p-ADIC HEIGHTS OF HEEGNER POINTS AND ANTICYCLOTOMIC ...

32 JENNIFER S. BALAKRISHNAN, MIRELA ÇIPERIANI, AND WILLIAM STEIN
So we see that we have b2 ≡ 0 (mod 3). Thus we go to the next coefficient; for σ ∈ Gal(K2/K) the
element of order p 2 fixed in Step 3 of Algorithm 8.1 we have:
Consequently, we find that
which gives b2 ≡ 6 (mod 9) and
〈z2, z2〉 K2 = 1 + 3 + 3 3 + 3 6 + 2 · 3 7 + O(3 8 )
〈z2, σz2〉 K2 = 2 + 2 · 3 2 + 2 · 3 3 + 3 6 + 3 7 + O(3 8 )
〈z2, σ 2 z2〉 K2 = 2 + 2 · 3 + 3 3 + 2 · 3 4 + 2 · 3 5 + 3 7 + O(3 8 )
〈z2, σ 3 z2〉 K2 = 1 + 3 + 3 2 + 2 · 3 3 + 2 · 3 6 + 3 7 + O(3 8 )
〈z2, σ 4 z2〉 K2 = 2 + 3 + 2 · 3 2 + 2 · 3 4 + 3 5 + 3 6 + 2 · 3 7 + O(3 8 )
〈z2, σ 5 z2〉 K2 = 〈z2, σ 4 z2〉 K2
〈z2, σ 6 z2〉 K2 = 〈z2, σ 3 z2〉 K2
〈z2, σ 7 z2〉 K2 = 〈z2, σ 2 z2〉 K2
〈z2, σ 8 z2〉 K2 = 〈z2, σz2〉 K2 .
〈c2, σc2〉 K2 ≡ 5 (mod 9)
〈c2, σ 2 c2〉 K2 ≡ 2 (mod 9)
〈c2, σ 3 c2〉 K2 ≡ 7 (mod 9)
〈c2, σ 4 c2〉 K2 ≡ 8 (mod 9),
b3 ≡ 2〈c2, σc2〉 K2 + 8〈c2, σ 2 c2〉 K2 + 3〈c2, σ 3 c2〉 K2 + 5〈c2, σ 4 c2〉 K2
≡ 6 (mod 9).
So we compute b4 (mod 9):
Then we find that
and finally
b4 ≡ 7〈c2, σc2〉 K2 + 8〈c2, σ 2 c2〉 K2 + 6〈c2, σ 3 c2〉 K2 + 6〈c2, σ 4 c2〉 K2
≡ 6 (mod 9).
b5 ≡ 2〈c2, σc2〉 K2 + 3〈c2, σ 2 c2〉 K2 + 6〈c2, σ 3 c2〉 K2 + 〈c2, σ 4 c2〉 K2
≡ 3 (mod 9),
b6 ≡ 〈c2, σc2〉 K2 + 7〈c2, σ 2 c2〉 K2 + 〈c2, σ 3 c2〉 K2
≡ 8 (mod 9).
(mod 9)
(mod 9)
(mod 9)
(mod 9)
Hence, we have now found an example where the Λ-adic regulator R is the product of a unit and a
distinguished polynomial of degree 6 in Z3[[T ]].