Thursday, September 04, 2008

Abel's Lemmas on Irreducibility

In today's blog, I present a few lemmas that Niels Abel published on irreducibility. These lemmas are used in Abel's proof on the insolvability of the quintic equation.

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 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.

QED

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

Proof:

(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 = rpp)(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 = rαM*rαN = 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.

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) = 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).

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

2 comments:

david.foley said...

In Abel's lemma step 6 you use an a, b, and c step to show C = r^p. From step 3 we know that r is by definition a solution of x^p = C so that r^p = C. It was brought into existence to have the very property that when raised to the pth power it will equal C.

david.foley said...

In Abel's irreducibility theorem step 3 it is asserted that there exists a common divisor of at least the first order but is that so obvious without some explanation. One way to show this is to assume there is no common divisor and then inspect the equation of step 1 evaluated at r. The RHS is g0 while the LHS is 0. Since we know that g0 is nonzero our assumption must be wrong and a common divisor must exist.