The Basic Theory - Elliptic Curves: Number Theory and Cryptography - page 29

ord m 3 ,P 23 ( D ) 2, since D is a sum of terms, each of which vanishes to order

at least 2. But ord m 3 ,P 23 ( 1 )=1,sowehave

≥

ord m 3 ,P 23 ( m 1 )=ord m 3 ,P 23 ( D ) − ord m 3 ,P 23 ( 1 ) ≥ 1 .

Therefore m 1 ( P 23 ) ( P 23 )=0.
In both cases, we have m 1 ( P 23 ) ( P 23 )=0.
If m 1 ( P 23 ) =0,then ( P 23 ) = 0, as desired.
If m 1 ( P 23 )=0,then P 23 lies on m 1 , and also on 2 and m 3 , by definition.
Therefore, P 23 = P 21 ,since 2 and m 1 intersect in a unique point. By as-
sumption, 2 is therefore tangent to C at P 23 . Therefore, ord 2 ,P 23 ( C ) ≥ 2.

As above, ord 2 ,P 23 ( D ) ≥ 2, so

ord 2 ,P 23 ( 1 ) ≥ 1 .

If in this case we have 1 ( P 23 )=0,then P 23 lies on 1 , 2 ,m 3 . Therefore
P 13 = P 23 . By assumption, the line m 3 is tangent to C at P 23 .Since P 23 is a
nonsingular point of C , Lemma 2.5 says that 2 = m 3 , contrary to hypothesis.
Therefore, 1 ( P 23 ) =0,so ( P 23 )=0.
Similarly, ( P 22 )= ( P 32 )=0.

If ( x, y, z ) is identically 0, then D is identically 0. Therefore, assume that
( x, y, z ) is not zero and hence it defines a line .
First suppose that P 23 ,P 22 ,P 32 are distinct. Then and 2 are lines through
P 23 and P 22 . Therefore = 2 . Similarly, = m 2 . Therefore 2 = m 2 ,

contradiction.
Now suppose that P 32 = P 22 .Then m 2 is tangent to C at P 22 . As before,

ord m 2 ,P 22 ( 1 m 1 ) ≥ 2 .

We want to show that this forces to be the same line as m 2 .
If m 1 ( P 22 )=0,then P 22 lies on m 1 ,m 2 , 2 . Therefore, P 21 = P 22 .This
means that 2 is tangent to C at P 22 . By Lemma 2.5, 2 = m 2 , contradiction.
Therefore, m 1 ( P 22 ) =0.

If 1 ( P 22 ) =0,thenord m 2 ,P 22 ( ) ≥ 2. This means that is the same line as

m 2 .
If 1 ( P 22 )=0,then P 22 = P 32 lies on 1 , 2 , 3 ,m 2 ,so P 12 = P 22 =
P 32 . Therefore ord m 2 ,P 22 ( C ) ≥ 3. By the reasoning above, we now have

ord m 2 ,P 22 ( 1 m 1 ) ≥ 3. Sincewehaveprovedthat m 1 ( P 22 ) =0,wehave

ord m 2 ,P 22 ( ) ≥ 2. This means that is the same line as m 2 .
So now we have proved, under the assumption that P 32 = P 22 ,that is the
same line as m 2 . By Lemma 2.9, P 23 lies on , and therefore on m 2 .Italso
lies on 2 and m 3 . Therefore, P 22 = P 23 . This means that 2 is tangent to C
at P 22 .Since P 32 = P 22 means that m 2 is also tangent to C at P 22 ,wehave
2 = m 2 , contradiction. Therefore, P 32 = P 22 (under the assumption that

=0).