If \(p=13\), there is just one value of \(j\) for which the Hasse invariant of the corresponding curve is 0. Find it. Answer: \(j=5 \pmod{13}\).
Solution
Work over the algebraically closed field \(k\) of characteristic \(13\).
1. For the Legendre curve\[ E_\lambda: \qquad y^2=x(x-1)(x-\lambda), \qquad \lambda\ne0,1, \]the Hasse invariant vanishes exactly when\[ S_{13}(\lambda)=0, \]where\[ \boxed{ S_{13}(T) = 1+10T+4T^2+10T^3+4T^4+10T^5+T^6 } \]in \(\FF_{13}[T]\).
For an odd prime \(p\) and \[ E:y^2=f(x) \] with \(f\) a monic cubic, the Hasse invariant is the coefficient of \(x^{p-1}\) in \[ f(x)^{(p-1)/2}. \] For \[ f(x)=x(x-1)(x-\lambda) \] this coefficient is, up to the sign \((-1)^{(p-1)/2}\), \[ S_p(\lambda) = \sum_{i=0}^{(p-1)/2} \binom{(p-1)/2}{i}^2\lambda^i. \] Here \((p-1)/2=6\), so the sign is \(1\). Reducing \[ \binom60^2,\binom61^2,\ldots,\binom66^2 \] modulo \(13\) gives \[ 1,10,4,10,4,10,1, \] which is the displayed polynomial.
2. In \(\FF_{13}[T]\) one has the identity\[ \boxed{ 256(T^2-T+1)^3 -5T^2(T-1)^2 = -4S_{13}(T). } \]
Since \(256\equiv9\pmod{13}\), direct expansion of the left-hand side gives \[ -4T^6-T^5-3T^4-T^3-3T^2-T-4. \] On the other hand, \[ -4S_{13}(T) = -4T^6-T^5-3T^4-T^3-3T^2-T-4 \] in \(\FF_{13}[T]\).
3. Every elliptic curve in characteristic \(13\) whose Hasse invariantvanishes has\[ \boxed{j=5}. \]
Over the algebraically closed field \(k\), every elliptic curve is isomorphic to a Legendre curve \(E_\lambda\): its three nonzero points of order \(2\) give three distinct branch points, which can be moved to \(0,1,\lambda\).
For such a curve, step 1 gives \[ S_{13}(\lambda)=0. \] Also \[ S_{13}(0)=S_{13}(1)=1, \] so a root of \(S_{13}\) is never \(0\) or \(1\). Dividing the identity in step 2 by \[ \lambda^2(\lambda-1)^2 \] and using Hartshorne’s Legendre formula \[ j(E_\lambda) = 256\frac{(\lambda^2-\lambda+1)^3} {\lambda^2(\lambda-1)^2} \] therefore gives \[ j(E_\lambda)=5. \]
4. There exists an elliptic curve with vanishing Hasse invariant incharacteristic \(13\).
The polynomial \(S_{13}\) has degree \(6\), so it has a root \(\lambda\in k\) because \(k\) is algebraically closed. Step 3 shows that \(\lambda\ne0,1\), so \(E_\lambda\) is smooth, and step 1 says its Hasse invariant vanishes.
5. Q.E.D.
Step 3 shows that no \(j\)-value other than \(5\) can occur, while step 4 shows that this value is attained.