MORDELL’S THEOREM: THE ‘IRREDUCIBLE’ CASE.
IAN KIMING
The following exposition is based mainly on [1], pp. 66–70.
1.
Let E : y 2 = f (x) be an elliptic curve over Q, f (x) = x3 +. . .. To finish the proof
of Mordell’s theorem that E(Q) is a finitely generated abelian group it is enough
to prove the ‘Weak Mordell’ theorem:
Theorem 1. The abelian group E(Q)/2E(Q) is finite.
We have already proved this theorem in case that E has a rational point of order
2, i.e., in case the polynomial f has a rational root. So we assume now that this is
not the case, i.e., that f is irreducible. Via a simple linear change of variables we
can and will assume that f has form x3 + bx + c with b, c ∈ Z, i.e., that E is given
by a Weierstraß equation of form:
E : y 2 = f (x) := x3 + bx + c ,
b, c ∈ Z ,
with f (x) irreducible.
Denote by θ = θ1 , θ2 , θ3 the roots of f . Consider the field extension K/Q
obtained by adjoining the root θ of f (x) to Q:
K := Q(θ) ∼
= Q[x]/(f (x)) .
So ‘physically’ we have K = fuθ2 + vθ + w j u, v, w ∈ Qg, — in particular a
vector space of dimension 3 over Q.
Definition 1. Let α ∈ K. The map ξ 7→ αξ is a linear map of the Q-vector
space K into itself. We define the norm of α, denoted by N (α), as the determinant of this linear map.
So we have that N (α) is a rational number which is non-zero if α is, and
that
N (αβ) = N (α)N (β)
for α, β ∈ K.
Lemma 1. If q ∈ Q then N (q − θ) = f (q).
Proof. Consider 1, θ, θ2 as a Q-basis for K. As
(q − θ) · 1 = q − θ , (q − θ) · θ = qθ − θ2 , (q − θ) · θ2 = c + bθ + qθ2
1
2
IAN KIMING
(where we used that −θ3 = bθ + c) we see that the determinant of the linear map
ξ 7→ (q − θ) · ξ is:


q −1 0
det 0 q −1 = q 3 + bq + c = f (q) .
c b
q
We will now define an abelian group A and a map µ : E(Q) → A that will turn
out to be a homomorphism. The abelian group A is the following subgroup of
K × /(K × )2 :
A := fα ∈ K × /(K × )2 j N (α) ∈ (Q× )2 g ;
that A is a subgroup is a consequence of the property N (αβ) = N (α)N (β) of the
norm map.
We define a map µ : E(Q) → A as follows: µ(O) = 1 mod (K × )2 , and for
(x0 , y0 ) ∈ E(Q):
µ(x0 , y0 ) := (x0 − θ) mod (K × )2 .
Notice that µ(x0 , y0 ) actually lands in A if (x0 , y0 ) ∈ E(Q): For according to
Lemma 1 we have:
N (x0 − θ) = f (x0 ) = y02 ,
so (x0 − θ) mod (K × )2 is really in A.
Theorem 1 will follow when we prove below that µ is a homomorphism with
kernel 2E(Q), and that µ has finite image.
Note: For the proof of these assertions it is not really necessary to know explicitly
that µ has its values in the subgroup A of K × /(K × )2 . I.e., we could work with µ
simply as a homomorphism to K × /(K × )2 . However, this property of µ will appear
implicitly in the arguments via the relation (∗) below. Also, if one wants to work
concretely with some examples, it is of value to know that µ lands in the smaller
subgroup A.
Lemma 2. µ is a homomorphism.
Proof. We have µ(O) = 1. Let P ∈ E(Q). From the definition of µ we clearly
have µ(P ) = µ(−P ). Notice also that η 2 = 1 for any element η ∈ A; so µ(−P ) =
µ(P ) = µ(P )−1 .
Thus, to show that µ is a homomorphism it suffices to show:
P1 + P2 + P3 = O ⇒ µ(P1 )µ(P2 )µ(P3 ) = 1
for points P1 , P2 , P3 ∈ E(Q).
So let P1 , P2 , P3 ∈ E(Q) with P1 + P2 + P3 = O. If at least 2 of the Pi are O
then so is the 3’rd, and so trivially µ(P1 )µ(P2 )µ(P3 ) = 1. If exactly 1 of the Pi ,
say P3 , is O, then P2 = −P1 , and again the claim trivially follows.
