Expressing field inclusions by polynomial equalities on coefficients - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T15:58:59Z http://mathoverflow.net/feeds/question/8160 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/8160/expressing-field-inclusions-by-polynomial-equalities-on-coefficients Expressing field inclusions by polynomial equalities on coefficients Ewan Delanoy 2009-12-08T04:08:49Z 2009-12-08T10:53:24Z <p>Let $A$ be the set of all quadruples $(a_0,a_1,a_2,a_3) \in {\mathbb Q}^4$ such that the polynomial $P=X^4+a_3X^3+a_2X^2+a_1X+a_0$ is irreducible and if $z$ is any root of $P$, then ${\mathbb Q}(z)$ contains $\sqrt{2}$. Is there a nontrivial polynomial relation $R(a_0,a_1,a_2,a_3)=0$ satisified by all $(a_0,a_1,a_2,a_3) \in A$ ? </p> http://mathoverflow.net/questions/8160/expressing-field-inclusions-by-polynomial-equalities-on-coefficients/8164#8164 Answer by Ben Weiss for Expressing field inclusions by polynomial equalities on coefficients Ben Weiss 2009-12-08T05:24:11Z 2009-12-08T05:35:53Z <p>This may ramble a bit much, but I hope it provides some help in how to think about the problem.</p> <p>Let's see what your extension of fields looks like. We have 4 possible extensions (perhaps the same) So that any of them is</p> <p>$\mathbb Q(z_i)$</p> <p>$|$</p> <p>$\mathbb Q\left(\sqrt2\right)$</p> <p>$|$</p> <p>$\mathbb Q$</p> <p>Where $z_i$ ranges of the 4 possible roots $z_1,...,z_4.$ Then $\mathbb Q(z_1)$ is degree 4 (since the polynomial is irreducible), but this polynomial factors into a product of quadratics over $\mathbb Q\left(\sqrt2\right).$ So indeed we've reduced to having only two possible extensions, in that the two roots of the same quadratic generate the same extension over $\mathbb Q(\sqrt2).$</p> <p>However, except for this restriction, I don't see anything else to lead to a relation on the coefficients. Hopefully this will help you or someone else get a start on the problem.</p> <p>One further thought:</p> <p>Since the roots appear in pairs (say $z_1$ and $z_2$ are conjugate over $\mathbb Q\left(\sqrt 2\right)$) then one can generate $\sqrt 2$ with either pair, and subtract them. However, I don't immediately see a way to gather that information from the symmetric polynomials of the roots (a.k.a. the coefficients $a_1, \ldots, a_4.$)</p> http://mathoverflow.net/questions/8160/expressing-field-inclusions-by-polynomial-equalities-on-coefficients/8168#8168 Answer by Zev Chonoles for Expressing field inclusions by polynomial equalities on coefficients Zev Chonoles 2009-12-08T07:04:15Z 2009-12-08T07:04:15Z <p>So, equivalently, suppose we have a symmetric function $S( , , , )$ such that $S(z_1,z_2,z_3,z_4)=0$ whenever $z_1,z_2,z_3,z_4$ are conjugates over $\mathbb{Q}$ and such that $\mathbb{Q}(z_i)\supset\mathbb{Q}(\sqrt{2})$ for each $i$. As described above, we can show that (WLOG) $z_1,z_2$ are roots of some $x^2+b_1x_1+b_0\in\mathbb{Q}(\sqrt{2})[x]$ and $z_3,z_4$ are roots of some $x^2+c_1x+c_0\in\mathbb{Q}(\sqrt{2})[x]$. I'm wondering, can we then say that $S(z_1,z_2,z_3,z_4)=T(b_0,b_1)(c_0,c_1)=0$ for a function $T$ (which would clearly not be symmetric, but would always be a function of things in $\mathbb{Q}(\sqrt{2})$?</p> http://mathoverflow.net/questions/8160/expressing-field-inclusions-by-polynomial-equalities-on-coefficients/8171#8171 Answer by Thorny for Expressing field inclusions by polynomial equalities on coefficients Thorny 2009-12-08T09:33:59Z 2009-12-08T10:53:24Z <p>If there was a nontrivial polynomial relation between the coefficients, it would be true for a dense subset (<s>reducibility is a nowhere dense condition</s> see comment below) of all polynomials of the form $(x^2+(\alpha +\beta\sqrt{2})x+\gamma+\delta\sqrt{2})(x^2+(\alpha -\beta\sqrt{2})x+\gamma-\delta\sqrt{2})$ with rational $\alpha,\beta,\gamma,\delta$, which would mean the same relation would be true for all real $\alpha,\beta,\gamma,\delta$ as well. But all quartic polynomials are of the form above with real $\alpha,\beta,\gamma,\delta$, so there are no nontrivial relations.</p>