# Polynomials for which roots can be expressed as polynomials in a single root

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.

• 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 $\alpha$.
– RP_
Mar 29 at 9:04
• 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? Mar 29 at 9:08
• @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 $K$. Mar 29 at 9:14
• @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 $\mathbb{Q}$. Mar 29 at 9:20
• 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 Mar 29 at 18:30

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.

• 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.) Mar 29 at 16:13
• 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. Mar 29 at 17:39
• 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! Mar 29 at 17:49
• 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 $\alpha$. Mar 30 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.

• I think this answer, and not mine, is the practical answer. Mar 29 at 19:41
• 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! Mar 29 at 20:57