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$ \def \Q {\mathbb Q} \def \x {\boldsymbol x} \def \y {\boldsymbol y} \def \a {\boldsymbol a} \def \b {\boldsymbol b} $This question is inspired by the post "Is there any theorem like implicit function theorem in $ \Q $?" and the discussion thereafter.

Let $ p ( \x , \y ) \in \Q [ \x , \y ] $, where $ \x $ and $ \y $ are finite sequences of indeterminates of lengths $ r $ and $ s $, repectively. Assume that for every $ \a \in \Q ^ r $ there exists $ \b \in \Q ^ s $ such that $ p ( \a , \b ) = 0 $. Can we conclude that there exists $ q ( \x ) \in \Q [ \x ] ^ s $ such that $ p \bigl ( \x , q ( \x ) \bigr ) = 0 $? (EDIT: as noted by Mueller, the requirement should be changed to $ q ( \x ) \in \Q ( \x ) ^ s $; i.e., $ q ( \x ) $ should be required to be a sequence of rational expressions rather than polynomials.)

The question is answered positively by Laurent Berger's post for the case $ r = s = 1 $, as follows:

Write $ p ( x , y ) $ as a product of irreducible polynomials $ p _ i ( x , y ) $. Hilbert's irreducibility theorem implies that there are infinitely many $ a $'s such that $ p _ i ( a , y ) $ is irreducible for every $ i $. If one of them has a solution, it is therefore of degree $ 1 $ in $ y $. Some $ p _ i $ is therefore of degree $ 1 $ in $ y $, which shows that there is some polynomial $ q $ such that $ p \bigl ( x , q ( x ) \bigr ) = 0 $.

Now, since Hilbert's irreducibility theorem also holds for the multivariate case, I wondered whether the statement holds for the more general case. But even if it does, the above argument needs some modifications, since for example the Hilbert set being infinite is not sufficient for concluding that $ p _ i $ is of degree $ 1 $ in the case of more than a single indeterminate. For example, I don't find the argument convincing when $ r = 2 $ and $ p ( x _ 1 , x _ 2 , y ) = ( x _ 1 - x _ 2 ) y ^ 3 + y $. In this example, while there are infinitely many specifications for $ ( x _ 1 , x _ 2 ) $ that make the irreducible factors of $ p $ irreducible in $ \Q [ y ] $ (namely those of the form $ ( a , a ) \in \Q ^ 2 $), one of them is of degree $ 2 $ in $ y $.

I've asked this question as a follow-up comment under the mentioned post, but haven't got any answers after eight months. Since that post is about twelve years old, I think it's unlikely that I get any response to that comment now. So I decided to post a separate question in that regard.

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  • $\begingroup$ Can this even be done over $\mathbb R$, that is, to find a polynomial (rather than just smooth function) on whose graph $p$ vanishes? $\endgroup$ Commented May 10 at 23:52
  • $\begingroup$ @LSpice Of course not. Consider $ p ( x , y ) = x ^ 2 + 1 - y ^ 2 $ for example, which requires $ q ( x ) = \pm \sqrt { x ^ 2 + 1 } $. Yet, Peter Mueller's answer below states that the statement is true for $ \mathbb Q $, if one has $ s = 1 $ and changes the requirement for $ q ( \boldsymbol x ) $ to being a rational function rather than a polynomial, which is in line with my motivation for asking the question. $\endgroup$ Commented May 11 at 13:34
  • $\begingroup$ Re, ah, sure. I thought such a result, if true for $\mathbb Q$, should be all the more true for $\mathbb R$, but I wasn't thinking about how hard it is for equations to have rational, as opposed to real, solutions. $\endgroup$ Commented May 11 at 18:16

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Berger's answer concerning the case $r=s=1$, quoted in the OP's question, actually gives a weaker result. It only shows that $q(\mathbf x)$ is a rational function in $\mathbf x$. This is unavoidable in view of $p(x,y)=(x^2+1)y-1$.

So the OP's question was changed to ask whether under the given assumptions $q(\mathbf x)$ can be chosen in $\mathbb{Q}(\mathbf x)^s$, where Berger's argument (see below) gives a positive answer for $s=1$ and any $r$.

However, for $s=4$ this relaxation does not hold: Set $p(\mathbf x,\mathbf y)=A(\mathbf x)-B(\mathbf y)$ where \begin{align} A(\mathbf x) &= 1+x_1^2+x_2^2+x_3^2+x_4^2\\ B(\mathbf y) &= y_1^2+y_2^2+y_3^3+y_4^2. \end{align} By Lagrange's four-square theorem, for any $a\in\mathbb Q^4$ there is $b\in\mathbb Q^4$ such that $A(a)=B(b)$. However, in On the representation of rational functions as sums of squares Cassels proves that $A(\mathbf x)$ is not a sum of the squares of $4$ rational functions in $x_1, x_2, x_3, x_4$ over $\mathbb R$.

For completeness, we give more details on Berger's argument for the case $s=1$ and any $r$. Let $c(\mathbf x)=c(x_1,\dots,x_r)$ be the leading coefficient of $p(\mathbf x, y)$ as a polynomial in $y$. Let $p=p_1\cdots p_n$ with $p_i(\mathbf x, y)\in\mathbb Q[\mathbf x][y]$ irreducible over $\mathbb Q(\mathbf x)$. Note that each $p_i$ has positive $y$-degree. Now a version of Hilbert's irreducibility theorem (see e.g. Section 4.4 in Schinzel's book

Polynomials with special regard to reducibility or Chapter 13 in Fried, Jarden: Field arithmetic) yields $a\in\mathbb Q^r$ such that each $p_i(a, y)$ is irreducible and $c(a)\ne0$. So each $p_i(a, y)$ has the same $y$-degree as $p_i(\mathbf x, y)$. However, as there is $b$ with $p(a,b)=0$, we get $p_i(a, b)=0$ for some $i$. Hence $p_i(\mathbf x, y)$ has $y$-degree $1$, and the result follows.

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There exists a negative answer for $s=2$ conditional on the parity conjecture (a consequence of the BSD conjecture). So this shows (conditionally) that the positive statement in the $s=1$ case by Peter Mueller, building on Berger's argument, is sharp.

Cassels and Schinzel showed that the elliptic curve

$$y^2 = x^3- (7+7 t^4)^2 x$$

has no points over $\mathbb Q(t)$ except for 2-torsion points, but has root number $-1$ when $t$ is specalized to every rational value. The 2-torsion points are exactly the points with $y=0$, which we can get rid of by considering the equation

$$ (1 + (x^3 - (7 + 7 t^4 )^2 x )y)^2 = x^3- (7+7 t^4)^2 x$$

which has no solutions in $\mathbb Q(t)$ so the conclusion of the implicit function theorem fails, taking $r=1$, $s=2$, $t$ the first set of indeterminates and $x$ and $y$ the second set of indeterminates.

However, since the root number of the original curve is $-1$, conditionally on the parity conjecture the rank of its group of rational points is odd, and in particular nonzero, so it has infintely many rational points. Hence the modified curve, which is the same except removing finitely many rational points, also has infintely many rational points. So, conditionally on this conjecture, the hypothesis of the implicit function theorem is satisfied, giving the desired contradiction.

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