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§2. Minimal Normal Forms for an Elliptic Curve 107

Since the valuation v takes positive integer values on the discriminant of a given
equation in normal form over R, it is clear that every elliptic curve has a minimal
model. In the next proposition we see in what sense they are unique. Following the
literature, we use the terminology minimal model, minimal normal form, and mini-
mal Weierstrass model interchangeably.

(2.3) Proposition. Let E and E be two elliptic curves over K with minimal models

having coefficients a j and a , respectively. Let f : E → E be an isomorphism with
j
2
2
3

∗
xf = u x + r and y f = u y + su x + t. Then v( ) = v( ), u is in R , and r,
s, and t are in R. The differential ω is unique upto a unit in R.
Proof. The equality v( ) = v( ) follows from the definition of minimal, and hence

12

v(u) = 0. So u is a unit in R from u = . The relation u b = b 8 +· · · in R
8
8
6
implies that 3r is in R, and the relation u b = b 6 +· · · in R implies that 4r is in R.
6
2
Hence the difference r is in R. The relation u a = a 2 + ··· in R implies that s is in
2
6
R and the relation u a = a 6 +· · · implies that t is in R. The last assertion follows
6
ω .
from the formula ωf = u −1
Now we look for estimates on v( ) ≥ 0 which hold for minimal models and
which further might characterize those equations over R corresponding to minimal
2
3
models. With f : E → E as above xf = u ¯x +· · · and yf = u ¯u +· · · , we can

change the valuation of the discriminant by
12
¯
v( ) = v(u ) = 12 · v(u) + v( ¯ ).
Therefore, we see that the following assertion holds.
(2.4) Proposition. If all a j are in R, and if 0 ≤ v( ) < 12, then the model is
minimal.
4
3
6
2
Observe that if π | A and π | B in R, then the equation y = x + Ax + B is
not minimal. Since all elements of K equal j(E) for some E over K, we will not
have j(E) in R for all E,thatis, v( j(E)) may be arbitrary, even with coefficients a i
in R. In terms of v( j) we can obtain estimates on v( ) for a minimal model.
(2.5) Proposition. Let E be an elliptic curve over K, and assume that the charac-
teristic of K is not 2 or 3. For a minimal model the valuation of the discriminant
satisfies
v( ) + min{v( j), 0} < 12 + 12v(2) + 6v(3).
In addition, assuming that the residue class characteristic is different from 2 and 3,
it follows that a model over R is minimal if and only if v( ) + min{v( j), 0} < 12.
3 2 3
Proof. Since c = j and c = ( j − 12 ), we have the relations v( ) + v( j) =
4 6
3
3v(c 4 ) and v( ) + V ( j − 12 ) = 2v(c 6 ). By 3(3.4) the equation of the cubic can be
transformed into the form