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Is there an example of an irreducible polynomial $f(x) \in \mathbb{Q}[x]$ with a real root expressible in terms of real radicals and another real root not expressible in terms of real radicals?

David Callan
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    But $X^3-2$ has no real root other than $\root 3 \of 2$. – Noam D. Elkies Sep 30 '14 at 17:40
  • There was an answer here that said similarly, but I think (following a reference from Dummit) that Isaacs notes that if a totally real polynomial (with solvable Galois group) requires only real radicals, then the Galois group is a 2-group. Though you don't require all the roots to be real as they do. http://www.jstor.org/stable/2323164 – NAME_IN_CAPS Sep 30 '14 at 19:35
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    @NAME_IN_CAPS: A jstor-free link: http://planetmath.org/proofofcasusirreducibilisforrealfields . – Emil Jeřábek Sep 30 '14 at 21:53
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    One root to real them all. – Asaf Karagila Oct 01 '14 at 00:06
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    @AsafKaragila And my $x$ – Felipe Voloch Oct 01 '14 at 01:13
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    @Gerry Myerson & NAME_IN_CAPS: I think there can't be an example involving only 2-power roots. In that case $\alpha\in\mathbb R$ is expressible in terms of real radicals iff there is a chain of real fields ${\mathbb Q}(\alpha)\supset K_n\supset\cdots\supset K_1\supset {\mathbb Q}$, each one quadratic over the next one. But if these conditions hold for $\alpha$ and $\beta=\sigma(\alpha)$ is a Galois conjugate root which is real, then hitting the chain with $\sigma$ produces a real chain for $\beta$, so $\beta$ is also expressible in terms of real radicals. – Tim Dokchitser Oct 01 '14 at 07:34

2 Answers2

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The answer to the question is yes (so the answer to the title is no) and I will give an example later.

Let me first recall a couple of results. The first one is the following, that can be found in [Cox, Galois Theory, Theorem 8.6.5].

Theorem 1. Let $F$ be a subfield of $\mathbb{R}$ and let $f \in F[x]$ be an irreducible polynomial with splitting field $F \subset L \subset \mathbb{R}$. Then the following conditions are equivalent.

(1) Some root of $f$ is expressible by real radicals over $F$.

(2) All roots of $f$ are expressible by real radicals over $F$ in which only square roots appear.

(3) $[L:F]$ is a power of $2$.

So, if $f$ splits completely over $\mathbb{R}$, the existence of a root expressible by real radicals forces all the roots to be so.

On the other hand, when $f$ does not split completely in $\mathbb{R}$ this is no longer true. Let us state our second result, that can be found in

[A. Loewy, Über die Reduktion algebraischer Gleichungen durch Adjunktion insbesondere reeller Radikale, Math. Zeitschr. 15, 261-273 (1922)],

see also Cox book, Theorem 8.6.12.

Theorem 2. Let $F$ be a subfield of $\mathbb{R}$ and $f \in F[x]$ irreducible of degree $2^mn$, with $n$ odd. Then $f$ has at most $2^m$ roots expressible by real radicals over $F$.

In particular, when $f \in \mathbb{Q}[x]$ is irreducible and of odd degree, Theorem 2 implies that at most one root of $f$ is expressible by real radicals. Note that if the degree is $3$ then Theorem 2 is consistent with Cardano's formulas, and if the degree is a power of $2$ then it is consistent with Theorem 1.

Finally, let us give the following example answering the question, that can be found in Loewy's paper quoted above, page 272. Let us consider the polynomial $$x^6+6x^4-234x^2-54x-3 =(x^3+(3+9 \sqrt{3})x+ \sqrt{3})(x^3+(3-9 \sqrt{3})x- \sqrt{3}).$$ It is irreducible over $\mathbb{Q}$ by Eisenstein's criterion and it has one real root expressible by real radicals, three real roots not expressible by real radicals and two complex roots.

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Francesco Polizzi
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  • The question does not demand that the polynomial splits in $\mathbb R$, that’s the whole problem. – Emil Jeřábek Sep 30 '14 at 18:49
  • ah, ok. I did not realize that. Thanks. – Francesco Polizzi Sep 30 '14 at 18:52
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    I added an example showing that the answer to the question is yes and one reference. I hope that this is now correct. – Francesco Polizzi Oct 01 '14 at 09:58
  • This sounds great! I was beginning to worry this question will turn into an entry in http://meta.mathoverflow.net/questions/1820 . – Emil Jeřábek Oct 01 '14 at 11:32
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    Very nice. So Loewy considered this very question long ago. The first polynomial factor in the counterexample, $x^3+(3+9\sqrt{3})x +\sqrt{3}$, has a real root $\big((\sqrt{331 +360 \sqrt{3}}- \sqrt{3})/2\big)^{1/3} - \big((\sqrt{331 +360 \sqrt{3}}+\sqrt{3})/2\big)^{1/3}$, patently expressible in terms of real radicals. The second polynomial has 3 real roots and, since it is over a real field, Theorem 1 of the answer says none of its roots are expressible in terms of real radicals. – David Callan Oct 01 '14 at 18:17
  • this is a deleted comment – David Callan Oct 01 '14 at 20:53
  • Oops, re previous comment, better check that the second polynomial, $x^3+(3-9\sqrt{3})x -\sqrt{3}$, is irreducible over $\mathbb{Q}[\sqrt{3},]$ before applying Theorem 1. This just requires showing it does not have a root in $\mathbb{Q}[\sqrt{3},]$. Plugging in a prospective root $a+b\sqrt{3},\ a,b \in \mathbb{Q}$, leads to $64 a^9+288 a^7+3240 a^5+972 a^4+3105 a^3-56862 a-6561=0$, and it is straightforward to show the latter equation has no rational root ($6561=3^8$). – David Callan Oct 01 '14 at 21:06
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    The original polynomial is irreducible of degree 6, hence it cannot have a root in a quadratic extension like $\mathbb Q[\sqrt3]$. – Emil Jeřábek Oct 02 '14 at 10:34
  • Emil, yes, of course. We get the required irreducibility over $\mathbb{Q}[\sqrt{3}]$ for free. – David Callan Oct 02 '14 at 18:59
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WARNING: Wrong answer follows, as per comments. Kept here to deter others from making the same mistake.

Perhaps $$x_1=\sqrt{2+2^{1/4}}+\sqrt{2-2^{1/4}},\quad x_2=\sqrt{2+i2^{1/4}}+\sqrt{2-i2^{1/4}}$$ are real numbers, solutions of $\bigl((x^2-4)^2-16\bigr)^2=32$, one expressed using real radicals, the other, not.

Gerry Myerson
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