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.
(1) Let a0xn + a1xn-1 + ... + an-1x + an be an n degree polynomial.
(2) By the Fundamental Theorem of Algebra (see Theorem, here), we know that there are n roots such that:
a0xn + a1xn-1 + ... + an-1x + an = (x - r1)*(x - r2)*...*(x - rn)
(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.
Lemma 2: Abel's Lemma
The equation xp = 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
(1) Assume that xp = C is reducible into rational functions.
(2) Then, there exists f(x),g(x) such that xp - C = f(x)g(x) where f(x),g(x) are rational functions of lower degree.
(3) We know that the roots to xp - C=0 are r, rα, rα2, ... rαp-1 where r is one of the roots and α is a pth root of unity since:
(rαi)p = (rp)(αi)p = (rp)(αp)i = rp*(1)i = rp
(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)*(rα2)*...*(rαp-1) = ± C
(6) We can see that C = rp since:
(a) If p=2, then α=-1 and the roots are r*(-r) = -r2
(b) If p is a prime ≥ 3, then using the summation formula (see Corollary 2.1, here), we have:
C = rpα[(1/2)(p)(p-1)]
(c) Since p-1 is even, we have:
C = rp(αp)(1/2)(p-1) = rp(1)(1/2)(p-1) = rp
(7) Likewise, there exists μ, M, ν, N such that:
A = rμαM
B = rναN
(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 = Ah*Bk
(12) So that K= Ah*Bk = rhμαM*rkναN = r(hμ + kν)αhM+kN = rαhM+kN
(13) But then Kp = (rαhM+kN)p = rp*(αp)hM+kN = rp
(14) But this is impossible since we selected an integer C that is not a p-th power.
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
All the roots of f(x) are roots of F(x) and F(x) can be divided by f(x) without a remainder.
(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) = g0 where g0 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, f1(x) must equal a constant and f(x)=g(x)*f1(x) = g(x)*f10.
(5) Then, F(x)=F1(x)*g(x) = F1(x)*f(x)*f10
(6) Thus, F(x) is divisible by f(x) and vanishes for every zero point of f(x).
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.
(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.
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.
(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).
- 100 Great Problems of Elementary Mathematics (Dover, 1965)