TITLE: Is there any theorem like implicit function theorem in $\mathbb{Q}$ ?
QUESTION [5 upvotes]: My qeustion is that,
is there any theorem like implicit function theorem in $\mathbb{Q}$ ?
More precisely, let $p(\bar{x},\bar{y})$ be in $\mathbb{Z}[\bar{x},\bar{y}]$ such that in $\mathbb{Q}$, for any $\bar{a}$, there is a solution of $p(\bar{x},\bar{a})$. Then for some polynomial(or rational polynomial) $q(\bar{y})$ with $\mathbb{Q}$ coefficients, $p(q(\bar{y}),\bar{y})=0$ holds in the rational polynomial fields over $\mathbb{Q}$.
For example, $x^2+y^2=1$ does not satisfy the condition
but for $x+y=0$ it holds.
And how about the same question in p-adic field $\mathbb{Q}_{p}$?

REPLY [5 votes]: The same question in $\mathbb Q_p$ is false. For instance, if $p \neq 2$, let $\alpha \in \mu_{p-1}$ be a primitive root of unity. Then $(x^2-y)(x^2-py)(x^2-\alpha y)(x^2-p \alpha y)$ has a solution for each $y$, but you cannot make that solution a polynomial in $y$.