SELMER’S EXAMPLE

SELMER’S EXAMPLE
KEITH CONRAD
The standard example of an irreducible polynomial with no nontrivial rational zero and
a nontrivial real and p-adic zero for all p is Selmer’s cubic 3x3 + 4y 3 + 5z 3 : the trivial zero
here is (0, 0, 0).
Theorem 1 (Selmer [4]). The equation 3x3 + 4y 3 + 5z 3 = 0 has only the zero solution over
Q, but there is a nonzero solution over every completion Qv .
Proof. Over Q∞ = R, it is obvious there is a solution besides (0, 0, 0). To handle p-adic
fields, roughly speaking the idea is to show there is a nonzero solution modulo p and then
lift it p-adically by Hensel’s lemma, following a method suggested by K. Buzzard. We will
separately treat the cases p = 3, p = 5, and all other p.
For a 3-adic solution, set x = 0 and z = −1, making Selmer’s equation 4y 3 − 5 = 0, or
y 3 = 5/4. Although 5/4 ≡ −1 mod 9 and −1 is a 3-adic cube, this congruence isn’t sharp
enough to conclude with Hensel’s lemma that 5/4 is a 3-adic cube: to use Hensel’s lemma
3
3
we seek a β ∈ Z×
3 such that |β − 5/4|3 < 1/9, i.e., β ≡ 5/4 ≡ 8 mod 27. The choice β = 2
works, so 5/4 is a 3-adic cube and we can solve Selmer’s equation with (0, y, −1) where
y 3 = 5/4 in Z3 . Note that we needed to work mod 27, not mod 3.
If p 6= 3 and α is a nonzero cube mod p then α is a cube in Z×
p by Hensel’s lemma for
3
X − α. In particular, for p = 5, set x = 1 and z = 0 to make Selmer’s equation 3 + 4y 3 = 0,
or y 3 = −3/4. Every element of (Z/5Z)× is a cube (why?), so every element of Z×
5 is a
cube. Thus −3/4 is a 5-adic cube. We get the 5-adic solution to Selmer’s equation (1, y, 0)
where y 3 = −3/4 in Z5 .
From now on let p be a prime other than 3 or 5.
If 3 is a cube in (Z/pZ)× then 3 is a cube in Zp and we can solve Selmer’s equation using
(x, 1, −1) where x3 = 1/3 in Qp .
If 3 is not a cube in (Z/pZ)× then the subgroup of cubes in (Z/pZ)× has index 3, so a
set of representatives for (Z/pZ)× modulo the cubes is {1, 3, 9}.
• If 5 = 1 mod cubes in (Z/pZ)× then 5 is a cube mod p, hence in Zp , and we can
solve Selmer’s equation using (−y, y, −1) where y 3 = 5 in Zp .
• If 5 = 3 mod cubes in (Z/pZ)× then 5/3 is a cube in Zp and we can solve Selmer’s
equation using (x, 0, −1) where x3 = 5/3.
• If 5 = 9 mod cubes in (Z/pZ)× then 15 is a cube in Zp and we can solve Selmer’s
equation using (3t/7, 5/7, −1) where t3 = 15.
This concludes the proof that Selmer’s equation has solutions everywhere locally.
Now we turn to the global case: 3x3 + 4y 3 + 5z 3 = 0 has no rational solution besides
(0, 0, 0). Our proof is based on [3, pp. 220-222], with a few minor changes and (typographical) corrections.
If there is a solution, then by multiplying through by 2 and rearranging terms we can
write (2y)3 + 6x3 = 10z 3 . We will show the only rational solution of the equation
(1)
X 3 + 6Y 3 = 10Z 3
1
2
KEITH CONRAD
is (0, 0, 0), which implies the theorem (X = 2y, Y = x, Z = z).
By clearing denominators, we can assume in equation (1) that X, Y , and Z are integers.
Assume they are not all 0, so in fact none are 0. If a prime p divides two of X, Y , or Z
then it also divides the third since the coefficients 6 and 10 in (1) aren’t divisible by the
cube of any prime. Then we can divide through all the terms in (1) by p3 to get a smaller
integral solution of the same equation. Hence without loss of generality X, Y , and Z are
pairwise relatively prime. Since X is even, Y and Z are odd. We can also conclude from
(1) that X and Z are not divisible by 3√and X and Y are not divisible by 5.
Factor the left side of (1) using α = 3 6:
(2)
(X + Y α)(X 2 − XY α + Y 2 α2 ) = 10Z 3 .
Claim 1: Z[α] is the ring of integers in Q[α]. √
√
Proof of claim: For any non-cube d ∈ Z, disc(Z[ 3 d])√= −27d2 , so disc(Z[ 3 6]) √
= −22 · 35 .
3
3
Since T − 6 is Eisenstein at 2 and 3, the index of Z[ 6] in the integers of Q( 3 6) is not
divisible by 2 or 3, so the index is 1.
