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

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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? – Zev Chonoles Dec 8 '09 at 5:07
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. – Ben Weiss Dec 8 '09 at 5:13
Ah, of course. Perhaps I would think more clearly if I got some sleep occasionally... – Zev Chonoles Dec 8 '09 at 5:47
I know the feeling. – Ben Weiss Dec 8 '09 at 5:47
Another example is (3x^2+ax+b)^2-2(2x^2+cx+d)^2 (for "nondegenerate" a,b,c,d) – Ewan Delanoy Dec 8 '09 at 6:23
up vote 1 down vote accepted

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.

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Why is reducibility a nowhere dense condition ? – Ewan Delanoy Dec 8 '09 at 10:21
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. – Thorny Dec 8 '09 at 10:52
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. – Ewan Delanoy Dec 8 '09 at 11:56

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 $a_1, \ldots, a_4.$)

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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? – Zev Chonoles Dec 8 '09 at 6:04
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. – Ben Weiss Dec 8 '09 at 6:10
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. – Ben Weiss Dec 8 '09 at 6:11

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

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(I posted separately because it would be a pain to read all that unrendered) – Zev Chonoles Dec 8 '09 at 7:05

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