CHAPTER 5 Fermat’s Last Theorem for Regular Primes

1. The theorem

Let p be an odd prime and let K = ℚ(ζ_{p}). We will write ζ for ζ_{p} for this section. It was observed early in the 19^{th} century that this field is intimately connected with Fermat’s last theorem. Specifically, if one has an equality

x^{p} + y^{p} = z^{p}

with x, y, z ∈ ℤ, one can use the factorization

x^{p} + y^{p} = (x + y)(x + ζy)(x + ζ^{2}y) ⋯ (x + ζ^{p−1}y)

to conclude that

(x + y)(x + ζy)(x + ζ^{2}y) ⋯ (x + ζ^{p−1}y) = z^{p}.

From here, one shows (with appropriate conditions on x, y, z) that the factors on the left side are pairwise relatively prime. If O_{K} is a UFD, it follows that each x + ζ^{i}y is a p^{th} power in O_{K}, since their product is. From here one can easily obtain a contradiction, which shows that Fermat’s equation has no non-trivial solution in this case.

This argument was first successfully carried out by Kummer in the mid 19^{th} century. He realized that his proof applied to not only those p for which ℤ[ζ_{p}] is a UFD, but also to a much larger class of primes. The key property turned out to be that p not divide the class number h_{ℚ(ζ_{p})}. Kummer called such primes regular; if a prime is not regular, then it is said to be irregular.

We will prove Kummer’s theorem with the additional simplifying hypothesis that p not divide xyz; this is classically referred to as Case I. Case I contains most of the interesting content of the general case and has the advantage of being far simpler technically.

Theorem 1.1 (Kummer). Let p ≥ 5 be a regular prime. Then the equation

x^{p} + y^{p} = z^{p}

has no solutions with x, y, z ∈ ℤ and p not dividing xyz.

Proof. To begin, note that by Exercise 5.1 we can assume that x and y are not congruent modulo p.

Let K = ℚ(ζ_{p}). Suppose that there is a solution x^{p} + y^{p} = z^{p}. As above we write

(x + y)(x + ζy) ⋯ (x + ζ^{p−1}y) = z^{p}.

We first show that the principal ideals (x + ζ^{i}y) and (x + ζ^{j}y) have no common factors for i ≠ j.

Lemma 1.2. Suppose x^{p} + y^{p} = z^{p} and p does not divide xyz. Then the ideals (x + ζ^{i}y) are pairwise relatively prime for i = 0, …, p − 1.

Proof. Let i and j be distinct integers between 0 and p − 1 and suppose that there is some prime ideal 𝔮 of O_{K} which divides both (x + ζ^{i}y) and (x + ζ^{j}y). 𝔮 therefore also divides the principal ideals

((x + ζ^{i}y) − (x + ζ^{j}y)) = ((ζ^{i} − ζ^{j})y)

and

((x + ζ^{i}y) − ζ^{i−j}(x + ζ^{j}y)) = ((1 − ζ^{i−j})x).

(See Exercise 5.2. Note that ζ^{i−j}(x + ζ^{j}y) generates the same ideal as x + ζ^{j}y since ζ^{i−j} is a unit.) Recall that since i ≠ j, ζ^{i} − ζ^{j} = ζ^{i}(1 − ζ^{j−i}) and 1 − ζ^{i−j} are both associate to 1 − ζ. We conclude that 𝔮 divides the ideals (1 − ζ)(x) and (1 − ζ)(y). However, since x and y are relatively prime in ℤ it follows that they can have no prime ideal factors in common in O_{K}; therefore, the only possibility is 𝔮 = (1 − ζ).

Suppose, then, that (1 − ζ) divides (x + ζ^{i}y) and (x + ζ^{j}y) as ideals. This implies immediately that 1−ζ divides x+ζ^{i}y and x+ζ^{j}y as elements of O_{K}. Thus

x + ζ^{i}y ≡ 0 (mod 1 − ζ).