Passing from (2) to an equation of principal ideals in Z[α],
(3)
(X + Y α)(X 2 − XY α + Y 2 α2 ) = (10)(Z)3 .
To derive information about the prime ideal factorization of (X + Y α) from this equation,
we need to determine how the ideal (10) factors.
The way a prime p factors in Z[α] matches how T 3 − 6 factors mod p. Since T 3 − 6 ≡
3
T mod 2 and T 3 − 6 ≡ (T − 1)(T 2 + T + 1) mod 5, we have (2) = p32 and (5) = p5 p25 ,
so there are unique prime ideals of norm 2 and 5, and (10) = p32 p5 p25 . For any integer k,
N(α + k) = k 3 + 6, so N(α − 1) = 5 and N(α − 2) = −2. Thus p2 = (α − 2) and p5 = (α − 1).
A consequence of equation (3) is
(4)
(X + Y α) = p2 p5 b3 = (α − 2)(α − 1)b3
for some ideal b. The derivation of this key equation will be postponed until later.
Claim 2: Z[α] has class number 1.
Proof of claim: The Minkowski bound for Q(α) is
√
r2
16 3
4
n! p
| disc(Z[α])| =
≈ 8.82.
π
nn
π
Therefore the class group is generated by the ideal classes of primes with norm at most
8. We have already seen that there are unique primes of norm 2 and 5: p2 = (α − 2)
and p5 = (α − 1). To factor (3), from T 3 − 6 ≡ T 3 mod 3 we obtain (3) = p33 . Since
N(α) = 6, we must have (α) = p2 p3 , so p3 is principal. It remains to factor (7). Since
T 3 − 6 ≡ (T − 3)(T − 5)(T − 6) mod 7, we have (7) = p7 p07 p007 . Computing N(α + k) at
small k which make the norm a multiple of 7, we find N(α + 1) = 7, N(α + 2) = 14, and
N(α + 4) = 70. Since 1, 2, and 4 are incongruent mod 7, we can choose the prime factors of
7 so that (α + 1) = p7 , (α + 2) = p2 p07 , and (α + 5) = p2 p5 p007 . Having already seen that p2
and p5 are principal, the prime ideals of norm 7 are principal. Thus the class number is 1.
By Claim 2 the ideal b is principal, say b = (β), so equation (4) leads to an equation of
elements:
(5)
X + Y α = (α − 2)(α − 1)β 3 u
SELMER’S EXAMPLE
3
for some unit u in Z[α]. In this equation the unit u only matters modulo multiplication by
unit cubes since any unit cube can be absorbed into β.
Claim 3: The units in Z[α] modulo unit cubes are represented by (1 − 6α + 3α2 )k for
k = 0, 1, or 2.
Proof of claim: Since Q(α) has r1 = 1 and r2 = 1, the unit group has rank 1: Z[α]× = ±εZ
for some ε, so Z[α]× /(Z[α]× )3 is cyclic of order 3. Therefore any unit which is not a cube
generates the units modulo cubes. To find a noncube unit, we observe that (2) = p32 =
(2 − α)3 , so the ratio
(2 − α)3
= 1 − 6α + 3α2 ≈ .00306
2
is a unit. To check this is not a cube of a unit, we verify it is not a cube in a suitably chosen
residue field. Specifically, the ideal p7 = (1 + α) has norm 7 and in Z[α]/p7 ∼
= Z/7Z
1 − 6α + 3α2 ≡ 1 − 6(−1) + 3(1) = 10 ≡ 3 mod p7 ,
and this is not a cube since 3 is not a cube in Z/7Z.
Remark 1. It is true that 1 − 6α + 3α2 is a generator of Z[α]× (modulo ±1), but that
takes more effort to prove and Claim 3 is sufficient information about the units of Z[α].
Since
(2 − α)3
,
2
by Claim 3 we can write u in equation (5) as ((2 − α)3 /2)k v 3 = ((2 − α)k v)3 /2k where
v ∈ Z[α]× and k is 0, 1, or 2. Multiplying through equation (5) by 2k , we absorb ((2−α)k v)3
into β 3 to get
1 − 6α + 3α2 =
(6)
2k X + 2k Y α = (α − 2)(α − 1)γ 3
for some γ ∈ Z[α]. Write γ = A + Bα + Cα2 , where A, B, and C are integers that are not
all 0.
Compute (α − 2)(α − 1)γ 3 as a Z-linear combination of 1, α, and α2 and then equate the
coefficients of α2 on both sides of equation (6) to get
(7) 0 = A3 + 6B 3 + 36C 3 + 36ABC − 9(A2 B + 6AC 2 + 6B 2 C) + 6(AB 2 + A2 C + 6BC 2 ).