So we may assume Pi 6= O. Write then Pi = (xi , yi ). Since P1 + P2 + P3 = O the
points Pi are precisely the points of intersection between E and some rational line
MORDELL’S THEOREM: THE ‘IRREDUCIBLE’ CASE.
3
`, counting multiplicities. For the line ` we have an equation αx + βy + γz = 0,
and in this equation we must have β 6= 0: For otherwise the line would apparently
pass through the point O = (0, 1, 0). But then the point O would have to be one
of the points Pi , contrary to assumption.
For the line ` we may thus an equation of form:
` : y = λx + ν .
Then we have that x1 , x2 , x3 are the roots of the polynomial f (x) − (λx + ν)2 ,
i.e.,
f (x) − (λx + ν)2 = (x − x1 )(x − x2 )(x − x3 ) .
Then:
(x1 − θ)(x2 − θ)(x3 − θ) = (λθ + ν)2 − f (θ) = (λθ + ν)2 ∈ (K × )2 ,
because f (θ) = 0. Thus, µ(P1 )µ(P2 )µ(P3 ) = 1.
Lemma 3. The kernel of µ is 2E(Q).
Proof. Since η 2 = 1 for any η ∈ A (A ‘has exponent 2’) it is clear that 2E(Q) is
contained in the kernel of µ.
Suppose that P = (x0 , y0 ) ∈ E(Q) is in the kernel of µ, i.e., that x0 − θ ∈ (K × )2 .
There are then u, v, w ∈ Q such that
x0 − θ = (uθ2 + vθ + w)2 .
Here we must have u 6= 0 since this equation would otherwise imply either that θ
were rational or root of a polynomial of degree 2 with rational coefficients (which
is not possible since θ is root of f , and f is irreducible of degree 3).
If we put:
v2
vw
v
, s :=
+ ub − w , t :=
+ uc ,
u
u
u
then r, s, t ∈ Q, and we check that
r :=
(r − θ)(uθ2 + vθ + w) = sθ + t
(use that −θ3 = bθ + c). Then:
(sθ + t)2 = (r − θ)2 (uθ2 + vθ + w)2 = (r − θ)2 (x0 − θ) .
So if we consider the polynomial
g(x) := (sx + t)2 − (r − x)2 (x0 − x)
we have g(θ) = 0. Since g is a monic polynomial of degree 3 with rational coefficients
it follows then that
g(x) = f (x) .
Let us now look at the intersection between E and the line y = sx + t. The
x-coordinates of the 3 intersection points are the roots of the polynomial
f (x) − (sx + t)2 = g(x) − (sx + t)2 = −(r − x)2 (x0 − x) ,
4
IAN KIMING
i.e., there is 1 point with x-coordinate x0 , and 2 points with x-coordinate r.
The point with x-coordinate x0 must be (x0 , ±y0 ) = ±P . The 2 points with
x-coordinate r must be equal, since the y-coordinate is uniquely determined by
the x-coordinate (because the points are on the line y = sx + t). So we have
±P + 2 · Q = O for some point Q ∈ E(Q). Thus P ∈ 2E(Q), as desired.
We can now begin the proof of Theorem 1. Since we now know that µ : E(Q) → A
is a homomorphism with kernel 2E(Q), Theorem 1 follows if we can show that the
image of µ is finite. Showing this is the real core of the matter.
Let P = (x0 , y0 ) ∈ E(Q). We know that we can write
n
m
x0 = 2 , y0 = 3
e
e
with integers m, n, e, where e ∈ N, gcd(m, e) = gcd(n, e) = 1.
Since (x0 , y0 ) is on E we have then
n2 /e6 = f (m/e2 ) = (m/e2 − θ1 )(m/e2 − θ2 )(m/e2 − θ3 ) ,
i.e.,
n2 = (m − θ1 e2 )(m − θ2 e2 )(m − θ3 e2 ) ,
(∗)
and we have
µ(x0 , y0 ) = (m − θ1 e2 )
mod (K × )2 .
So one way or the other we will have to use the information contained in (∗) to
deduce that there are only a finite number of possibilities for the quantity (m−θ1 e2 )
mod (K × )2 . We will use algebraic number theory in the algebraic number field
L := Q(θ1 , θ2 , θ3 )
to do that.
2. Interlude: Some results from algebraic number theory.
An algebraic number field is a field obtained by adjoining to Q a finite number
of roots of polynomials with rational coefficients. So the fields K and L of the
previous section are algebraic number fields. Our application of the review below
will be exclusively to the field L, but they hold generally.
