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    The primes where the Fermat cubic has Hasse invariant $0$ have density $\frac{1}{2}$

    P-AGH4414FERMATCUBICHASSE

    Area
    Algebraic Geometry
    Topics
    Curves, Elliptic Curves
    Status
    draft

    The Fermat curve \(X: x^3+y^3=z^3\) gives a nonsingular curve in characteristic \(p\) for every \(p \neq 3\). Determine the set \(\mfp=\ts{p \neq 3 \st X_{(p)} \text{ has Hasse invariant } 0}\), and observe (modulo Dirichlet’s theorem) that it is a set of primes of density \(\frac{1}{2}\).

    Solution

    Write the homogeneous equation as \[ F=x^3+y^3-z^3. \]

    1. By Proposition IV.4.21, the Hasse invariant of \(X_{(p)}\) is nonzeroif and only if the coefficient of\[ (xyz)^{p-1} \]in \(F^{p-1}\) is nonzero modulo \(p\).

    This is exactly Proposition IV.4.21 applied to the plane cubic \(F=0\).

    2. If \(p\equiv2\pmod3\), then the coefficient in step 1 is zero.

    Every monomial occurring in \[ (x^3+y^3-z^3)^{p-1} \] has each exponent divisible by \(3\). For the monomial \[ x^{p-1}y^{p-1}z^{p-1} \] to occur, one must therefore have \[ 3\mid p-1. \] If \(p\equiv2\pmod3\), this is impossible. Hence the required coefficient is zero, so the Hasse invariant vanishes.

    3. If \(p\equiv1\pmod3\), then the coefficient in step 1 is nonzero.

    Put \[ m=\frac{p-1}{3}. \] There is exactly one way to obtain \[ x^{p-1}y^{p-1}z^{p-1}=x^{3m}y^{3m}z^{3m} \] from the multinomial expansion: choose \(x^3\), \(y^3\), and \(-z^3\) each \(m\) times. Its coefficient is \[ (-1)^m\frac{(p-1)!}{(m!)^3}. \] Since \(0<m<p\), neither \(m!\) nor \((p-1)!\) is divisible by \(p\). Therefore this coefficient is nonzero in \(\FF_p\), and so the Hasse invariant is nonzero.

    4. Consequently\[ \boxed{ \mfp=\{p\text{ prime}:p\equiv2\pmod3\}. } \]

    The excluded prime is \(p=3\). Every other prime is congruent to \(1\) or \(2\) modulo \(3\). Steps <1>2–<1>3 show that the Hasse invariant vanishes exactly in the second residue class. This includes \(p=2\).

    5. The set \(\mfp\) has density \(1/2\).

    By Dirichlet’s theorem on primes in arithmetic progressions, the primes are equidistributed between the two reduced residue classes \(1\) and \(2\) modulo \(3\). The single excluded prime \(3\) has density zero. Hence \[ \delta(\mfp)=\frac{1}{\varphi(3)}=\frac12. \]

    6. Q.E.D.

    Steps <1>4–<1>5 determine the required set of primes and its density.