Fermat's Last Theorem: Kummer's Proof for Regular Primes

Today's blog is on the proof first presented by Ernst Kummer for regular primes. Before Andrew Wiles, this was the highpoint of the work done on Fermat's Last Theorem.

Kummer himself claimed that he worked on the proof not out of an interest in Fermat's Last Theorem but in generalizing the ideas implicit in quadratic reciprocity (see H. M. Edwards' book). Still, this proof would later serve as an important foundation to the work done by Richard Dedekind in establishing algebraic number theory.

Implicit in the concept of regular primes (see here) is the concept of class number (see here), ideal numbers (see here), and cyclotomic integers (see here). Through out the proof, I have included links to help people review the necessary concepts.

The details of the proof are based on the work done by H. M. Edwards in his book: Fermat's Last Theorem: A Genetic Introduction to Algebraic Number Theory.

Theorem: if x,y,z,n are integers and n is a regular prime, then xⁿ + yⁿ = zⁿ → xyz = 0.

Proof:

(1) From the given, we know that n is a regular prime. [See Definition 1, here]

(2) We can assume that x,y,z are relatively prime. [See here for details]

(3) Let λ = n. [We use λ to represent n just as Kummer did in his original paper, see Definition 1, here for details]

(4) Let α be a primitive root of unity such that αⁱ ≠ 1 when i is less than λ and α^λ = 1 [See here for details on primitive root of unity]

(5) Then, using α, we can refactor xⁿ + yⁿ to get:

z^λ = x^λ + y^λ = (x + y)(x + αy)(x + α²y)*...*(x + α^λ-1y). [See Lemma 1 here for details on this refactoring]

(6) We can assume that all elements (x + αⁱy) in step #4 are either relatively prime or have only α - 1 as a common factor since:

(a) Assume that there exists some prime p that divides both x + αⁱy and x + α^i+ky.

(b) Now, we note that:

(x + α^i+ky) - (x + αⁱy) = α^i+ky - αⁱy = αⁱ(α^k - 1)y

(x + α^i+ky) - α^k(x + αⁱy) = x - α^kx = (α^k - 1)*(-1)*x

(c) So, if p divides x + αⁱy and x + α^i+ky, then p divides both αⁱ(α^k - 1)y and (α^k - 1)*(-1)*x

This is true since if a factor a divides b and c, then a divides (b - c) and a also divides (b-dc).

(d) Now, we note that αⁱ is a unit since αⁱ * α^λ-i = 1 and since -1 is a unit.

This gives us that p divides (α^k - 1)y and (α^k - 1)*x [Since only a unit divides a unit, see here for details on cyclotomic units]

(e) Now, we know that p cannot divide both x,y (since x,y are relatively prime) so p must divide α^k - 1 which means that p = α - 1. [See Lemma 3, here]

(7) If α - 1 divides any of the element of the form (x + αⁱy), then it divides all the other factors of this form since:

(a) if (α - 1) divides (x + αⁱy), then (α-1) divides (x + αⁱ⁺¹y) since:

(x + αⁱ⁺¹y) - (x + αⁱy) = αⁱ⁺¹y - αⁱy = αⁱ(α - 1)y.

In other words, if a factor a divides c and d and b + c = d, then a must divide b.

(b) if (α - 1) divides (x + αⁱy), then (α-1) divides (x + α^i-1y) since:

(x + αⁱy) - (x + α^i-1y) = α^i-1(α - 1)y.

(8) Since there are λ - 1 factors of the form (x + αⁱy), then (α - 1)^λ-1 = λ*unit (see Corollary 3.2, here) gives us that λ divides z^λ which using Euclid's Generalized Lemma (see here) gives us that λ divides z.

(9) Even if λ divides x or y, we can assume that it divides z.

(a) Let's assume that λ divide x. [if λ divides y, we can use the same argument substituting y for x]

(b) Since λ is odd, we know that -(-x)^λ = x^λ and -(-z)^λ = z^λ.

(c) This gives us:

(-x)^λ = yλ + (-z)^λ

(d) Then, if we label (-x)^λ as z' and (-z)^λ as x', then we have shown that if λ divides x,y,or z, we can assume an equation of the form z'^λ = x'^λ + y^λ where λ divides z'^λ

(e) Finally, if λ divides z^λ, then by Euclid's Generalized Lemma, it divides z.

(10) Step #9 is useful because it means that we only have to consider two cases:

Case I: x,y,z,λ are coprime and therefore each factor (a+x^jy) are relatively prime to each other.

Case II: λ divides z.

(11) We can now conclude that Fermat's Last Theorem holds for regular primes since:

(a) Fermat's Last Theorem holds for Case I. [See Theorem, here]

(b) Fermat's Last Theorem holds for Case II. [See Theorem, here]

QED