# ag.algebraic geometry – Can a quadratic polynomial in \$ mathbb {F} _ {p}[x_{1},dots,x_{d}]\$ a linear divide?

To let $$2 leq d ll p$$ and $$p$$ is a prime number. Consider the (non-trivial) polynomials $$M (x_ {1}, dots, x_ {d}) = x ^ {2} _ {1} + dots + x ^ {2} _ {d} -c$$ and $$L (x_ {1}, dots, x_ {d}) = a_ {1} x_ {1} + dots + a_ {d} x_ {d} + a_ {0}$$ in the $$mathbb {F} _ {p} (x_ {1}, dots, x_ {d})$$, Is it possible that $$M vert L$$in that sense that $$M (x) = 0 Rightarrow L (x) = 0$$ for all $$x in mathbb {F} _ {p} ^ {d}$$?

Note 1: The answer to this question is wrong when $$d = 1$$, since $$(1 + x ^ {2}) (x-x ^ {3} + x ^ {5}) = 2x mod mathbb {F} _ {7}$$, In fact, this can be shown by direct calculation $$(x ^ {2} -c) f (x) = 1$$ for some $$f$$ then and only if $$c neq 0$$ and $$x ^ {2} = c mod p$$ has a solution.

Note 2: I have an analytical approach to this question. To let $$A$$ be the sentence of $$x in mathbb {F} _ {p} ^ {d}$$ so that $$M (x) = 0$$, Consider the exponential sum
$$frac {1} { vert A vert} sum_ {x in A} e ^ { frac {2 pi i L (x)} {p}},$$
Where $$L (x)$$ is considered as an embedding of $$mathbb {F} _ {p}$$ to $${0,1, dots, p-1 }$$, You can calculate that this sum is $$O (p ^ {- frac {d-2} {2}})$$ when $$L$$ is not trivial (see, for example, Lemma 3.3 of https://arxiv.org/abs/math/0703504v2). This shows that the answer to the question is negative $$d geq 3$$ (and $$d ll p$$). But is there a more direct way to answer this question, since it is an algebraic geometry question (I'm not an expert in this field)?