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Example: Computing the Matrix of Frobenius for $ 37A$ at $ p = 5$

Let $ p = 5$ and consider the elliptic curve $ 37A$ , with minimal model $ y^2+y=x^3-x$ .

Step 1.
Put the elliptic curve into Weierstrass form $ y^2 = x^3 + a_4x + a_6$ , via the transformation

$\displaystyle a_4$ $\displaystyle = -\frac{c_4}{2^4\cdot 3},$    
$\displaystyle a_6$ $\displaystyle = -\frac{c_6}{2^5 \cdot 3^3}.$    

In our case, we obtain the curve

$\displaystyle y^2 = x^3 - x + \frac{1}{4}.$

Let

$\displaystyle Q(x)= x^3 - x+ \frac{1}{4}.$

Step 2.
Fix the precision $ N$ and compute $ M$ . In our case, $ N = 2$ and $ M = 3$ .
Step 3.
Compute the action of Frobenius on the two differentials $ z dx$ and $ xz dx$ as an element of $ \mathbb{Z}_p[x,y,z]/(y^2-Q(x),yz-1)$ , with a precision of $ N$ digits. Furthermore, group the terms of $ {\mathrm{Frob}}_p(x^iz dx)$ as $ \sum
(p^{k+1}c_{i,k,j}z^j)zdx$ , where the $ c_{i,k,j}$ are in $ \mathbb{Z}_p[x]$ of degree less than 3.

In our case, we compute

$\displaystyle {\mathrm{Frob}}_5(zdx)$ $\displaystyle \equiv (5xz^2 + (5x +5x^2)z^4) zdx \pmod{25}$    
$\displaystyle {\mathrm{Frob}}_5(xz dx)$ $\displaystyle \equiv (10 + 10x + 5x^3 + (20 + 5x + 15x^2)z^2 + (10 + 20x + 15x^2)z^4) zdx \pmod{25}.$    

Step 4.
Now we must reduce the differentials. We want to write each of the

$\displaystyle F_{5,i}={\mathrm{Frob}}_5(x^izdx)$

as

$\displaystyle \left(a_{0i} + a_{1i}x\right)zdx + \sum d(x^iy^j) = \left(a_{0i} + a_{1i}x\right)zdx$

in $ H^1_{\textrm{MW}}(C'_Q)^-$ . We begin with

$\displaystyle F_{5,0} \equiv (5xz^2 + (5x
+5x^2)z^4) zdx \pmod{25}$

and compute the appropriate list of differentials:
$ i$ $ j$ $ d(x^iz^j)\pmod{25}$
0 $ 1$ $ (13z^2 + 11z^2x^2) zdx$
1 $ 1$ $ (12 + 16z^2 + 24z^2x) zdx$
2 $ 1$ $ (13x + 16z^2x + 24z^2x^2)zdx$
0 $ 3$ $ (14z^4 + 8z^4x^2)zdx$
1 $ 3$ $ (9z^2 + 23z^4 + 22z^4x)zdx$
2 $ 3$ $ (10z^2x + 23z^4x + 22z^4x^2)zdx$

Thus we wish to write $ (5x+5x^2)z^4$ as a linear combination of $ 14z^4 + 8z^4x^2$ , $ 23z^4 + 22z^4x$ , and $ 23z^4x + 22z^4x^2$ , all modulo 25 (we may ignore the lower powers of $ z$ present in the differentials, as we will take care of them in the steps to come). We find that taking

$\displaystyle F_{5,0} -
5d(z^3)-10d(xz^{3})-20d(x^2z^{3})\pmod{25}$

leaves us with

$\displaystyle (10+5x)z^2\; zdx.$

Now we wish to write $ (10+5x)z^2$ as a linear combination of $ 13z^2 + 11z^2x^2$ , $ 16z^2 + 24z^2x$ , and $ 16z^2x + 24z^2x^2$ , modulo 25. We find that taking

$\displaystyle (10+5x)z^2\;zdx -
10d(z)-5d(xz)-10d(x^2z)$

leaves us with

$\displaystyle (15+20x)zdx.$

Next, we reduce

$\displaystyle F_{5,1}\equiv (10 + 10x +
5x^3 + (20 + 5x + 15x^2)z^2 + (10 + 20x + 15x^2)z^4) zdx
\pmod{25}.$

Note that this has an $ x^3 zdx$ term, so we take care of this first:

$\displaystyle F_{5,1} - \frac{1}{3}d(x^4z) = (13 + 2x +
(13 + 10x + 7x^2)z^2 + (10 + 20x + 15x^2)z^4)\;zdx.$

Now we proceed as in the case of $ F_{5,0}$ , and we wish to write $ (10 + 20x + 15x^2)z^4$ as a linear combination of $ 14z^4 + 8z^4x^2$ , $ 23z^4 + 22z^4x$ , and $ 23z^4x + 22z^4x^2$ , all modulo 25. We find that taking

$\displaystyle (13 + 2x + 13z^2 + 10z^2x + 7z^2x^2 +
10z^4 + 20z^4x + 15z^4x^2)z dx -
10d(z^{3})-15d(xz^{3})-5d(x^2z^{3})$

leaves us with

$\displaystyle (13 + 2x +
(3 + 10x + 7x^2)z^2)\;zdx.$

Finally, we wish to write $ (3 + 10x + 7x^2)z^2$ as a linear combination of $ 13z^2 + 11z^2x^2$ , $ 16z^2 + 24z^2x$ , and $ 16z^2x + 24z^2x^2$ , all modulo 25. We find that taking

$\displaystyle (13 + 2x + (3 +
10x + 7x^2)z^2\;zdx-20d(z)-23d(xz)-13d(x^2z)$

leaves us with

$\displaystyle (12+8x)zdx.$

Step 5.
Now we form the matrix $ F$ of the reduced differentials, where each reduced differential gives us a column in the matrix of absolute Frobenius. In our case, we have $ F=\left( \begin{array}{cc}
15 & 12 \\
20 & 8 \end{array} \right)$ .
As a consistency check, we have that $ F$ has trace 23, which is $ a_5$ modulo 25 and determinant $ -120$ , which is $ p = 5$ modulo 25.


next up previous contents
Next: Source Code Up: An Example: Computing Previous: Introduction   Contents
William Stein 2006-10-20