We also have ζ^{i} ≡ 1 (mod 1 − ζ), so we conclude that

x + y ≡ 0 (mod 1 − ζ).

However, x + y is a rational integer, so if it is divisible by 1 − ζ, then it must be divisible by p. (See Lemma II.4.1.)

We have now that p divides x + y in ℤ. Since

x^{p} + y^{p} ≡ x + y (mod p),

it follows that p divides x^{p} + y^{p}, and therefore that p divides z. This contradicts our assumption that p does not divide xyz (or our assumption that x and y are relatively prime), so we conclude that (x + ζ^{i}) and (x + ζ^{j}y) are relatively prime ideals, as claimed.

Let

(z) = 𝔮_{1}^{n_{1}} ⋯ 𝔮_{r}^{n_{r}}

be the ideal factorization of (z) in O_{K}. The equality of ideals

(x + y)(x + ζy) ⋯ (x + ζ^{p−1}y) = (z)^{p}.

shows that

(x + y)(x + ζy) ⋯ (x + ζ^{p−1}y) = 𝔮_{1}^{pn_{1}} ⋯ 𝔮_{r}^{pn_{r}}.

Since the ideals (x + ζ^{i}y) are pairwise relatively prime, each 𝔮_{i} must occur in the factorization of exactly one of them. As each 𝔮_{i} occurs with multiplicity divisible by p, it follows that every prime factor of each (x + ζ^{i}y) occurs with multiplicity divisible by p. Put differently, each (x + ζ^{i}y) is the p^{th} power of some ideal 𝔞_{i} of O_{K}:

(x + ζ^{i}y) = 𝔞_{i}^{p}.

We now use the hypothesis that p is regular to conclude that the 𝔞_{i} are all principal. Specifically, note that 𝔞_{i}^{p} is trivial in C_{K}, since it is just the principal ideal (x + ζ^{i}y). Since p does not divide the order of C_{K}, this implies that 𝔞_{i} itself must be trivial in C_{K} (since if C_{K} had an element of order p then it would have order divisible by p), and thus principal. Therefore we can write 𝔞_{i} = (α_{i}) for some α_{i} ∈ O_{K}, and we have the equality of principal ideals.

(x + ζ^{i}y) = (α_{i})^{p}.

This implies that

x + ζ^{i}y = uα_{i}^{p}

for some u ∈ O_{K}^{∗}. The next step is to get a little more information on the unit u.

Lemma 1.3. Let u be a unit of O_{K}. Then u can be written as ζ^{a}ε with ε a unit of the maximal real subfield of K.

Proof. By Exercise II.2.17 we know that u/ū = ζ^{b} for some b, where ū is the complex conjugate of u. Now choose a ∈ ℤ such that 2a ≡ b (mod p) and set ε = ζ^{−a}u. Then u = ζ^{a}ε, and

ε̄ = ζ^{a}ū = ζ^{a}ζ^{−b}u = ζ^{−a}u = ε,

so ε is real and thus lies in the maximal real subfield of K.

We now take i = 1; by our results to this point we can write

x + ζy = ζ^{a}εα^{p}

for some integer a, some real unit ε and some α = α_{1} ∈ O_{K}. By Exercise 5.3 we have that α^{p} ≡ b (mod p) for some rational integer b, so we conclude that

x + ζy ≡ ζ^{a}εb (mod p).

Since ε, b and p are all real, taking complex conjugates yields

\overline{x + ζy} ≡ ζ^{−a}εb (mod p).

As \overline{x + ζy} = x + ζ^{−1}y, we find that

x + ζ^{−1}y ≡ ζ^{−a}εb (mod p).

Combining these equations we conclude that

ζ^{−a}(x + ζy) ≡ ζ^{a}(x + ζ^{−1}y) (mod p)

which simplifies to

x + ζy − ζ^{2a−1}y − ζ^{2a}x ≡ 0 (mod p).

