## Thursday, September 24, 2009

### Kronecker's Theorem: Lemmas on Complex Conjugates

The content in today's blog is taken directly from David Antin's translation of Heinrich Dorrie's 100 Great Problems of Elementary Mathematics.

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