the torsion subgroup of an elliptic curve injects in the reduction of the curve

the torsion subgroup of an elliptic curve injects in the reduction of the curve

Let

E

be an

elliptic curve

defined over

ℚ

and let

p

∈

ℤ

be a prime. Let

y

2

+

a

1

⁢

x

⁢

y

+

a

3

⁢

y

=

x

3

+

a

2

⁢

x

2

+

a

4

⁢

x

+

a

6

be a

minimal

Weierstrass equation

for

E

/

ℚ

,
with coefficients

a

i

∈

ℤ

. Let

E

~

be the

reduction

of

E

modulo

p

(see

bad reduction

) which is a curve defined
over

𝔽

p

=

ℤ

/

p

⁢

ℤ

. The curve

E

/

ℚ

can also be considered as a curve over the

p

-adics,

E

/

ℚ

p

, and, in fact, the group of

rational points

E

⁢

(

ℚ

)

injects into

E

⁢

(

ℚ

p

)

. Also, the groups

E

⁢

(

ℚ

p

)

and

E

⁢

(

𝔽

p

)

are related via the reduction
map:

π

p

:

E

⁢

(

ℚ

p

)

→

E

~

⁢

(

𝔽

p

)

π

p

⁢

(

P

)

=

π

p

⁢

(

[

x

0

,

y

0

,

z

0

]

)

=

[

x

0

⁢

mod

⁡

p

,

y

0

⁢

mod

⁡

p

,

z

0

⁢

mod

⁡

p

]

=

P

~

Recall that

E

~

might be a

singular

curve at some
points. We denote

E

~

ns

⁢

(

𝔽

p

)

the set of

non-singular points

of

E

~

. We also define

E

0

⁢

(

ℚ

p

)

=

{

P

∈

E

⁢

(

ℚ

p

)

∣

π

p

⁢

(

P

)

=

P

~

∈

E

~

ns

⁢

(

𝔽

p

)

}

E

1

⁢

(

ℚ

p

)

=

{

P

∈

E

⁢

(

ℚ

p

)

∣

π

p

⁢

(

P

)

=

P

~

=

O

~

}

=

Ker

⁡

(

π

p

)

.

Proposition 1

.

There is an exact sequence of

abelian groups

0

⟶

E

1

⁢

(

ℚ

p

)

⟶

E

0

⁢

(

ℚ

p

)

⟶

E

~

ns

⁢

(

𝔽

p

)

⟶

0

where the right-hand side map is

π

p

restricted to

E

0

⁢

(

Q

p

)

.

Notation: Given an abelian group

G

, we denote by

G

⁢

[

m

]

the

m

-

torsion

of

G

, i.e. the points of order

m

.

Proposition 2

.

Let

E

/

Q

be an elliptic curve (as above) and let

m

be a

positive integer

such that

gcd

⁡

(

p

,

m

)

=

1

. Then:

1.

E

1

⁢

(

ℚ

p

)

⁢

[

m

]

=

{

O

}

2.

If

E

~

⁢

(

𝔽

p

)

is a

non-singular

curve, then the
reduction map, restricted to

E

⁢

(

ℚ

p

)

⁢

[

m

]

, is

injective

. This is

E

⁢

(

ℚ

p

)

⁢

[

m

]

⟶

E

~

⁢

(

𝔽

p

)

is injective.

Remark

: Part

2

of the

proposition

is quite useful when
trying to compute the torsion subgroup of

E

/

ℚ

. As we mentioned above,

E

⁢

(

ℚ

)

injects into

E

⁢

(

ℚ

p

)

. The proposition can be reworded as follows: for all primes

p

which do not
divide

m

,

E

⁢

(

ℚ

)

⁢

[

m

]

⟶

E

~

⁢

(

𝔽

p

)

must be injective and therefore the
number of

m

-torsion points divides the number of points defined
over

𝔽

p

.

Example

:

Let

E

/

ℚ

be given by

y

2

=

x

3

+

3

The

discriminant

of this curve is

Δ

=

-

3888

=

-

2

4

⁢

3

5

. Recall that
if

p

is a prime of bad reduction, then

p

∣

Δ

. Thus the
only primes of bad reduction are

2

,

3

, so

E

~

is
non-singular for all

p

≥

5

.

Let

p

=

5

and consider the reduction of

E

modulo

5

,

E

~

. Then we have

E

~

⁢

(

ℤ

/

5

⁢

ℤ

)

=

{

O

~

,

(

1

,

2

)

,

(

1

,

3

)

,

(

2

,

1

)

,

(

2

,

4

)

,

(

3

,

0

)

}

where all the coordinates are to be
considered modulo

5

(remember the point at infinity!). Hence

N

5

=

∣

E

~

⁢

(

ℤ

/

5

⁢

ℤ

)

∣

=

6

. Similarly, we can
prove that

N

7

=

13

.

Now let

q

≠

5

,

7

be a

prime number

. Then we claim that

E

⁢

(

ℚ

)

⁢

[

q

]

is

trivial

. Indeed, by the remark above we have

∣

E

⁢

(

ℚ

)

⁢

[

q

]

∣

⁢

divides

⁢

N

5

=

6

,

N

7

=

13

so

∣

E

⁢

(

ℚ

)

⁢

[

q

]

∣

must be 1.

For the case

q

=

5

be know that

∣

E

⁢

(

ℚ

)

⁢

[

5

]

∣

divides

N

7

=

13

. But it is easy to see that if

E

⁢

(

ℚ

)

⁢

[

p

]

is
non-trivial, then

p

divides its order. Since

5

does not divide

13

, we conclude that

E

⁢

(

ℚ

)

⁢

[

5

]

must be trivial. Similarly

E

⁢

(

ℚ

)

⁢

[

7

]

is trivial as well. Therefore

E

⁢

(

ℚ

)

has trivial
torsion subgroup.

Notice that

(

1

,

2

)

∈

E

⁢

(

ℚ

)

is an

obvious

point in the curve.
Since we have proved that there is no non-trivial torsion, this
point must be of

infinite order

! In fact

E

⁢

(

ℚ

)

≅

ℤ

and the group is generated by

(

1

,

2

)

.

Title

the torsion subgroup of an elliptic curve injects in the reduction of the curve

Canonical

name

TheTorsionSubgroupOfAnEllipticCurveInjectsInTheReductionOfTheCurve

Date of creation

2013-03-22 13:55:47

Last modified on

2013-03-22 13:55:47

Owner

alozano (2414)

Last modified by

alozano (2414)

Numerical id

7

Author

alozano (2414)

Entry type

Theorem

Classification

msc 14H52

Related topic

EllipticCurve

Related topic

BadReduction

Related topic

MazursTheoremOnTorsionOfEllipticCurves

Related topic

NagellLutzTheorem

Related topic

ArithmeticOfEllipticCurves

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