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Suppose $K$ is a finite field and $q \in K[x_1,\dots,x_n]$. In general, is there anything we can say about which extension fields of $K$ the polynomial $q$ has roots?

In the special case where $n = 1$, we could write $q$ as a product of irreducible polynomials in $K$. Suppose $a_1,\dots,a_m$ are the degrees of these polynomials. Then, since a degree $a$ irreducible polynomial over $K$ has a root in a degree $b$ extension of $K$ if and only if $a | b$, the polynomial $q$ has roots precisely in extension fields whose degrees are a multiple of $a_i$, for some $i$.

I don't believe this logic extends to the multi-variant case, particularly the irreducible degree argument.

As an example, take $K = GF(2)$ and $q(x,y,z,w) = 1 + w (x+y) (x+z) (y+z)$. Then in order to have a root in a field $F$, we must have $x \ne y$, $x \ne z$, $y \ne z$, and $w = (y+z)^{-1} (x+z)^{-1} (x+y)^{-1}$. In particular, $x,y,$ and $z$ must be distinct, so $q$ has no roots in $GF(2)$, but any other field of characteristic $2$ would work. However, this set of extension fields does not fit the proposed form in the single-variable case, i.e. there is not a finite set of integers $a_1,\dots,a_m$ such that $q$ has roots precisely in extension fields whose degrees are a multiple of some $a_i$.

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  • $\begingroup$ The set of roots of your polynomial is an affine algebraic variety $\subset \overline{K}^n$. Fix $L/K$ and $a_j \in L$ as you want, then $q(a_1,\ldots,a_{n-1},t)$ is a polynomial of $L[t]$ and it has its roots in some finite extension of $L$. See discussions about counting points on elliptic curves over finite fields. $\endgroup$ – reuns Sep 5 '17 at 1:09
  • $\begingroup$ Unless something strange and degenerate happens, a multivariable polynomial defined over a finite field will have roots in all large enough extension fields. If the variety (see reuns' comment) is irreducible, then the number of points on a d-dimensional variety differs with coordinates in $\Bbb{F}_{q^n}$ differs from $q^{dn}$ by at most $Kq^{n(d-1/2)}$. Here $K$ is a constant (depending on the variety but not on $n$). In particular the number of solutions is positive when $q^{n/2}>K$. $\endgroup$ – Jyrki Lahtonen Sep 11 '17 at 18:26

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