Is there any theorem like implicit function theorem in $\mathbb{Q}$ ? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T11:41:55Z http://mathoverflow.net/feeds/question/110544 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/110544/is-there-any-theorem-like-implicit-function-theorem-in-mathbbq Is there any theorem like implicit function theorem in $\mathbb{Q}$ ? Luke 2012-10-24T14:23:04Z 2012-11-08T06:31:43Z <p>My qeustion is that, is there any theorem like implicit function theorem in $\mathbb{Q}$ ?</p> <p>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 $\mathbb{Q}$.</p> <p>For example, $x^2+y^2=1$ does not satisfy the condition but for $x+y=0$ it holds.</p> <p>And how about the same question in p-adic field $\mathbb{Q}_{p}$?</p> http://mathoverflow.net/questions/110544/is-there-any-theorem-like-implicit-function-theorem-in-mathbbq/110546#110546 Answer by Margaret Friedland for Is there any theorem like implicit function theorem in $\mathbb{Q}$ ? Margaret Friedland 2012-10-24T14:59:37Z 2012-10-24T14:59:37Z <p>Look up Hensel's lemma. E.g., in the following form ($K$ is a field, without further assumptions):</p> <p>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 $h(0,Y)=q(Y)$.</p> <p>(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) </p> <p>More information, including the $p$-adic version, can be found here:</p> <p><a href="http://mathoverflow.net/questions/15673/an-unfamiliar-to-me-form-of-hensels-lemma" rel="nofollow">http://mathoverflow.net/questions/15673/an-unfamiliar-to-me-form-of-hensels-lemma</a></p> <p>(especially Wanderer's answer).</p> http://mathoverflow.net/questions/110544/is-there-any-theorem-like-implicit-function-theorem-in-mathbbq/110554#110554 Answer by Laurent Berger for Is there any theorem like implicit function theorem in $\mathbb{Q}$ ? Laurent Berger 2012-10-24T16:02:45Z 2012-10-24T17:41:50Z <p>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 ( <a href="http://en.wikipedia.org/wiki/Hilbert%27s_irreducibility_theorem" rel="nofollow">http://en.wikipedia.org/wiki/Hilbert%27s_irreducibility_theorem</a> ) 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.</p> <p>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.</p> http://mathoverflow.net/questions/110544/is-there-any-theorem-like-implicit-function-theorem-in-mathbbq/111784#111784 Answer by Will Sawin for Is there any theorem like implicit function theorem in $\mathbb{Q}$ ? Will Sawin 2012-11-08T06:31:43Z 2012-11-08T06:31:43Z <p>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 $y$.</p>