Let $p(x)$ be a polynomial with rational coefficients with a root of the form $a + \sqrt[n]b$, with $a, b \in \mathbb Q, n \in \mathbb N$. What other roots must be there? For $n = 2$, this is trivial; what about higher values, even and odd, of $n$?

And what about nested radicals?


I was surprised to learn that $x^3 - 12x^2 + 48x - 71$ has exactly one real irrational root, namely $4 + \sqrt[3]7$, with the other two roots being $4 - \frac {\sqrt[3]7}{2}(1 \pm \sqrt 3 i)$. This is of course different behavior than what happens with square roots. This is consistent with the fact that radicals eliminate when multiplying $a + \sqrt b$ by $a - \sqrt b$, but not with cube roots, and that every square root admits two values, whereas cubic roots admit only one value.

Still, the roots above show a clear relationship, reminiscent of, but not identical to, those of radical conjugates, which I'd like to understand.

I'm interested both in defining the relationship, and the intuition behind it (what can a rational polynomial "distinguish" and what can't it?). As explained here, radical conjugates (i.e. $a \pm \sqrt b$) are indistinguishable by polynomials with rational coefficients, in that if one is a root, the other is. Clearly, that only applies to square roots, not cube roots. What else can rational polynomials distinguish or not? Is this somehow connected to roots of unity (which the roots above seem to resemble, at least superficially)?

To turn this into a solid question, I came up with the question at the top.


(Inspired by comment from dxiv): A better way to list the roots is $$4 + \sqrt[3]7 \\ 4 + \frac {-1 \pm \sqrt 3 i}{2}{\sqrt[3]7} $$

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    $\begingroup$ Hint: for $n=2$ the roots are $a + \omega \sqrt{b}$ where $\omega^2 = 1$, for $n=3$ the roots are $a + \omega \sqrt[3]{b}$ where $\omega^3 = 1$. $\endgroup$
    – dxiv
    Jun 27, 2023 at 22:18
  • $\begingroup$ @dxiv Aha, the roots of unity! I thought I spotted them.... but only the irrational $\sqrt[n]b$ is multiplied by them, not the rational $a$, need to understand that better... $\endgroup$ Jun 27, 2023 at 22:23
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    $\begingroup$ @RobertShore Applying Galois theory requires knowing Galois theory, which I've yet to be able to learn (even superficially). $\endgroup$ Jun 27, 2023 at 22:27
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    $\begingroup$ (All that you need to know is that the minimal polynomial for a root of unity is the corresponding cyclotomic polynomial.) If you have a polynomial $f(x)$ with $ x = a + \sqrt[n]b$, do a change of variables and consider the polynomial with the root $ y = x-a$. $g(x) = f(y+a)$ has a root of $ \sqrt[n]b$. What can you say from there? Hence, what are the roots of $g(x)$ and $f(x)$? $\endgroup$
    – Calvin Lin
    Jun 27, 2023 at 22:33
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    $\begingroup$ An expression like $\sqrt[3]7$ is, algebraically, just a cube root of 7 and can't be distinguished from any of the other cube roots of 7 by a rational polynomial. So if $4 + \sqrt[3]7$ is a root of a rational polynomial, then this will be true no matter which of the three cube roots of 7 we interpret $\sqrt[3]7$ to be. $\endgroup$
    – Ted
    Jun 27, 2023 at 23:27

2 Answers 2


Suppose that $f(a + \sqrt[n]{b}) = 0$. Suppose that $f(x) \neq 0$. Instead of $f(x)$, I would like to consider $g(x) = f(a+x)$, which is still a polynomial with rational coefficients and whose degree is the same as that of $f(x)$. Note that $\beta := \sqrt[n]{b}$ is a root of $g(x)$.

I denote the set of all polynomials (in $x$) with rational coefficients as $\mathbb{Q}[x]$. I define a subset: $$ S = \{\, h(x) \in \mathbb{Q}[x] \mid \text{$h(\beta) = 0$ and $h(x) \neq 0$} \,\}. $$ It is not empty, since $g(x) \in S$. Among the members of $S$, there exists a unique polynomial $m(x) \in S$ such that:

  • the coefficient of the highest-degree term in $m(x)$ is $1$;
  • the degree of $m(x)$ is not higher than that of any member of $S$.