So in this section we let L be an arbitrary algebraic number field. We will not
prove any of the statements in this section.
Definition 2. The set of algebraic integers in L, denoted by OL , is the set of
elements of L that are roots of some monic polynomial with integral coefficients
(the polynomial depending of course on the element).
One can show that OL is a ring with respect to addition and multiplication in
the field L.
Example: L = Q, OL = Z. Another example: L = Q(i), OL = Z[i].
For every number λ ∈ L× there is an integer z 6= 0 such that z · λ ∈ OL . In
particular, L is then the field of fractions of OL .
MORDELL’S THEOREM: THE ‘IRREDUCIBLE’ CASE.
5
Since OL is a ring we can consider ideals in it.
One can show that non-zero ideals can be written uniquely as (finite) products
of prime ideals (OL is a so-called Dedekind ring).
One obtains from this the possibility of computing gcd (greatest common divisor)
of ideals: If
Y a
Y b
a=
pi i ,
b=
pi i
i
i
with the pi prime ideals, then
(])
gcd(a, b) := a + b =
Y
minfai ,bi g
pi
.
i
In particular on has
([)
b ⊆ a ⇒ (bi ≥ ai , ∀i) .
If λ ∈ L× we can consider the set:
(λ) := fν · λ j ν ∈ OL g .
This is visibly a set closed under addition and under multiplication by elements
of OL . So, if λ ∈ OL then obviously (λ) is an ideal of OL : The principal ideal
generated by λ. If λ 6∈ OL then (λ) is not an ideal, but still there may exist an
ideal a such that the set
a · (λ) := fν · λ j ν ∈ ag
is an ideal; in fact, there will exist such an a (why?).
The so-called ‘Theorem on finiteness of class number’ implies the following statement:
There exist finitely many ideals v1 , . . . vτ of OL such that: For any ideal a of OL
there exist j ∈ f1, . . . , τ g and an element λ ∈ L× such that:
a = vj · (λ)
The invertible elements of OL (i.e. elements u ∈ OL s.t. u 6= 0 and u−1 also
belongs to OL ) are called units of L (abuse of notation). Denote the set of units
by UL . Then UL is obviously an abelian group with respect to multiplication.
Theorem: (Dirichlet). UL is finitely generated.
In fact, one can also give the rank of UL , and there is an algorithm that will
produce a system of generators for this group.
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IAN KIMING
3. End of proof of Theorem 1
Let us continue the reasoning from the end of section 1. Retain the notation:
We have P = (x0 , y0 ) ∈ E(Q), and write x0 = em2 , y0 = en3 with integers m, n, e,
where e ∈ N, gcd(m, e) = gcd(n, e) = 1. So we have the equation:
(∗)
n2 = (m − θ1 e2 )(m − θ2 e2 )(m − θ3 e2 ) ,
and
mod (K × )2 .
µ(x0 , y0 ) = (m − θ1 e2 )
We will work in the algebraic number field
L := Q(θ1 , θ2 , θ3 ) .
The numbers θj −θk and m−θj e2 are all in OL . We shall work with the principal
ideals generated by these. Recalling the facts mentioned in section 2 we may choose
the following:
• A finite number of prime ideals p1 , . . . , pσ such that: For j 6= k the ideal
(θj − θk ) is a power product of the pi .
• Ideals v1 , . . . , vτ as in section 2 above.
• Numbers δ1 , . . . , δζ ∈ L× such that: Any one of the finitely many products
Y a
pi i
with ai ∈ f0, 1g
i
equals vr · (δs ) for some r, s.
• Numbers η1 , . . . ηξ ∈ OL such that: If α, β ∈ f1, . . . , τ g, and if vα v2β is
principal then for some γ we have
vα v2β = (ηγ ) .
• Units u1 , . . . uπ ∈ UL that are a system of representatives for the finite
quotient UL /UL2 .
Now, for each j we can write:
!
2
(m − θj e ) =
Y
a
qi ij
· a2j ,
i
where the qi are prime ideals, aij ∈ f0, 1g, and aj is just some ideal.
We may and will assume for each i that aij = 1 for at least one j (since, if this
is not the case we have aij = 0 for this i and j = 1, 2, 3; but then in the above we
can just leave out the ideal qi for this particular i).
Now, the relation (∗):
(∗)
n2 = (m − θ1 e2 )(m − θ2 e2 )(m − θ3 e2 ) ,
MORDELL’S THEOREM: THE ‘IRREDUCIBLE’ CASE.
