Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $\mathbb{Q}_{p}$?

share|improve this question
3  
Why the calligraphy? Why not \mathbb to produce $\mathbb{Q}$, etc.? –  Todd Trimble Oct 24 '12 at 15:07
    
Thank you:) I changed them. –  Luke Oct 25 '12 at 2:22

3 Answers 3

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 $y$.

share|improve this answer
    
Thank you for your notice. It is good. But now I cann't understand how your example works because of my limiting knowledge about $\mathbb(Q)_p$. It looks like related with $\mathbb{Q}_p/\mathbb{Q}_p^2$. –  Luke Nov 22 '12 at 8:47
    
Yes, this is about that quotient of multiplicative groups. For $p$ odd, the quotient is the Klein four group, with the coset representatives described in this answer. For a proof, see an appropriate algebraic number theory textbook. –  Will Sawin Nov 22 '12 at 19:25

Here's what I think happens over $\mathbb{Q}$. Write your polynomial $P(X,Y)$ as a product of irreducible polynomials $P_i(X,Y)$. Hilbert's irreducibility theorem ( http://en.wikipedia.org/wiki/Hilbert%27s_irreducibility_theorem ) tells you that there are infinitely many $a$'s such that $P_i(X,a)$ is irreducible for every $i$. If one of them has a solution, it is therefore of degree $1$ in $X$. Some $P_i$ is therefore of degree $1$ in $X$, which answers your question.

EDIT: it does not answer the question but rather shows that there is some polynomial $Q$ such that $P(Q(Y),Y)=0$ which is more reasonable, since then $P(Q(a)),a)=0$. This should have been the question.

share|improve this answer
    
You are right. I have to change my question in the form P(Q(Y),Y)=0. Thank you! –  Luke Oct 25 '12 at 2:31

Look up Hensel's lemma. E.g., in the following form ($K$ is a field, without further assumptions):

Let $f \in K[[X]][Y]$ be monic and such that $f(0,Y)=p(Y)q(Y)$, where $p(Y),q(Y) \in K[Y]$ are relatively prime and non-constant, of degrees respectively $r$ and $s$. Then there exist two uniquely determined polynomials $g,h \in K[[X]][Y]$, of degrees respectively $r$ and $s$, such that $f=gh$, with $g(0,Y)=p(Y)$ and $h(0,Y)=q(Y)$.

(after Hefez, Abramo: Irreducible plane curve singularities. Real and complex singularities, 1–120, Lecture Notes in Pure and Appl. Math., 232, Dekker, New York, 2003)

More information, including the $p$-adic version, can be found here:

An unfamiliar (to me) form of Hensel's Lemma

(especially Wanderer's answer).

share|improve this answer
    
Hmm, thank you. I'll check Hensel's lemma! –  Luke Oct 25 '12 at 2:29
    
This is an attempt to answer your general question, ``is there any theorem like implicit function theorem in $\mathbb{Q}$ ?", rather than the specific one that follows. –  Margaret Friedland Oct 25 '12 at 14:56

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.