Saturday, June 24, 2006

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.


(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 αλ = 1 [See here for details on primitive root of unity]

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

zλ = xλ + yλ = (x + y)(x + αy)(x + α2y)*...*(x + αλ-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 + αi+ky.

(b) Now, we note that:

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

(x + αi+ky) - αk(x + αiy) = x - αkx = (αk - 1)*(-1)*x

(c) So, if p divides x + αiy and x + αi+ky, then p divides both αik - 1)y and 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 * αλ-i = 1 and since -1 is a unit.

This gives us that p divides k - 1)y and 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 αk - 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 + αi-1y) since:

(x + αiy) - (x + αi-1y) = αi-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 zλ 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 = xλ and -(-z)λ = zλ.

(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 + yλ where λ divides z'λ

(e) Finally, if λ divides zλ, 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+xjy) 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]



Anonymous said...

I have greatly enjoyed reading various entries in your blog since I happened upon it a few weeks ago (especially your thorough debunking of that Escultura guy). I have heard about something called Wieferich's Criterion, which was proven with Kummer's Criteria: If p is a prime exponent in a counterexample to the first case of FLT, then 2^(p-1) = 1 mod p^2. There are a number of variations as well, with 3, 5, etc. replacing 2. If you could go through the proofs of these results, I would greatly appreciate it - I have been unable to find any explanation of these beautiful theorems so far.

Larry Freeman said...

Hi Zev,

Thanks for your comments! I am glad to hear that you are enjoying my blog.

I have not yet reviewed Wieferich's Criterion but I am glad to look into it when my blog gets to twentieth century developments regarding Fermat's Last Theorem.

Perhaps, I will cover it after I talk about Dedekind.

Right now, I am focusing on the history of group theory from Cardano to Galois.



dkc said...

I've just started reading your articles and find them very helpful for the amateur. A "simple" way to derive Wieferich's criterion is to use the "pth power with respect to" concept. This is explained in detail at my website, "". I don't have a proof; this is just empirically derived.