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)?