7
together with the fact that we have unique factorization of ideals as products of
prime ideals implies:
ai1 + ai2 + ai3 ≡ 0
(\)
(2) ,
for every i.
Claim: qi ∈ fp1 , . . . , pσ g for each i.
Proof of the claim: Fix an i. We first notice that there are j, k ∈ f1, 2, 3g with
j 6= k and such that
aij = aik = 1 .
For because of our assumption there is j such that aij = 1; because of (\) (and
since the a’s are all 0 or 1) there is then a k 6= j such that aik = 1.
Consider then the exponent of qi in the factorization into prime ideals of the
ideal (m − θj e2 ) + (m − θk e2 ): According to (]) this exponent is the minimum of
two (non-negative) integers congruent mod 2 to aij = 1 and aik = 1, respectively.
So, the exponent is odd, and hence in particular non-zero.
I.e., we have that qi occurs with a non-zero exponent in the prime factorization
of the ideal
(m − θj e2 ) + (m − θk e2 ) .
Now we notice that this ideal contains the (non-zero) ideal (θj − θk ): For, the
equality of numbers:
−(m − θj e2 ) + (m − θk e2 ) = (θj − θk )e2 ,
shows that the ideal (m − θj e2 ) + (m − θk e2 ) contains the number (θj − θk )e2 ; but
it also contains the number n2 , and hence the number (θj − θk )n2 , because of (∗);
as (n, e) = 1 the ideal contains the number θj − θk , and the statement follows.
Now ([) implies that qi occurs with non-zero exponent in the prime factorization
of the ideal (θj −θk ); by definition however, the only prime ideals with that property
are p1 , . . . , pσ . So, qi must be one of these which was the claim.
By the choices we made above, we can now conclude that
(m − θe2 ) = (m − θ1 e2 ) = vr · (δs ) · a21 ,
for some r, s. We can also write:
a1 = vt · (λ)
for some t and some λ ∈ L× . Consequently,
(m − θe2 ) = (δs λ2 ) · vr v2t .
From this equality we may conclude that vr v2t is a principal ideal. So according to
the choices we have made, we may write
vr v2t = (ηγ )
for some γ, and accordingly
(m − θe2 ) = (δs ηγ λ2 ) .
8
IAN KIMING
This equality of ideals implies that δs ηγ λ2 ∈ OL , and that
m − θe2 = δs ηγ λ2 · u
for some unit u ∈ UL . There is some ρ such that u ≡ uρ mod UL2 . But then we
have:
m − θe2 ≡ δs ηγ uρ mod (L× )2
for some m ∈ f1, . . . ζg, γ ∈ f1, . . . ξg, ρ ∈ f1, . . . πg.
We have now shown that there are only finitely many possibilities for
µ(x0 , y0 ) mod (L× )2 = m − θe2 mod (L× )2 ,
where obviously µ(x0 , y0 ) mod (L× )2 means the image of µ(x0 , y0 ) under the homomorphism
K × /(K × )2 −→ L× /(L× )2
induced by the injection K × ,→ L× .
The proof of finiteness of µ(E(Q)), and hence of Theorem 1, may now be concluded by the following lemma (which is essentially what is classically called ‘Kummer theory’):
Lemma 4. Let L/K be a finite extension of (algebraic number) fields. Then
there is a 1–1 correspondence between the kernel of the natural homomorphism
K × /(K × )2 −→ L× /(L× )2
coming from the injection K × ,→ L× , and the set of quadratic extensions of
K contained in L.
In particular, this kernel is finite.
Proof. The kernel in question is seen to be
(†)
K × ∩ (L× )2 /(K × )2 .
√
Attach to any element ξ of the group (†) the quadratic extension K( ξ); this
extension is contained in L, as ξ ∈ (L× )2 . Conversely, any quadratic extension
√
of K has the form K( ξ) for some ξ ∈ K × , and this extension is contained in L
precisely if ξ ∈ (L× )2 .
√
The claim now follows at once when we remind ourselves that we have K( ξ1 ) =
√
K( ξ2 ) for 2 elements ξ1 , ξ2 ∈ K × , precisely if ξ1 and ξ2 differ by a square in
K ×.
References
[1] J. W. S. Cassels: ‘Lectures on elliptic curves.’ London Mathematical Society Student Texts
24, Cambridge University Press 1991.
Department of Mathematics, University of Copenhagen, Universitetsparken 5, DK2100 Copenhagen Ø, Denmark.
E-mail address: [email protected]