) + (b + e)x + (c + f )x2 + x3 − 7
For multiplication
φ a + b(7)1/3 + c(7)2/3 d + e(7)1/3 + f (7)2/3
= φ (ad + 7bf + 7ce) + (ae + bd + 7cf )71/3 + (af + be + cd)72/3
= (ad + 7bf + 7ce) + (ae + bd + 7cf )x + (af + be + cd)x2 + x3 − 7
φ a + b(7)1/3 + c(7)2/3 φ d + e(7)1/3 + f (7)2/3
and
a + bx + cx2 + x3 − 7
d + ex + f x2 + x3 − 7
= ad + aex + af x2 + bdx + bex2 + bf x3 + cdx2 + cex3 + cf x4 + x3 − 7
= (ad + 7bf + 7ce) + (ae + bd + 7cf )x + (af + be + cd)x2 + x3 − 7