Here K = \mathbb{Q}(\sqrt{5}), a = \frac{1+\sqrt{5}}{2}, and R = \mathbb{Z}[a].
We start with an elliptic curve E in Weierstrass form:
y^2+a_1 xy +a_3y = x^3+a_2x^2+a_4x+a_6
with coefficients a_1,a_2,a_3,a_4, and a_6 in R (for our purposes.)
|
|
|
|
Here's a curve over R:
Elliptic Curve defined by y^2 + (8*a+2)*x*y + (80*a-128)*y = x^3 + (-176*a+232)*x^2 + (-14976*a+24224)*x + (-436480*a+706240) over Number Field in a with defining polynomial x^2 - x - 1 Elliptic Curve defined by y^2 + (8*a+2)*x*y + (80*a-128)*y = x^3 + (-176*a+232)*x^2 + (-14976*a+24224)*x + (-436480*a+706240) over Number Field in a with defining polynomial x^2 - x - 1 |
with a-invariants (these are the coefficients):
(8*a + 2, -176*a + 232, 80*a - 128, -14976*a + 24224, -436480*a + 706240) (8*a + 2, -176*a + 232, 80*a - 128, -14976*a + 24224, -436480*a + 706240) |
We can apply a linear change of variables \eta = y + (a_1 x+a_3)/2) to get:
\eta^2 = x^3 + \frac{b_2}{4}x^2 + \frac{b_4}{2}x + \frac{b_6}{4}
and by setting \zeta = x+b_2/12
\eta^2 = \zeta^3 - \frac{c_4}{48}\zeta - \frac{c_6}{864}.
All these b_i's and c_i's are refered to as the b-invariants and covariants of an elliptic curve.
Another invariant of the curve is the discriminant \Delta. \Delta is defined in terms of the b-invariants, but the relationship that is important to us is:
c_4^3-c_6^2 = 1728 \Delta.
Once we have isomorphisms of elliptic curves, we will be concerned with finding curves in an isomorphism class which have minimal discriminants.
Additionally, as elliptic curves are singular iff \Delta = 0, and we're only concerned with non-singular elliptic curves, we will only be considering curves with \Delta \neq 0.
(-608*a + 996, -30176*a + 48832, -1760000*a + 2847744, -94447616*a + 152819456) (-608*a + 996, -30176*a + 48832, -1760000*a + 2847744, -94447616*a + 152819456) |
(-117248*a + 189712, 44219392*a - 71548480) (-117248*a + 189712, 44219392*a - 71548480) |
(102334155*a - 165580141) * 2^12 * (5*a - 2) (102334155*a - 165580141) * 2^12 * (5*a - 2) |
Let E and E' be elliptic curves over K, where E' is y'^2+a_1' x'y' +a_3'y' = x'^3+a_2'x'^2+a_4'x'+a_6'.
Nifty Theorem: If E and E' are isomorphic over K, then there exist r,s,t \in K and u \in K^* such that:
x = u^2x'+r
y = u^3y'+su^2x'+t
thus
ua_1' = a_1 + 2s
u^2a_2' = a_2 - sa_1 + 3r -s^2
u^3a_3' = a_3 +ra_1 + 2t
u^4a_4' = a_4 - sa_3 + 2ra_2 - (t+rs)a_1+3r^2 - 2st
u^6a_6' = a_6 + ra_4 +r^2a_2 + r^3 -ta_3 - t^2 -rta_1
and after a bit more work:
u^4 c_4' = c_4, u^6c_6' = c_6, and u^{12} \Delta' = \Delta.
We write such transformation as \tau = [r,s,t,u].
((102334155*a - 165580141) * (5*a - 2), (102334155*a - 165580141) * 2^12 * (5*a - 2)) ((102334155*a - 165580141) * (5*a - 2), (102334155*a - 165580141) * 2^12 * (5*a - 2)) |
Lemma: Let E be defined over K with covariants c_4,c_6. Then E is K-isomorphic to its c-form:
E_c: y^2 = x^3 - \frac{c_4}{48}x - \frac{c_6}{864}.
