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

$$a_4 = -\frac{c_4}{2^4\cdot 3},$$

$$a_6 = -\frac{c_6}{2^5 \cdot 3^3}.$$

In our case, we obtain the curve

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

Let

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

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

$${\mathrm{Frob}}_5(zdx) \equiv (5xz^2 + (5x +5x^2)z^4) zdx \pmod{25}$$

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

Step 4.

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

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

as

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

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

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

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

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

leaves us with

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

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

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

leaves us with

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

Next, we reduce

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

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

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

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

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

leaves us with

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

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

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

leaves us with

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

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.