# An algebraic version of the implicit function theorem for integers

$$\def \x {\boldsymbol x} \def \a {\boldsymbol a} \def \Z {\mathbb Z}$$ The famous version of the implicit function theorem (IFT) starts with the assumption of continuous differentiability on the conditions and ends with claiming the same for the solution. There are versions replacing continuous differentiability with varying degrees of smoothness, and show that the smoothness of the conditions results in the smoothness of the solution. I'm looking for a version of the theorem with some sort of algebraic restriction on the conditions and the solution. Also, I'm only interested in the case where we're looking for integer values for our variables.

So, let's say we have $$k$$ independent variables $$x _ 1 , \dots , x _ k$$, together abbreviated as $$\x$$, and a dependent variable $$y$$. We have $$l$$ conditions that are given by $$p _ i ( \x , y ) = 0$$ for $$1 \le i \le l$$, where $$p _ i ( \x , y ) \in \Z [ \x , y ]$$ for each $$i$$. For more simplicity, we may assume that $$\gcd _ { \Z [ \x , y ] } \bigl ( p _ 1 ( \x , y ) , \dots , p _ l ( \x , y ) \bigr ) = 1$$. We also have a condition that guarantees the single-valuedness of the solution (and more): for every $$\a \in \Z ^ k$$, there are $$c , d \in \Z$$ (depending on $$\a$$) such that $$\gcd ( c , d ) = 1$$ and $$\gcd _ { \Z [ y ] } \bigl ( p _ 1 ( \a , y ) , \dots , p _ l ( \a , y ) \bigr ) = c y + d$$. The domain of the function we're looking for is the algebraic set $$D = \left \{ \a \in \Z ^ k \bigm | \exists b \in \Z \forall i \ p _ i ( \a , b ) = 0 \right \}$$. Let $$I$$ be the ideal (in $$\Z [ \x , y ]$$) generated by all the $$p _ i$$ and $$V$$ be the vanishing set (in $$\Z ^ { k + 1 }$$) of $$I$$. The question is whether the following statement must hold, which is saying that the solution can be explicitly given as a rational function restricted to $$D$$.

There are $$r ( \x ) , s ( \x ) \in \Z [ \x ]$$ such that $$s ( \a ) \ne 0$$ for all $$\a \in D$$ and $$V = \left \{ \left ( \a , \frac { r ( \a ) } { s ( \a ) } \right ) \Bigm | \a \in D \right \}$$.

I considered many examples, and the above statement did hold in all of them which satisfied the assumptions, so I conjectured that it must always be true. Of course, in all of my examples, $$k$$ and $$l$$ weren't that large, and I don't know whether the statement may fail for larger $$k$$ and $$l$$. Yet, I think that the condition on existence of $$c$$ and $$d$$ for each $$\a$$ seems to be very strong, as is shows that the polynomials $$p _ i ( \a , y )$$ can only have a single common rational root, let alone a common integer root.

As is noted here, when $$k = 1$$, $$D$$ must be finite, and then one can easily find suitable candidates for $$r ( x )$$ and $$s ( x )$$. As an example for larger $$k$$, consider $$k = 2$$, $$l = 2$$, $$p _ 1 ( \x , y ) = ( y - x _ 1 ) ^ 2 - 1$$ and $$p _ 2 ( \x , y ) = ( y - x _ 2 ) ^ 2 - 4$$. Here, we have $$D = \left \{ ( a _ 1 , a _ 2 ) \in \Z ^ 2 \bigm | a _ 2 - a _ 1 \in \{ \pm 1 , \pm 3 \} \right \}$$, and one can see that $$r ( \x ) = x _ 2 ^ 2 - x _ 1 ^ 2 - 3$$ and $$s ( \x ) = 2 ( x _ 2 - x _ 1 )$$ works (because $$p _ 1 ( \x , y ) - p _ 2 ( \x , y ) = s ( \x ) y - r ( \x ) \in I$$). Incidentally, in all the examples I checked, I could choose suitable $$r$$ and $$s$$, so that $$s ( \x ) y - r ( \x ) \in I$$. I don't know if this would always be possible, even if the original statement is true.

I tried searching the literature for something similar to the above problem. It seems that there hasn't been much interest in problems like this; or maybe it is too difficult of a problem and I can't see it. Please let me know if there is something related that would help me. Also, feel free to edit the tags, since while I found them related to the problem at hand, I'm not sure whether I've chosen them well enough.

• Looking at your conjecture I suspect it is false, for a very general choice of the polynomials $p_i$. Rational functions are far more rigid than the functions that are admissible from the implicit function theorem. I do not have a proof however, though I suspect someone more well-read in algebraic geometry than I am will be able to answer your question in the negative. Dec 13, 2023 at 15:59
• @StanleyYaoXiao At first, I thought so, too. But investigating more and more examples, I came to realize that the assumption $\gcd \bigl ( p _ 1 ( \boldsymbol a , y ) , \dots , p _ l ( \boldsymbol a , y ) \bigr ) = c y + d$ is really restrictive, and became more confident that the conjecture may hold. In any case, refuting the conjecture would be nice as well, and might give me some insight on the problem, and what might be true instead. Dec 13, 2023 at 20:51
• related(?): mathoverflow.net/a/52711/137249 Jan 2 at 7:56
• @HypatiaduBois-Marie I must confess that I couldn't see how the link you provided is related to this problem. I tried to find the relation several times, but failed. Of course, I'm not surprised, as neither algebraic geometry nor number theory is my area of expertise. Feb 12 at 19:32