$ \let \ovr \overline \def \Z {\mathbb Z} \def \C {\mathbb C} \def \F {\mathbb F} \def \P {\mathcal P} \def \x {\boldsymbol x} \def \a {\boldsymbol a} $ Let $ \P = \{ p _ i ( \x , y ) \} _ { i = 1 } ^ l \subseteq \Z [ \x , y ] $ be a finite set of polynomials, where $ \x = ( x _ 1 , \dots , x _ k ) $. Given any ring $ R $ and any index $ i $, let $ p _ i ^ R : R ^ { k + 1 } \to R $ denote the polynomial function corresponding to $ p _ i $. We say that $ \P $ defines a partial function over $ R $, or $ \P $ has the implicit function property (IFP) over $ R $ whenever for every $ \a \in R ^ k $ there is at most one $ b \in R $ such that $ p _ i ^ R ( \a , b ) = 0 $ for all indices $ i $. An instance of the implicit function theorem (IFT) is a statement of the form "if $ \P $ is subject to certain conditions then it has IFP over the specified ring $ R $" (here I'm not concerned with specifying a domain for the implicit function beforehand, and the domain can be taken as the set of those $ \a \in R ^ k $ for which there exists a corresponding $ b \in R $ as above). Fixing the notation that $ q $ will always denote a (positive) power of a prime and $ \F _ q $ being the corresponding Galois field, I'm interested in the following IFT in this post:
Assuming that $ \P $ has IFP over $ \Z _ n $ for every positive integer $ n $, $ \P $ also has IFP over $ \F _ q $.
I still don't know whether the above IFT holds or not, and my attempts only include direct calculations in $ \F _ q $ for small $ q $, mostly $ q = 4 $. As of yet, my attempts have been inconclusive.
The motivation for studying whether the above IFT holds comes from the following observations. Each observation is itself an instance of IFT, with the condition posed on $ \P $ being that $ \P $ has IFP over the rings in a certain class. If the above IFT holds, then, combining with the following observations, one can get $ \P $'s IFP over rings in a wide class only from its IFP over the finite rings $ \Z _ n $.
Observation 1. If $ \P $ defines a partial function over $ \Z _ q $ for all $ q $ then it has IFP over $ \Z $.
This can be proven using a method similar to the answer posted by Eric Wofsey to a related question of mine on MSE, which goes as follows. Fix $ \a \in \Z $ and let $ \hat p _ i ( y ) = p _ i ( \a , y ) $ for all $ i $. If $ b $ is a common root of the $ \hat p _ i $, then it must be a root of $ \hat p ( y ) = \gcd _ { i = 1 } ^ l \hat p _ i ( y ) $. Let $ f ( y ) $ be an irreducible factor of $ \hat p ( y ) $ in $ \Z [ y ] $. By Chebotarev density theorem (or the elementary method proposed by Aphelli in a comment), there are infinitely many primes $ p $ modulo which $ f ( y ) $ splits as a product of distinct linear factors. Since $ \P $ has IFP over $ \Z _ p $ for each such $ p $, we can conclude that $ f ( y ) $ itself must be linear. In other words, $ \hat p ( y ) $ must be a product of linear factors in $ \Z [ y ] $; i.e., all of its roots are rational. If $ \hat p ( y ) $ has two distinct rational roots, choosing a prime $ p $ neither dividing their denominators nor the numerator of their difference, $ \hat p ( y ) $ will have two distinct roots over $ \Z _ p $, which contradicts $ \P $'s IFP over $ \Z _ p $. Therefore, $ \hat p ( y ) $ must be of the form $ e ( c y + d ) ^ m $ for some $ c , d , e \in \Z $ and some positive integer $ m $, with $ \gcd ( c , d ) = 1 $ (since $ \hat p ( y ) $ is defined as a GCD, we can assume, without loss of generality, that $ e , c \ge 0 $, and in case $ c = 0 $, also that $ d = m = 1 $). We show that $ e = m = 1 $ by showing that other cases contradict $ \P $'s IFP over certain $ \Z _ q $, which will complete the proof of the observation. If $ e \ne 1 $, then $ \hat p ( y ) $ will have distinct roots over $ \Z _ p $ with $ p $ being a prime factor of $ e $, as every element of $ \Z _ p $ will be a root. If $ d = 0 $ and $ m \ge 2 $, then $ 0 $ and $ 2 $ will give rise to distinct roots of $ \hat p ( y ) $ in $ \Z _ { 2 ^ m } $ (note that we must have $ c = 1 $ in this case). Also, if $ d \ne 0 $ and $ m \ge 2 $, then choosing a prime $ p $ not dividing $ c $, we can see that $ - d c ^ { p - 1 } $ and $ ( p - d ) c ^ { p - 1 } $ give rise to distinct roots of $ \hat p ( y ) $ in $ \Z _ { p ^ m } $.
Observation 2. If $ \P $ defines a partial function over $ \F _ q $ for all sufficiently large $ q $ then it has IFP over $ \C $.
This can be proven using Lefschetz principle, particularly its first-order version in the usual language for rings, as in such a language, there is a sentence asserting $ \P $ has IFP. Note that for any prime $ p $, $ \ovr { \F _ p } = \bigcup _ { m = 1 } ^ \infty \F _ { p ^ m } $ is an algebraically closed field of characteristic $ p $, and $ \P $'s IFP over sufficiently large $ p ^ m $ implies $ \P $'s IFP over $ \ovr { \F _ p } $. By Lefschetz principle, $ \P $ has IFP over $ \C $, which is an algebraically closed field of characteristic $ 0 $.
Observation 3. In the previous observation, $ \C $ can be replaced by an arbitrary integral domain.
The proof is almost immediate. Given an integral domain $ R $, one can consider the algebraic closure of $ R $'s field of fractions, which being an algebraically closed field, is elementarily equivalent to $ \C $ or $ \ovr { \F _ p } $ for a prime $ p $, as the first-order theory of algebraically closed fields with fixed characteristic is complete. Since $ \P $'s IFP over any ring is inherited by its subrings and it is a first-order property, we're done.
I want to add that I have previously asked two questions related to observation 1, which don't have conclusive answers as of yet: the first question is about a conjecture regarding the explicit form of the implicitly given function, and the second question (which is on MSE) is about some sort of "converse" to observation 1.
