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Consider a homogeneous polynomial, f, of total degree n in n variables, with coefficients in a prime order finite field, GF(p).

Are there any general rules regarding the existence of nontrivial roots of f in an extension field of GF(p)? I conjecture that there will exist nontrivial roots in GF(p)^k if k and n are not relatively prime, but I can't find any real proof.

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    $\begingroup$ Take a linear form L in four variables whose coefficients form a basis of GF(p^4)/GF(p) and take the form of degree four which is the product of the conjugates of L. This won't have points in GF(p^2) and is a counterexample to your conjecture. $\endgroup$
    – Felipe Voloch
    Commented May 16, 2013 at 20:25
  • $\begingroup$ My answer given yesteray is wrong. One can choose $p$ variables to have non-zero value, and set all others to be zero to get a non-trivial solution. Should I delete it? $\endgroup$
    – P Vanchinathan
    Commented May 19, 2013 at 2:58
  • $\begingroup$ I don't understand how Felipe's suggestion would work. Those four coefficients can't be linearly independent over $GF(p^2)$, so won't any of the factors of the quartic form have zeros over $GF(p^2)$? $\endgroup$
    – Jyrki Lahtonen
    Commented May 24, 2013 at 10:58

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For a prime $p\ne2$, take the homogeneous polynomial $\sum_{j=1}^{p^2-1} X_j^{p^2-1}$ of degree $p^2-1$ in as many variables. For any non-zero element of $\mathbf{F}_{p^2}$ each term evaluates to 1 and so the given polynomial is the constant function $-1$ and has no zero. The number of variable $p^2-1$ is a multiple of 2, and yet has no solution in the quadratic extension of the prime field, providing another counterexample.

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    $\begingroup$ The above is wrong. One can choose $p$ variables to have non-zero value, and set all others to be zero to get a non-trivial solution. I am sorry to have rushed in without fully verifying. $\endgroup$
    – P Vanchinathan
    Commented May 18, 2013 at 10:32

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