Download IRREDUCIBILITY OF ELLIPTIC CURVES AND INTERSECTION

IRREDUCIBILITY OF ELLIPTIC CURVES AND INTERSECTION
WITH LINES.
IAN KIMING
1. Non-singular points and tangents.
Suppose that k is a field and that F (x1 , . . . , xn ) is a homogeneous polynomial
in n variables (n ≥ 2) with coefficients in k, and of degree d, say. The equation:
F (x1 , . . . , xn ) = 0
defines a projective variety V defined over k.
If (ξ1 , . . . , ξn ) ∈ Pn−1 (k̄) is a point on V (k̄), we say that (ξ1 , . . . , ξn ) is non∂F
singular if the partial derivatives ∂x
do not all vanish at the point.
i
If so, we have a well-defined ‘hyperplane’ in Pn given by the equation:
∂F
∂F
(∗)
· x1 + . . . +
· xn = 0 ;
∂x1 x1 =ξ1
∂xn xn =ξn
this hyperplane is called the tangent (space) at the point (ξ1 , . . . , ξn ). This tangent space does indeed pass through the point (ξ1 , . . . , ξn ): For F is a sum with
coefficients in k of monomials xi11 · · · xinn where i1 + . . . + in = d; such a monomial
is seen to give a contribution
(i1 + . . . + in ) · xi11 · · · xinn = d · xi11 · · · xinn
to the left hand side of the tangent equation (∗), whence:
∂F
∂F
· ξ1 + . . . +
· ξn = d · F (ξ1 , . . . , ξn ) = 0 .
∂x1 x1 =ξ1
∂xn xn =ξn
We call V smooth if all points in V (k̄) are non-singular.
It is clear that the question of whether V contains any singular points is independent of whether we apply a linear change of variables in the equation of V .
2. Elliptic curves.
We define (in this course ...) an elliptic curve over a field k as a non-singular
projective curve given by a general Weierstraß equation:
y 2 z + a1 xyz + a3 yz 2 = x3 + a2 x2 z + a4 xz 2 + a6 z 3 .
Proposition 1. (i). Assume char(k) 6= 2.
Then, using a linear change of variables, a general Weierstraß equation
may be put in shape:
(∗∗)
y 2 z = x3 + ax2 z + bxz 2 + cz 3 .
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IAN KIMING
If additionally char(k) 6= 3, we can also attain a = 0.
A Weierstraß equation (∗∗) defines an elliptic curves if and only if the
polynomial x3 + ax2 + bx + c does not have any multiple roots.
(ii). Assume char(k) = 2.
Then, using a linear change of variables, a general Weierstraß equation
may be put in shape
(])
y 2 z + a1 xyz = x3 + ax2 z + bxz 2 + cz 3 ,
a1 6= 0 ,
y 2 z + a3 yz 2 = x3 + ax2 z + bxz 2 + cz 3 ,
a3 6= 0 .
or
(]])
An equation (]) defines an elliptic curve if and only if b2 + a1 c 6= 0. An
equation (]]) always defines an elliptic curve.
Proof. (i). Since char(k) 6= 2 we can consider the linear change of variables:
a1
a3
(x, y, z) 7→ (x, y − x − , z)
2
2
that will bring the Weierstraß equation in shape (∗∗). If also char(k) 6= 3, we can
further consider the change x 7→ x − a3 , y 7→ y, z 7→ z.
By definition, a Weierstraß equation (∗∗) defines an elliptic curve if and only
if the projective curve defined by it is without singularities. Writing the equation
(∗∗) as F (x, y, z) = 0 where:
F (x, y, z) := y 2 z − x3 − ax2 z − bxz 2 − cz 3 ,
we find that:
∂F
= −3x2 − 2axz − bz 2 ,
∂x
∂F
= 2yz ,
∂y
∂F
= y 2 − ax2 − 2bxz − 3cz 2 .
∂z
∂F ∂F
Hence, we see that ( ∂F
∂x , ∂y , ∂z )(0, 1, 0) = (0, 0, 1), and so O = (0, 1, 0) is never
a singular point.
Suppose that (x0 , y0 , 1) is an affine singular point. Then 2y0 = 0, and hence
y0 = 0 since char(k) 6= 2. We conclude that x0 is a root of x3 + ax2 + bx + c.
Since we have −3x20 − 2ax0 − b = 0 we conclude that x0 is a multiple root of
x3 + ax2 + bx + c.
If conversely the polynomial x3 +ax2 +bx+c has a multiple root x0 then (x0 , 0, 1)
∂F
is a point on the projective curve defined by (∗∗). We also see that ∂F
∂x and ∂y
both vanish at this point. The same holds for the partial derivative ∂F
∂z :
−ax20 − 2bx0 − 3c = (3x20 + 2ax0 + b)x0 − 3(x30 + ax20 + bx0 + c) = 0 ,
so that (x0 , 0, 1) is in fact a singular point.
