Lemma 1:
Let η be a primitive nth root of unity.
Let f(x) be an nth degree polynomial that irreducible over Q where n is an odd prime number and Q is the set of rational numbers extended with η. That is: Q(η).
Let K be a rational number such that λ = K(1/l) and λ is not a rational number and l is a prime and λ is a real number
Let f(x) be reducible over Q(λ) such that:
f(x) = g(x,λ)*g(x,λη)*...*g(x,ληn-1)
Then:
Then the roots of f(x) have the following form:
ω0 = a0 + a1λ0 + ... + an-1λ0n-1
ω1 = a0 + a1λ1 + ... + an-1λ1n-1
...
ωn-1 = a0 + a1λn-1 + ... + an-1λn-1n-1
where ai ∈ Q and λi = ληi
Proof:
(1) Since n is odd, we know that f(x) has at least one real root. [see Theorem 3, here]
(2) Let us use λi to denote the different parameters of g(x,y) so that we have:
λ0 = λη0 = λ
λ1 = λη1 = λη
...
λn-1 = ληn-1
(3) Let:
g(x,λi) =a0 + a1λi + ... + an-1λin-1
where ai ∈ Q
(4) From f(x) = g(x,λ)*g(x,λη)*...*g(x,ληn-1), we can see that f(x) has n roots:
ω0 = a0 + a1λ0 + ... + an-1λ0n-1
ω1 = a0 + a1λ1 + ... + an-1λ1n-1
...
ωn-1 = a0 + a1λn-1 + ... + an-1λn-1n-1
QED
Lemma 2:
Let η be a primitive nth root of unity.
Let f(x) be an nth degree polynomial that irreducible over Q where n is an odd prime number and Q is the set of rational numbers extended with η. That is: Q(η).
Let K be a rational number such that λ = K(1/l) and λ is not a rational number and l is a prime and λ is a real number
Let f(x) be reducible over Q(λ) such that:
f(x) = g(x,λ)*g(x,λη)*...*g(x,ληn-1)
Then:
All the roots of f(x) but one are complex and not real.
Proof:
(1) Using Lemma 1 above, we know that f(x) has the following roots:
ω0 = a0 + a1λ0 + ... + an-1λ0n-1
ω1 = a0 + a1λ1 + ... + an-1λ1n-1
...
ωn-1 = a0 + a1λn-1 + ... + an-1λn-1n-1
where ai ∈ Q and λi = ληi
(2) Since n is odd, we know that f(x) has at least one real root. [see Theorem 3, here]
(3) Let ω be the real root so that we have:
ω = a0 + a1λ + ... + an-1λn-1
where ai ∈ Q
(4) We assume that ω = ω0. (We can make the same argument regardless of which ωi is real.)
(5) Since ω is real, we know that ω = ω [see Theorem 1, here] which means that:
a0 + a1λ + ... + an-1λn-1 =a0 + a1λ + ... + an-1λn-1
so that:
(a0 - a0) + ... + (an-1 - an-1) = 0
(6) So, that for all i, ai = ai
(7) This means that all ai are real numbers.
(8) Since λ is real, it follows that ληi is not real because i ≠ 0.
(9) λv and λ-v are complex conjugates since (see Theorem 2, here):
λv = λ*ηv = λη-v = λ-v
(10) So it follows that we have the following (n-1)/2 complex conjugate pairs:
ωn-1 and ω1
ωn-2 and ω2
...
(11) This shows that all but one of the roots of f(x) are not real.
QED
Lemma 3:
Let η be a primitive nth root of unity.
Let f(x) be an nth degree polynomial that irreducible over Q where n is an odd prime number and Q is the set of rational numbers extended with η. That is: Q(η).
Let K be a rational number such that λ = K(1/l) and λ is not a rational number and l is a prime and λ is not a real number and f(x) is not reducible over norm(λ)
Let f(x) be reducible over Q(λ) such that:
f(x) = g(x,λ)*g(x,λη)*...*g(x,ληn-1)
Then:
All the roots of f(x) are real.
Proof:
(1) Using Lemma 1 above, we know that f(x) has the following roots:
ω0 = a0 + a1λ0 + ... + an-1λ0n-1
ω1 = a0 + a1λ1 + ... + an-1λ1n-1
...
ωn-1 = a0 + a1λn-1 + ... + an-1λn-1n-1
where ai ∈ Q and λi = ληi
(2) Since n is odd, we know that f(x) has at least one real root. [see Theorem 3, here]
(3) Let ω be the real root so that we have:
ω = a0 + a1λ + ... + an-1λn-1
where ai ∈ Q
(4) We assume that ω = ω0. (We can make the same argument regardless of which ωi is real.)
(5) Since ω is real, we know that ω = ω [see Thereom 1, here] which means that:
a0 + a1λ + ... + an-1λn-1 =a0 + a1λ + ... + an-1λn-1
(6) Let Λ = the norm(λ) so that Λ = λ*λ which then is a real number. [see Lemma 1, here]
(7) So that we have:
a0 + a1λ + ... + an-1λn-1 =a0 + a1(Λ/λ) + ... + an-1(Λ/λ)n-1
(8) Now we can define an equation h(x) such that:
h(x) = a0 + a1x + ... + an-1xn-1 - a0 + a1(Λ/x) + ... + an-1(Λ/x)n-1
(9) From step #7 above, it is clear that λ is a root of this equation.
(10) But λ is also a root of xn = K which is irreducible over Q(Λ) [see Lemma 1, here]
(11) So, by Abel's Lemma [see Theorem 3, here], all the roots of xn = K are also solutions to the equation in step #8.
(12) Now, for every λv, we have:
Λ/λv = Λ/(ληv) = λ/(ηv) = ληn = λv
(13) Thus, for all λv, we have [from step #11 above]:
h(x) = a0 + a1x + ... + an-1xn-1 - a0 + a1(Λ/x) + ... + an-1(Λ/x)n-1
which means:
a0 + a1λv + ... + an-1λvn-1 =a0 + a1(Λ/λv) + ... + an-1(Λ/λv)n-1
which means:
a0 + a1λv + ... + an-1λvn-1 =a0 + a1λv + ... + an-1λvn-1
so that for all ωv:
ωv = ωv
(14) But this can only be true if all the roots are real [see Theorem 1, here]
QED
References
- 100 Great Problems of Elementary Mathematics (Dover, 1965)
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