Hence if E' is also defined over K, with covariants c_4', c_6', then E and E' are K-isomorphic iff there exists u \in K^* such that
c_4 = u^4c_4' and c_6 = u^6 c_6'.
|
|
Generic morphism: From: Abelian group of points on Elliptic Curve defined by y^2 + (8*a+2)*x*y + (80*a-128)*y = x^3 + (-176*a+232)*x^2 + (-14976*a+24224)*x + (-436480*a+706240) over Number Field in a with defining polynomial x^2 - x - 1 To: Abelian group of points on Elliptic Curve defined by y^2 = x^3 + (7328/3*a-11857/3)*x + (-1381856/27*a+2235890/27) over Number Field in a with defining polynomial x^2 - x - 1 Via: (u,r,s,t) = (1, 152/3*a - 83, -4*a - 1, 116/3*a - 167/3) Generic morphism: From: Abelian group of points on Elliptic Curve defined by y^2 + (8*a+2)*x*y + (80*a-128)*y = x^3 + (-176*a+232)*x^2 + (-14976*a+24224)*x + (-436480*a+706240) over Number Field in a with defining polynomial x^2 - x - 1 To: Abelian group of points on Elliptic Curve defined by y^2 = x^3 + (7328/3*a-11857/3)*x + (-1381856/27*a+2235890/27) over Number Field in a with defining polynomial x^2 - x - 1 Via: (u,r,s,t) = (1, 152/3*a - 83, -4*a - 1, 116/3*a - 167/3) |
Over \mathbb{Q}, we can find a canonical choice of elliptic curve for each isomorphism class. We would like to over \mathbb{Q}(\sqrt{5}). Over \mathbb{Q}, each isomorphism class has a "minimal model", this is the curve from that isomorphism class with minimal descriminant, and a few conditions on the a-invariants.
For two integers n and m, we define the symmetric residue of n mod m to be the unique integer satsifying mods(n,m) = n mod m and -m/2 < mods(n,m) \leq m/2.
Let b = \alpha + \beta a \in R and m be a positive integer. Then b \equiv 0 mod m iff m|\alpha and m|\beta.
We define mods(b,m) = mods(\alpha,m)+mods(\beta,m).
We say that b is m-restricted if mods(b,m) = b.
A Weierstrass equation E with coefficients in R is of restricted type if a_1 and a_3 are 2-restricted and a_2 is 3-restricted. Note that this property depends on the choice of integral basis.
Proposition: Given E with coefficients in R there exists a unique R-isomorphism of the form \tau = [r,s,t,1] such that E' = \tau E is of restricted type.
Notice that since u = 1, E and E' have the same discriminants.
So given an isomorphism class of elliptic curves, if we can find a curve with a minimal discriminant, we can pick the restricted model to be our representative.
Let V be a discrete valuation ring with quotient field K.
E is minimal (with respect to the valuation v) if E is defined over V and the nonnegative integer v(\Delta) is minimal among all K-isomorphic E' defined over V.
If E is minimal for all valuations v with v(\Delta)>0, then E is global minimal.
Basically, if E has integral coefficients and for each prime p dividing \Delta, E is v_{p}-minimal, then E is global minimal.
Note: if we are working in a ring that is not a PID, a global minimal model might not exists.
Back to K and R: Notice that if E has \Delta = (-1)^{e_{-1}}a^{e_0}p_1^{e_1}...p_n^{e_n} and e_i < 12 for i = 1,...,n, then E is global minimal.
However, since the unit group of R is infinite (generated by -1 and a), we only have a unique minimal model up to a unit. This is one use for Andrew's \eta function. This allows us to pick a basis for the unit group and give a (non-cannonical) choice for E, basically, we take mods(e_0,12) and e_{-1} = 0,1.
Since we have global minimal models over R and can pick a "unique" minimal discriminant (where "unique" depends on a choice of basis for the unit group) and since given a discriminant, we can use the restricted model to find a unique curve, we can pick a unique global minimal model for each isomorphism class of elliptic curves over R.
|
|
|
|
|
|