(ii). If a1 6= 0 the linear change of variables
(x, y, z) 7→ (x +
a3
, y, z)
a1
IRREDUCIBILITY OF ELLIPTIC CURVES AND INTERSECTION WITH LINES.
3
transforms (∗∗) to an equation of type (]).
The remaining statements are left as exercises.
Proposition 2. Suppose that E is an elliptic curve defined over a field k.
Then E is irreducible, i.e., does not contain a whole line in P2 (k̄).
Proof. We use the previous proposition and leave the proof as an exercise in case
char(k) = 2.
So assume char(k) 6= 2. According to the previous proposition we may then
assume the equation for E in form:
E : y 2 z = x3 + ax2 z + bxz 2 + cz 3 .
By definition E is a curve without singularities, and by the previous proposition
we know that this is equivalent to the polynomial
x3 + ax2 + bx + c
not having any multiple roots.
Now suppose that E contains the line:
` : ux + vy + wz = 0 ,
where (u, v, w) ∈ k̄\f(0, 0, 0)g. Suppose that w = 0. Then (u, v) 6= (0, 0). The
equation ux + vy = 0 has a solution (x0 , y0 ) 6= (0, 0). Then ` would be the set of
points (x0 , y0 , t), t ∈ k̄. But then we would have:
0 = ct3 + bx0 t2 + (ax20 − y02 )t + x30
for all t ∈ k̄; but that implies first x30 = 0 hence x0 = 0, and then −y02 = ax20 −y02 = 0
so y0 = 0 and contradiction.
So we may suppose that ` is given by an equation of form z = αx + βy. If now
` is contained in E we deduce that the polynomial:
g(x, y) := y 2 (αx + βy) − x3 − ax2 (αx + βy) − bx(αx + βy)2 − c(αx + βy)3
vanishes identically. Looking at the coefficients of x3 , xy 2 and x2 y, respectively,
we deduce:
(1)
cα3 + bα2 + aα + 1 = 0 ,
(2)
α − bβ 2 − 3cαβ 2 = 0 ,
(3)
−aβ − 2bαβ − 3cα2 β = 0 .
Then (1) implies α 6= 0; then (2) gives β 6= 0. Then (3) implies a+2bα+3cα2 = 0.
Together with (1) this implies that α−1 is (at least) a double root of the polynomial
x3 + ax2 + bx + c, contradiction.
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IAN KIMING
3. Intersection between lines and elliptic curves.
Let k be a field and let E be an elliptic curve over k given by a Weierstrass
equation:
y 2 z + a1 xyz + a3 yz 2 = x3 + a2 x2 z + a4 xz 2 + a6 z 3 .
We shall write this as:
F (x, y, z) = 0
where:
F (x, y, z) := y 2 z + a1 xyz + a3 yz 2 − (x3 + a2 x2 z + a4 xz 2 + a6 z 3 )
so that F is a homogeneous polynomial of degree 3.
Consider a projective line ` given by an equation:
αx + βy + γz = 0 .
Thus, (α, β, γ) =
6 (0, 0, 0). Notice that we do not necessarily assume that ` is
‘k-rational’ in the sense that α, β, γ ∈ k.
The statement we will now discuss is the following:
With an appropriately defined notion of ‘multiplicity’ of intersection points,
the line ` intersects E in precisely 3 points. Furthermore, if ` is k-rational,
and if 2 of the intersection points are k-rational then so is the 3’rd point of
intersection.
Actually, what we describe below is not only a proof of this statement but also
an algorithm for computing these intersection points.
With the above statement we will not need any reference to Bezout’s theorem
as in chap. 1 of [1].
The statement above holds of course in much greater generality: Without going
into Bezout’s theorem, you can at least easily see from the following how to generalize the statement to a statement about points of intersection between a line and
a projective curve of degree n (n ∈ N); you do need some kind of assumption on
the curve though, — specifically an irreducibility condition. For us, in connection
with elliptic curves it is enough that we showed that an elliptic curve does not
contain any line.
Now to the argument in favor of the above statement. As we said, the notion of
multiplicity will be defined in course of the analysis.
One needs to split the discussion up into 3 cases depending on which of α, β, γ
is 6= 0. We do only 1 case, assuming that:
γ 6= 0 .
We may thus assume an equation of form:
z = rx + sy
for our line `.
IRREDUCIBILITY OF ELLIPTIC CURVES AND INTERSECTION WITH LINES.
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A point P = (x, y, z) is a point of intersection between ` and E precisely if:
(x, y, z) = (x, y, rx + sy)
and:
0 = F (x, y, z) = F (x, y, rx + sy) .
Thus, if we put:
g(x, y) := F (x, y, rx + sy) ,
the points of intersection are precisely the points (x, y, rx+sy) where (x, y) 6= (0, 0)
is such that g(x, y) = 0 (notice that (x, y, rx + sy) is a point in the projective plane
exactly if (x, y) 6= (0, 0)).
