3
$\begingroup$

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 $(a_0,a_1,a_2,a_3) \in A$ ?

$\endgroup$
5
  • $\begingroup$ Can you explain why you think this might be true? (the motivation for the question may help find the answer) Also, off the top of my head I can't think of any polynomials other than (x^2-2)^2 = x^4-4x^2+4; can you provide some other examples for people to test their R's on? $\endgroup$ Dec 8, 2009 at 5:07
  • $\begingroup$ One example is x^4 - 2. The various z are the 4th roots of 2, and any field containing a 4th root of 2 has both square roots of 2 in it. $\endgroup$
    – Ben Weiss
    Dec 8, 2009 at 5:13
  • $\begingroup$ Ah, of course. Perhaps I would think more clearly if I got some sleep occasionally... $\endgroup$ Dec 8, 2009 at 5:47
  • $\begingroup$ I know the feeling. $\endgroup$
    – Ben Weiss
    Dec 8, 2009 at 5:47
  • $\begingroup$ Another example is (3x^2+ax+b)^2-2(2x^2+cx+d)^2 (for "nondegenerate" a,b,c,d) $\endgroup$ Dec 8, 2009 at 6:23

3 Answers 3

1
$\begingroup$

If there was a nontrivial polynomial relation between the coefficients, it would be true for a dense subset (reducibility is a nowhere dense condition 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.

$\endgroup$
3
  • $\begingroup$ Why is reducibility a nowhere dense condition ? $\endgroup$ Dec 8, 2009 at 10:21
  • $\begingroup$ Sorry, it isn't. However, the set of irreducible $x^2+ux+v$ is dense because it is equivalent to $v-(u/2)^2$ not being square in $\mathbb{Q}[\sqrt{2}]$. This latter condition is satisfied by all numbers of the form $\frac{1}{n^2}(r+s\sqrt{2})$ with $r$ even and $s$ odd, say. $\endgroup$
    – Thorny
    Dec 8, 2009 at 10:52
  • $\begingroup$ Your proof is complete now, Thorny. BTW, here is a way to see that reducibility is not a nowhere dense condition : The Pell equation x^2-8y^2=1 has integer solutions with y arbitrarily large. Then the polynomials X^2+(1/y^2)X-2 are all reducible, and converge to X^2-2. $\endgroup$ Dec 8, 2009 at 11:56
0
$\begingroup$

This may ramble a bit much, but I hope it provides some help in how to think about the problem.

Let's see what your extension of fields looks like. We have 4 possible extensions (perhaps the same) So that any of them is

$\mathbb Q(z_i)$

$|$

$\mathbb Q\left(\sqrt2\right)$

$|$

$\mathbb Q$

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 $\mathbb Q(\sqrt2).$

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.

One further thought:

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 $a_1, \ldots, a_4.$)

$\endgroup$
3
  • $\begingroup$ Would I be correct in saying that a polynomial relation in the $a_i$ is necessarily a symmetric function of the roots, because the $a_i$ are the elementary symmetric functions of the roots? $\endgroup$ Dec 8, 2009 at 6:04
  • $\begingroup$ Yes, that's exactly correct. The problem is trying to use that with our knowledge of the roots to say anything about a relation they satisfy. $\endgroup$
    – Ben Weiss
    Dec 8, 2009 at 6:10
  • $\begingroup$ It could be that because the roots are all quadratic over Q(\sqrt 2) that the coefficient a_2 which is the quadratic symmetric function plays a key role. $\endgroup$
    – Ben Weiss
    Dec 8, 2009 at 6:11
0
$\begingroup$

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 $\mathbb{Q}(\sqrt{2})$?

$\endgroup$
1
  • $\begingroup$ (I posted separately because it would be a pain to read all that unrendered) $\endgroup$ Dec 8, 2009 at 7:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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