We can use this congruence to obtain our desired contradiction. Suppose first that none of the p^{th} roots of unity 1, ζ, ζ^{2a−1} and ζ^{2a} are equal. Since p ≥ 5 this implies that these elements are part of an integral basis of O_{K}. Now the fact that

x + ζy − ζ^{2a−1}y − ζ^{2a}x

is divisible by p in O_{K} implies that x and y must be divisible by p in ℤ; this contradicts our assumption that p does not divide xyz, which finishes this case.

This leaves the cases where some of 1, ζ, ζ^{2a−1}, ζ^{2a} are equal. The possibilities are:

1. 1 = ζ^{2a−1}. Then ζ = ζ^{2a}, so we find that

(x − y) + (y − x)ζ ≡ 0 (mod p).

This p divides (x−y)(1−ζ). As we assumed that x and y were not congruent modulo p, x − y is relatively prime to p; since also p does not divide 1 − ζ (they aren’t relatively prime, but it doesn’t matter) this implies that p can not divide (x − y)(1 − ζ); this is the desired contradiction.

2. 1 = ζ^{2a}. Then ζ^{2a−1} = ζ^{−1}, so the congruence reduces to

ζy − ζ^{−1}y ≡ 0 (mod p).

This implies that p divides y(ζ − ζ^{−1}) = −yζ^{−1}(1 − ζ^{2}); the fact that p does not divide y now yields a contradiction as in the previous case.

3. ζ = ζ^{2a−1}. Then ζ^{2a} = ζ^{2} and the congreunce reduces to

(1 − ζ^{2})x ≡ 0 (mod p).

This time p divides x(1 − ζ^{2}); the fact that p does not divide x now yields the contradiction.

This completes the proof.

We used the fact that p does not xyz in an essential way, but Kummer was able to extend the theorem to the case p|xyz; see [20, Chapter 9] for a proof.

2. Regular primes

We have not yet given any methods for determining whether or not a prime is regular. In this section we will state some results of Kummer’s which give easily computible criteria for regularity.

Define the Bernoulli numbers B_{n} ∈ ℝ by the formula

\frac{t}{e^{t} − 1} = ∑_{n=0}^{∞} B_{n} \frac{t^{n}}{n!} .

Exercise 5.4 shows that B_{n} = 0 if n is odd and > 1. One also has the formula

∑_{k=0}^{n−1} \binom{n}{k}B_{k} = 0

of Exercise 5.5, which makes them easy to compute explcitly and also shows that they are actually in ℚ. We include a short table; for a more extensive table, see [20, pp. 407–409].

Kummer’s main results on regular primes are the following theorems. Let h_{p} be the class number of ℚ(ζ_{p}) and let h_{p}^{+} be the class number of the maximal real subfield ℚ(ζ_{p} + ζ_{p}^{−1}). Recall that h_{p}^{+} divides h_{p}, and we set h_{p}^{−} = h_{p}/h_{p}^{+}. In the theorems below, whenever we speak of an integer dividing the numerator of a rational number, we assume that the rational number is written in lowest terms.

Theorem 2.1 (Kummer). Let p be an odd prime. Then p divides h_{p}^{−} if and only if p divides the numerator of some Bernoulli number B_{j} with j = 2, 4, …, p−3.

Proof. See [8] for Kummer’s original proof or [20, Theorem 5.16] for a proof using p-adic L-functions. This theorem has been strengthened by Herbrand, Ribet and Kolyvagin; they have shown that which Bernoulli number p divides gives information on how the Galois group acts on the ideal class group.

Theorem 2.2 (Kummer). If p divides h_{p}^{+}, then p divides h_{p}^{−}.