Existence: Consider $$ T = \{\, \text{degree of $h(x)$} \mid h(x) \in S \,\}, $$ which is a nonempty subset of $\mathbb{N}$. Hence the minimal member exists, which means that there exists $M(x) \in S$ such that $$ \text{degree of $M(x)$} \leq \text{degree of $h(x)$} $$ for any $h(x) \in S$. Suppose that $$ M(x) = a_k x^k + a_{k-1} x^{k-1} + \dots + a_0 $$ with $a_k \neq 0$. Let $m(x) = \frac{1}{a_k} M(x)$. It is not hard to see that $m(x)$ satisfies the two requirements above.

Uniqueness: Suppose that $m(x)$, $z(x)$ both satisfy the two requirements above but are unequal. By the second point, the degree of $m(x)$ has to be equal to that of $z(x)$. By the first point, $d(x) := m(x) - z(x)$ is a polynomial with lower degree than that of $m(x)$. Since $d(\beta) = 0$ and $d(x) \neq 0$, $d(x)$ is a member of $S$. This is a contradiction, since the degree of $d(x)$ is lower than that of $m(x)$.

I call $m(x)$ the minimal polynomial of $\beta$.

$m(x)$ is irreducible. I prove the claim by assuming that there exist two polynomials $m_1 (x)$, $m_2 (x) \in \mathbb{Q}[x]$ such that $m(x) = m_1 (x) m_2 (x)$ and that neither of $m_1(x)$, $m_2 (x)$ is a polynomial of degree zero. Since $0 = m(\beta) = m_1 (\beta) m_2 (\beta)$, one of $m_1 (\beta)$, $m_2 (\beta)$ must be zero. Without loss of generality, suppose that $m_1 (\beta) = 0$. $m_1 (x)$ is not a zero polynomial; $\beta$ is a root of $m_1 (x)$. Hence $m_1 (x) \in S$. Hence the degree of $m_1 (x)$ is not lower than that of $m(x)$. Note that the degree of $m(x)$ is not lower than that of $m_1 (x)$. Hence the degree of $m_1 (x)$ is equal to that of $m(x)$. Hence $m_2 (x)$ is a polynomial of degree zero, which is a contradiction.

I claim that $m(x)$ divides $g(x)$. I use "reductio ad absurdum" again. Suppose that it is not the case. Then $m(x)$ and $g(x)$ are coprime. Hence there exist two polynomials $r(x)$, $s(x) \in \mathbb{Q}[x]$ such that $$ m(x) r(x) + g(x) s(x) = 1. $$ Hence $$ m(\beta) r(\beta) + g(\beta) s(\beta) = 1, $$ which is a contradiction.

Let me summarize what I have got:

Let $f(x)$ be a polynomial with rational coefficients. Suppose that $f(a + \sqrt[n]{b}) = 0$, in which $a$, $b$ are rational numbers. Let $g(x) = f(a + x)$. Let $\beta = \sqrt[n]{b}$. Let the minimal polynomial of $\beta$ be $m(x)$. Then $m(x)$ divides $g(x)$.

If $\beta$ is a rational number, what I can learn is not interesting, since the minimal polynomial of $\beta$ is simply $x - \beta$.

If $\beta$ is not a rational number, what I can learn is not boring, since the minimal polynomial of $\beta$ cannot be of degree $1$.

Example. It can be verified that the minimal polynomial of $\sqrt[3]{7}$ is $m(x) = x^3 - 7$. It can be verified that $4 + \sqrt[3]{7}$ is a root of $f(x) = x^3 - 12x^2 + 48x - 71$. Hence $m(x)$ must divide $g(x)$, in which $$ g(x) = f(x + 4) = x^3 - 7, $$ which is just the same as $m(x)$. The roots of $m(x)$ are well known, so it can be learned what other roots $g(x)$ must have, from which it can be learned what other roots $f(x)$ must have.

And what about nested radicals?

You are invited to explore this yourself.


This is more or less the beginnings of "Galois theory." For your particular question, the roots that must exist (under a few assumptions) are $a+\zeta^i\sqrt[n]{b}$, $0\leq i<n$ where $\zeta$ is a primitive $n$th root of unity, e.g. $\zeta=e^{2\pi i/n}$ (taking $\sqrt[n]{b}$ to be the real root of a positive rational number $b$).

The polynomial $f(x)=\prod_{i=0}^{n-1}(x-(a+\zeta^i\sqrt[n]{b}))=(x-a)^n-b$ has rational coefficients as it is invariant under the action of the Galois group of $\mathbb{Q}(\zeta,\sqrt[n]{b})$ (or by inspection; the $n$ roots of LHS are roots of RHS). This will be the minimial polynomial assuming $\mathbb{Q}(\sqrt[n]{b})$ has degree $n$, e.g. not considering something like $\sqrt[4]{4}=\sqrt{2}$.


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