From this equation, obviously 3 | A. This implies 0 ≡ 6B 3 mod 9, so 3 | B, which in turn
implies 0 ≡ 36C 3 mod 27, so 3 | C. The right side of (7) is homogeneous of degree 3 in A,
B, and C, so we can remove a common factor of 27 from all the terms and obtain another
equation (7) where A, B, and C are one-third as large. Repeating this infinitely often forces
A, B, and C to equal 0, which is a contradiction.
This completes the proof of the theorem except for the derivation of equation (4) from
equation (3), which involves a careful analysis of ideal factorizations in Z[α].
Let p be a prime ideal such that
p | (X + Y α)
and p | (X 2 − XY α + Y 2 α2 ),
so
X + Y α ≡ 0 mod p and X 2 − XY α + Y 2 α2 ≡ 0 mod p.
We will show that p must be p2 .
Since X 2 − XY α + Y 2 α2 = (X + Y α)2 − 3XY α, we get 3XY α ≡ 0 mod p. Thus
p | (3)(X)(Y )(α).
4
KEITH CONRAD
• If p | (3) then N(p) is a power of 3. Since p divides (Z)3 by equation (3), taking
norms implies Z is divisible by 3, which is false.
• If p | (X) then Y α ≡ 0 mod p, which implies p | (Y )(α). From relative primality of
X and Y , p can’t divide (Y ), so p | (α).
• If p | (Y ) then X ≡ 0 mod p, but that means p | (X), which contradicts the relative
primality of X and Y (they’d both be divisible by whatever prime number p divides).
The conclusion is that any common prime factor p of the two ideals on the left side of
equation (3) is a factor of (α) and not a factor of the ideal (3). Since (α)3 = (6) = (2)(3),
p must be a factor of the ideal (2), so p = p2 . Thus any common factor of the ideals on the
left side of equation (3) is a power of p2 . How high a power can work?
Since X is even and Y is odd and (α) is divisible by p2 just once, (X + Y α) is divisible
by p2 just once. Therefore
(X 2 − XY α + Y 2 α2 ) = p2 c0
(X + Y α) = p2 c,
where c and c0 are relatively prime ideals in Z[α] and p2 doesn’t divide c.
Plugging these factorizations into equation (3), p22 cc0 = (10)(Z)3 = p32 p5 p25 (Z)3 , so p2 is
a factor of c0 . Which of c or c0 is divisible by p5 and p25 ?
From X 3 + 6Y 3 = 10Z 3 , check X ≡ −Y mod 5, so X + Y α ≡ X + Y ≡ 0 mod p5 , which
means p5 | (X + Y α). If p25 | (X + Y α) then p5 p25 = (5) divides (X + Y α), so 5 is a factor
of X + Y α in Z[α], which implies X and Y are divisible by 5 in Z, and that is false. Thus
p25 is not a factor of (X + Y α), so p25 is a factor of (X 2 − XY α + Y 2 α2 ).
Write c = p5 m and c0 = p2 p25 m0 . Then
(X + Y α) = p2 p5 m and
(X 2 − XY α + Y 2 α2 ) = p22 p25 m0 .
Multiplying these together, we obtain (10)mm0 = (10)(Z)3 . The ideals m and m0 are relatively prime, so m (as well as m0 ) must be a cube. This finally leads to equation (4).
This treatment of Selmer’s equation is based on [3, pp. 220–222], where the analogue of
our equation (7) on the top of p. 222 has one incorrect coefficient on the right side.
Other examples of homogeneous cubics fitting the conditions of Selmer’s theorem are
x3 + 5y 3 + 12z 3 , x3 + 4y 3 + 15z 3 , x3 + 3y 3 + 20z 3 , x3 + 3y 3 + 22z 3 .
Other examples whose analysis does not require algebraic number theory are in [1].
Remark 2. Just as counterexamples to unique factorization in number fields can acquire a
positive interpretation as non-trivial elements in an ideal class group (that is, such phenomena are associated to non-principal ideals), Selmer’s example has a positive interpretation:
it represents a non-trivial element in the Tate-Shafarevich group of an elliptic curve over
Q, specifically the elliptic curve x3 + y 3 + 60z 3 = 0. The lack of rational solutions besides
(0, 0, 0) to 3x3 + 4y 3 + 5z 3 = 0 can be proved more simply using the theory of elliptic curves
instead of purely by algebraic number theory. See [2, pp. 86–87].
References
[1] W. Aitken and F. Lemmermeyer, Simple Counterexamples to the Local–Global Principle, at http://
public.csusm.edu/aitken html/m372/diophantine.pdf.
[2] J. W. S. Cassels, “Lectures on Elliptic Curves,” Cambridge Univ. Press, Cambridge, 1991.
[3] J. W. S. Cassels, “Local Fields,” Cambridge Univ. Press, Cambridge, 1986.
[4] E. Selmer, The Diophantine equation ax3 + by 3 + cz 3 = 0, Acta Mathematica 85 (1951), 203–362.