Classical Galois theory gives necessary and sufficient conditions for the roots of a polynomial in $k[x]$ to be expressible in terms of nested radicals of the coefficients.

Suppose instead that a single root $\alpha$ of $p(x)\in \mathbb{Q}[x]$ is known. Are there known necessary and sufficient conditions on $p(x)$ such that all remaining roots can be expressed as polynomial (or rational) functions of $\alpha$ and the coefficients of $p(x)$?

For example, the cyclotomic polynomials have this property, since every primitive $n^{\textrm{th}}$ root of unity can be written as a power of some fixed root.

  • $\begingroup$ This is only possible when $\alpha$ generates the splitting field, which means that the Galois group $G=\operatorname{Gal}(p)$ has order equal to $d=\deg(p)$. Furthermore, since the action of $G$ on the roots of $p$ is also transitive, it must be the cyclic group of order $d$. Conversely, if $G$ is (cyclic) of order $d$, then every root of $p$ is expressible as a polynomial in $\alpha$. $\endgroup$
    – R.P.
    Commented Mar 29, 2022 at 9:04
  • $\begingroup$ The coefficients of $p$ are in $\mathbb{Q}$, so I guess you just want the roots to be polynomial functions in $\alpha$ with coefficients in $\mathbb{Q}$? Otherwise, can you clarify? $\endgroup$ Commented Mar 29, 2022 at 9:08
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    $\begingroup$ @RP_ The Galois group is not necessarily cyclic, if $K/\mathbb{Q}$ is any finite Galois extension then any $\alpha$ generating $K/\mathbb{Q}$ will work. We can also choose $\alpha$ so that its conjugates form a $\mathbb{Q}$-basis of $K$. $\endgroup$ Commented Mar 29, 2022 at 9:14
  • $\begingroup$ @FrançoisBrunault To be clear, if $p(x) = a_{n}x^{2} + \ldots + a_{0}$ then I'd like an expression for the other roots as polynomial functions in the $a_{i}$ and $\alpha$ with coefficients in the base field. Of course, for a fixed $p(x)$ the other roots would just be polynomial in $\alpha$ with coefficients in $\mathbb{Q}$. $\endgroup$ Commented Mar 29, 2022 at 9:20
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    $\begingroup$ Just a remark. This is related to how Galois himself did Galois theory from the point of view of the theory of symmetric function: looking at symmetric functions of all roots except one. See Theorem 6 in the paper "The fundamental theorem on symmetric polynomials: History's first whiff of Galois theory" by Blum-Smith and Coskey tandfonline.com/doi/abs/10.4169/college.math.j.48.1.18 $\endgroup$ Commented Mar 29, 2022 at 18:30

2 Answers 2


Let $\alpha=\alpha_1$, $\alpha_2$, ..., $\alpha_n$ be the roots of $p(x)$. You want $\mathbb{Q}(\alpha_1,\alpha_2, \ldots, \alpha_n) = \mathbb{Q}(\alpha)$. If the Galois group is $G \subseteq S_n$, then $\mathbb{Q}(\alpha_1)$ corresponds to the stabilizer of $1$ in $G$, and $\mathbb{Q}(\alpha_1,\alpha_2, \ldots, \alpha_n)$ corresponds to the trivial subgroup. So the condition is that the stabilizer of $1$ in $G$ is trivial. In other words, the action of $G$ on the orbit of $\alpha_1$ should be regular.

All of this was basically said in comments above, but their seemed to be some confusion about the case where $p(x)$ has multiple factors, so here is an answer which doesn't assume that $p$ is irreducible.

As per discussion in comments, let $L/K$ be a Galois extension with Galois group $G$; put $N = |G|$. Let $\alpha \in L$ be an element with trivial stabilizer. Let $\beta$ be an other element of $L$. We want to write $\beta$ as a polynomial in $K(\alpha)$. Set $\gamma_j = \text{Tr}_{L/K}(\alpha^j \beta)$. Then the $\gamma_j$ are in $K$. If $K = \mathbb{Q}$ and $\alpha$ and $\beta$ are algebraic integers, then the $\gamma_j$ are integers.

For any nonnegative integer $j$, we have $$\text{Tr}_{L/K}(\alpha^j \beta) = \sum_{\sigma \in G} \sigma(\alpha)^j \sigma(\beta).$$ If, for some magic reason, we explicitly have floating point values for the $\sigma(\alpha)$ and $\sigma(\beta)$, and know the $G$-action on these values, we can use this formulato numerically compute the $\gamma_j$; if the $\gamma_j$ are then integers, we can round our computations to the nearest integer and get the result. In practice, I'm not sure how you'd get the $\gamma_j$, but I'll pretend you know them.