Now, g(x, y) is clearly a homogeneous polynomial of degree 3, so we can write:
g(x, y) = ax3 + bx2 y + cxy 2 + dy 3 .
(])
Notice that, as the coefficients of F are in k, we can conclude that a, b, c, d are
also in k if ` is k-rational, i.e., if r, s ∈ k.
We first claim that g splits into linear factors:
(∗)
g(x, y) = (u1 x + v1 y)(u2 x + v2 y)(u3 x + v3 y)
with ui , vi in an algebraic closure of k and (ui , vi ) 6= (0, 0) for each i = 1, 2, 3.
To see this, notice first that g is not identically 0: For otherwise, we would
conclude that E contained the whole line `.
So if a = d = 0 then (b, c) 6= (0, 0), and also:
g(x, y) = xy(bx + cy)
so that things are clear in this case.
Now, in principle we would have to discuss both of the cases a 6= 0 and d 6= 0.
Let us take just the second case: If d 6= 0 the polynomial:
(]])
g(1, y) = a + by + cy 2 + dy 3
is a polynomial of degree 3 in y. Hence:
g(1, y) = d(y − θ1 )(y − θ2 )(y − θ3 )
with θ1 , θ2 , θ3 in an algebraic closure of k. Then:
d(−θ1 − θ2 − θ3 ) = c , d(θ1 θ2 + θ1 θ3 + θ2 θ3 ) = b , d(−θ1 θ2 θ3 ) = a .
Using these formulas we see that:
(]]])
g(x, y) = d(y − θ1 x)(y − θ2 x)(y − θ3 x)
simply by multiplying the right hand side out.
Thus we have established (∗) in all cases.
Using the form (∗) we can now say that the points of intersection between `
and E are the points (x0 , y0 , rx0 + sy0 ) where (x0 , y0 ) 6= (0, 0) is a solution to an
equation:
(†)
ui x0 + vi y0 = 0
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IAN KIMING
for some i ∈ f1, 2, 3g. Now, if (x0 , y0 ), (x00 , y00 ) 6= (0, 0) are 2 solutions to (†) then
(x00 , y00 ) = (λx0 , λy0 )
for some non-zero constant λ: We can deduce this since we had (ui , vi ) 6= (0, 0).
Consequently, these 2 solutions to (†) (for a fixed i) give us the same intersection
point (x0 , y0 , rx0 + sy0 ) in the projective plane.
Or put in another way: We get 3 intersection points by considering the 3 linear
factors of g in (∗) one after the other. Of course we may get the same point 2 or
even 3 times, but it is now clear how to define multiplicity of an intersection point:
It is the number of times the point arises when we consider the linear factors of g
one after the other.
Concerning the last remark (about k-rational intersection points): Suppose that
our line ` is k-rational. Then r, s above are in k. Since F has coefficients in k (the
elliptic curve E being defined over k) we deduce that a, b, c, d in (]) and (]]) are in
k.
Let the 3 points of intersection be (xi , yi , rxi + syi ), i = 1, 2, 3, corresponding as
in the above to solutions to (†) for i = 1, 2, 3, respectively.
Assume now further that the first 2 of these points are k-rational, i.e., that
x1 , x2 , y1 , y2 ∈ k.
We claim that then the ui , vi in the decomposition (∗) can be chosen in k (the
decomposition (∗) is not unique, — the factors are only determined up to multiplication with a non-zero constant; clearly we could multiply one of the factors
with some non-zero λ and then some other factor with λ−1 ).
Let us argue in the case we considered above: If g(x, y) = xy(bx + cy) the claim
is clear as b, c ∈ k. Suppose then that d 6= 0. Considering (]]]) we may choose
(say) (ui , vi ) = (−θi , 1) for i = 1, 2, and (u3 , v3 ) = (−dθ3 , d). Since (xi , yi ) is a
solution 6= (0, 0) to ui x + vi y = 0 we see that we can not possibly have xi = 0.
Thus, θi = yi /xi ∈ k for i = 1, 2 by our assumption x1 , x2 , y1 , y2 ∈ k. We deduce:
θ3 = −c/d − θ1 − θ2 ∈ k
and hence the claim ui , vi ∈ k for i = 1, 2, 3.
But since now u3 , v3 ∈ k the equation u3 x + v3 y = 0 has a solution (x03 , y30 ) 6=
(0, 0) with x03 , y30 ∈ k. As r, s ∈ k the point (x03 , y30 , rx03 + sy30 ) is a k-rational point.
But this point is the 3’rd point of intersection between ` and E.
References
[1] J. H. Silverman, J. Tate: ‘Rational points on elliptic curves’, Undergraduate Texts in Mathematics. Springer, 1994.
Department of Mathematics, University of Copenhagen, Universitetsparken 5, DK2100 Copenhagen Ø, Denmark.
E-mail address: [email protected]