Proof. See [8] for Kummer’s original proof or [20, Theorem 5.34] for a proof using the p-adic class number formula. Although there are infinitely many primes for which p divides h_{p}^{−}, there are no known p for which p divides h_{p}^{+}. It has been conjectured by Vandiver that this never occurs, although this conjecture is not universally believed.

n  Numerator Denominator
0  1  1
1  −1  2
2  1  6
4  −1  30
6  1  42
8  −1  30
10  5  66
12  −691  2, 730
14  7  6
16  −3, 617  510
18  43, 867  798
20  −174, 611  330
22  854, 513  138
24  −236, 364, 091  2, 730
26  8, 553, 103  6
28  −23, 749, 461, 029  870
30 8, 615, 841, 276, 005  14, 322
32 −7, 709, 321, 041, 217  510
34 2, 577, 687, 858, 367  6

Corollary 2.3 (Kummer). p divides h_{p} if and only if p divides the numerator of some Bernoulli number B_{j} with j = 2, 4, …, p − 3.

Using these results we find that 37 is the first irregular prime; it divides the numerator of B_{32}. The next few irregular primes are 59, 67, 101, 103, 131, 149 and 157. For a longer list see [20, pp. 410–411].

We can give a heuristic argument for the percentage of primes which are irregular. Define the index of irregularity i(p) to be the number of Bernoulli numbers B_{j} with j = 2, 4, …, p−3 for which p divides the numerator of B_{j}; thus i(p) = 0 if and only if p is regular. Assuming that the Bernoulli numbers are randomly distributed modulo p (meaning that p divides B_{j} with probability 1/p), the probability that i(p) = k for some k is

\binom{(p − 3)/2}{k} (1 − \frac{1}{p})^{\frac{p−3}{2} −k} (\frac{1}{p})^{k} .

As p grows this approaches the Poisson distribution

\frac{(\frac{1}{2})^{k} e^{−1/2}}{k!} .

Taking k = 0 we find that the proportion of regular primes should be e^{−1/2}, which is approximately 60.65%. This result agrees very closely with numerical evidence.

Strangely, even though no one has been able to prove that there are infinitely many regular primes, Kummer did succeed in proving that there are infinitely many irregular primes. His proof is based on the following theorems.

Theorem 2.4 (von Staudt-Clausen). Let n be even and positive. Then

B_{n} + ∑_{(p−1)|n} \frac{1}{p}

is an integer.

Proof. See [20, Theorem 5.10].

Theorem 2.5 (Kummer). Let p be a prime and let m and n be even positive integers, not divisible by p − 1, with

m ≡ n (mod p − 1).

Then neither \frac{B_{m}}{m} nor \frac{B_{n}}{n} has any factors of p in the denominator, and

\frac{B_{m}}{m} ≡ \frac{B_{n}}{n} (mod p).

Proof. See [20, Corollary 5.14].

Corollary 2.6 (Kummer). There are infinitely many irregular primes.

Proof. We will suppose that there are only finitely many irregular primes p_{1}, p_{2}, …, p_{r} and obtain a contradiction. Set

m = (p_{1} − 1)(p_{2} − 1) ⋯ (p_{r} − 1).

By Exercise 5.9, |B_{2n}/n| goes to infinity as n goes to infinity, so there must be some multiple M of m such that

|B_{M}/M| > 1.

Thus there exists some prime p dividing the numerator of |B_{M}/M|. Since p_{i} − 1 divides M for all i, Theorem 2.4 shows that each p_{i} is in the denominator of B_{M}; this means that there is no way that p_{i} could be in the numerator of B_{M}/M, and thus that p ≠ p_{i} for any i. Similarly, if p − 1 were to divide M, then Theorem 2.4 would imply that p was in the denominator of B_{M}, which can not occur since p is in the numerator of B_{M}/M. Thus p − 1 does not divide M.

We can now apply Theorem 2.5. Specifically, choose M′ with 2 ≤ M′ ≤ p − 3 which is congruent to M modulo p − 1. Since p − 1 does not divide M we can apply Theorem 2.5 to conclude that

\frac{B_{M′}}{M′} ≡ \frac{B_{M}}{M} ≡ 0 (mod p),

since p divides the numerator of B_{M}/M by assumption. Thus p divides the numerator of B_{M′}, so by Corollary 2.3 it is irregular. This contradicts our assumption that there were finitely many irregular primes, and thus proves the corollary.