Let $A$ be the $N \times N$ matrix with entries $\sigma(\alpha)^j$ for $0 \leq j \leq N-1$. Let $\vec{b}$ be the vector with entries $\sigma(\beta)$ and let $\vec{c}$ be the vector with entries $\gamma_j$. So the displayed equation above states that $A \vec{b} = \vec{c}$, and thus $\vec{b} = A^{-1} \vec{c}$. In particular, $\beta$ is the dot product of the first row of $A^{-1}$ with $\vec{c}$.

The entries of $\vec{c}$ are in $K$, so it remains to show that the entries of the first row of $A^{-1}$ are in $K(\alpha)$. Let the Galois orbit of $\alpha$ be $\{ \alpha_1, \alpha_2, \ldots, \alpha_N \}$ with $\alpha = \alpha_1$. Then $A$ is a Vandermonde matrix in the $\alpha_i$'s, so the first row of its inverse is $$\pm \frac{e_i(\alpha_2, \alpha_3, \ldots, \alpha_n)}{\prod_{j=2}^N (\alpha_1 - \alpha_j)}. \qquad (\ast)$$

Let $p(x)$ be the polynomial $f(x)/(x-\alpha_1) = \prod_{j=2}^N (x-\alpha_j)$. Then the coefficients of $p$ are clearly in $K(\alpha_1)$. The numerator $e_i(\alpha_2, \alpha_3, \ldots, \alpha_n)$ of $(\ast)$ is (up to sign) the coefficient of $x^{n-i-1}$ in $p$, and the denominator is $p(\alpha_1) = f'(\alpha_1)$. So $(\ast)$ is in $K(\alpha_1)$ and we are done.

My memory is that I read that this was Galois's proof, but I couldn't find the source quickly.

  • $\begingroup$ Yes! It is now clear to me that this is necessary and sufficient. Is it possible to describe the other roots explicitly as polynomials in $\alpha_{1}$ under these assumptions? (Which of course is what I should have asked in the first place.) $\endgroup$ Commented Mar 29, 2022 at 16:13
  • $\begingroup$ So, at this point, I want to know in what form the data is given to me. If you give me a polynomial $p(x)$ of degree $N$, a subgroup $G$ of $S_N$ acting freely and transitively on the roots of $p$, another polynomial $q(x)$ and an action of $G$ on the roots of $q$, and you promise me that these actions extend to field automorphisms, I do know a way to get polynomial expressions for the roots of $q$ in terms of a root of $p$ from this data. But that is a pretty weird thing to promise me and not something you naturally get computationally. $\endgroup$ Commented Mar 29, 2022 at 17:39
  • $\begingroup$ I agree this is a weird set of demands, but in fact I think this is pretty close to my situation, actually. In any case, I'd like to see how it is done! $\endgroup$ Commented Mar 29, 2022 at 17:49
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    $\begingroup$ Regarding the history, I looked at the article "Galois for 21st century readers" by Edwards. He explains how Galois proved that (in modern language) each root of $p(x)$ can be expressed rationally in terms of a generator of the splitting field (Lemma 3 in his first memoir). The construction looks a bit different than the one from your answer, but apparently, Galois also uses the important fact that every symmetric function of the other roots $\alpha', \alpha'',\ldots$ of $p(x)$ can be expressed rationally in terms of $\alpha$. $\endgroup$ Commented Mar 30, 2022 at 7:01

From a computational point of view, one should not try to compute the Galois group. Assuming $p(x) \in \mathbb{Q}[x]$ is irreducible, and $\alpha$ is a root of $p(x)$, it is sufficient to factor $p(x)$ over the number field $\mathbb{Q}(\alpha) = \mathbb{Q}[x]/(p(x))$, and look whether all the irreducible factors have degree 1. In this way, you also get the expression of the roots in terms of $\alpha$. This is much less expensive than computing the Galois group, which is feasible only in relatively low degree.

  • $\begingroup$ I think this answer, and not mine, is the practical answer. $\endgroup$ Commented Mar 29, 2022 at 19:41
  • $\begingroup$ Yes! David's answer is wonderful, and I feel I learned something useful from it, but this is the way to get what I want computationally. Thank you both! $\endgroup$ Commented Mar 29, 2022 at 20:57

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