The content in today's blog is taken from 100 Great Problems of Elementary Mathematics by Heinrich Dorrie.

Definition 1: expressible in rationals

"A number or equation is expressible in rationals if it is expressible using only addition, subtraction, multiplication, and division of integers."

Definition 2: rational functions

A function f(x) is rational if its coefficients are all rational numbers.

Definition 3: degree of a polynomial

The degree of a polynomial is the highest power of the polynomial that has nonzero coefficients.

Definition 4: reducible over rationals

A function f(x) with coefficients in the rational numbers is said to be reducible over rationals if it can be divided into a product of polynomials with lower degree and rational coefficients.

Definition 5: free term or constant term of a polynomial

The free term or constant term of a polynomial is the value that is not bound to an unknown.

Lemma 1: The free term is equal to the product of roots.

Proof:

(1) Let a

_{0}x

^{n}+ a

_{1}x

^{n-1}+ ... + a

_{n-1}x + a

_{n}be an n degree polynomial.

(2) By the Fundamental Theorem of Algebra (see Theorem, here), we know that there are n roots such that:

a

_{0}x

^{n}+ a

_{1}x

^{n-1}+ ... + a

_{n-1}x + a

_{n}= (x - r

_{1})*(x - r

_{2})*...*(x - r

_{n})

(3) Now, it is clear that of the products in step #2, the only term that does not include x is the product of all the roots.

QED

Lemma 2: Abel's Lemma

The equation x

^{p}= C where p is a prime number is irreducible over rationals when C is a rational number not the pth power of a rational number

Proof:

(1) Assume that x

^{p}= C is reducible into rational functions.

(2) Then, there exists f(x),g(x) such that x

^{p}- C = f(x)g(x) where f(x),g(x) are rational functions of lower degree.

(3) We know that the roots to x

^{p}- C=0 are r, rα, rα

^{2}, ... rα

^{p-1}where r is one of the roots and α is a pth root of unity since:

(rα

^{i})

^{p}= (r

^{p})(α

^{i})

^{p}= (r

^{p})(α

^{p})

^{i}= r

^{p}*(1)

^{i}= r

^{p}

(4) Let A,B be the the free terms for f(x) and g(x) respectively.

(5) Since a free term is the product of a function's root (see Lemma 1 above), we know that A*B = (r)*(rα)*(rα

^{2})*...*(rα

^{p-1}) = ± C

(6) We can see that C = r

^{p}since:

(a) If p=2, then α=-1 and the roots are r*(-r) = -r

^{2}

(b) If p is a prime ≥ 3, then using the summation formula (see Corollary 2.1, here), we have:

C = r

^{p}α

^{[(1/2)(p)(p-1)]}

(c) Since p-1 is even, we have:

C = r

^{p}(α

^{p})

^{(1/2)(p-1)}= r

^{p}(1)

^{(1/2)(p-1)}= r

^{p}

(7) Likewise, there exists μ, M, ν, N such that:

A = r

^{μ}α

^{M}

B = r

^{ν}α

^{N}

(8) Since there are p instances of r in the product in step #5, we know that:

μ + ν = p

(9) We further know that gcd(μ,ν) = 1 since:

(a) Assume that gcd(μ,ν) = f which is greater than 1.

(b) So μ = mf and ν = nf

(c) Then p = mf + nf = f(m+n)

(d) But p is prime so this is impossible and f=1.

(10) Using Bezout's Identity (see Lemma 1, here), we know that there exists h,k such that:

μh + νk = 1

(11) Now let's define a rational number K as K = A

^{h}*B

^{k}

(12) So that K= A

^{h}*B

^{k}= r

^{hμ}α

^{M}*r

^{kν}α

^{N}= r

^{(hμ + kν)}α

^{hM+kN}= rα

^{hM+kN}

(13) But then K

^{p}= (rα

^{hM+kN})

^{p}= r

^{p}*(α

^{p})

^{hM+kN}= r

^{p}

(14) But this is impossible since we selected an integer C that is not a p-th power.

QED

Theorem 3: Abel's Irreducibility Theorem

Let f(x) be irreducible over rationals.

If one root of the equation f(x) is also a root of the rational equation F(x)=0

Then:

All the roots of f(x) are roots of F(x) and F(x) can be divided by f(x) without a remainder.

Proof:

(1) Using Euclid's Greatest Common Divisor Algorithm for Polynomials (see Theorem 3, here), we are left with:

V(x)F(x) + v(x)f(x) = g(x)

(2) If F(x) and f(x) have no common divisor, then g(x) is a constant. That is, g(x) = g

_{0}where g

_{0}is the free term.

(3) If f(x) is irreducible and a root r of f = 0 is also a root of F(x), then there exists a common divisor of at least the first degree (x-r)

(4) Since f(x) is irreducible, f

_{1}(x) must equal a constant and f(x)=g(x)*f

_{1}(x) = g(x)*f1

_{0}.

(5) Then, F(x)=F

_{1}(x)*g(x) = F

_{1}(x)*f(x)*f1

_{0}

(6) Thus, F(x) is divisible by f(x) and vanishes for every zero point of f(x).

QED

Corollary 3.1:

If a root of an equation f(x)=0 which is irreducible in rational numbers is also a root of F(x)=0 in rational numbers of lower degree than f, then all the coefficients of F are equal to zero.

Proof:

(1) Assume that at least one coefficient of F is not zero.

(2) Then F is a polynomial with a degree of at least 1.

(3) Since there is a root that divides both f(x) and F(x) and since f(x) is irreducible over rationals, we can use Theorem 3 above to conclude that every root of f(x) divides F(x).

(4) But F(x) has a lower degree than f(x) which is impossible.

(5) So we reject our assumption at step #1 and conclude that all coefficients of F must be 0.

QED

Corollary 3.2:

If f(x)=0 is an irreducible over rationals, then there is no other equation irreducible over rationals that has a common root with f(x)=0.

Proof:

(1) Let f(x) and g(x) be functions irreducible over rationals.

(2) Assume that f(x) and g(x) have a common root r.

(3) Then we can use Theorem 3 above to conclude that f(x) divides g(x) and g(x) divides f(x).

(4) But if f(x) divides g(x) and g(x) divides f(x), it is clear that f(x) = g(x).

QED

References

- 100 Great Problems of Elementary Mathematics (Dover, 1965)