# Relations betweens roots of polynomial equations

It does not seem to be so well-known that there can be only very few linear relations between the roots of an irreducible polynomial which has a “large” Galois group (in a certain sense, which is made clearer below). For instance, given a monic polynomial P in Q[T], written

$P(T)=T^d+a_{d-1}T^{d-1}+\cdots+a_1T+a_0,$

it is easy to create a polynomial Q of the form

$Q(T)=T^d+b_{d-2}T^{d-2}+\cdots+b_1T+b_0$

whose roots generate the same field. If βi are the roots of Q, the vanishing of the coefficient of Td-1 means that they satisfy the Q-linear relation

$\beta_1+\cdots+\beta_d=0.$

If the Galois group of the splitting field of Q (or P) is the full symmetric group on d letters, then in fact this is the only possible relation (up to multiplication by a fixed scalar)!

In other words, it is impossible, for instance, that the roots αi of P satisfy

$\alpha_1-\alpha_2+\cdots +(-1)^d\alpha_d=0,$

or

$\alpha_1+2\alpha_2+\cdots +d\alpha_d=0.$

Proving this type of fact can be seen either as a fun exercise in the representation theory of finite groups, or as a good way of motivating its usefulness. (The basic knowledge needed is the existence of decomposition into irreducibles, and unicity of the isotypical space corresponding to a given irreducible representation; the treatment by J-P. Serre is probably the clearest introduction).

Indeed, if the question was presented to a student in the form of showing that either of the previous two relations are incompatible with the αi being roots of a polynomial with symmetric Galois group G=Sd, it is likely that it would be natural to expand the hypothetical relation by stating that, for any σ in G, we have (for the last case, say):

$\sigma(\alpha_1)+2\sigma(\alpha_2)+\cdots +d\sigma(\alpha_d)=0.$

If the Galois group is indeed Sd, we can choose σ so that

$\sigma(\alpha_1)=\alpha_2,\ \sigma(\alpha_2)=\alpha_1$

and the other roots are fixed; then subtracting the new relation from the first one, we get

$-\alpha_2+\alpha_1=0,$

which is impossible.

This was particularly easy, but to refute the possibility of all relations (except the sum of the roots being 0), something more systematic is needed. The type of argument used suggests to consider the vector space of all possible relations:

$R=\{(\lambda_1,\ldots,\lambda_d)\,\mid\,\lambda_1\alpha_1+\cdots+\lambda_d\alpha_d=0\}$

and the basic idea is that G acts on R by its action on the roots, in other words, R is a linear representation of G (over C, or a smaller field if one wants).

If we then take the point of view of trying to identify this representation among the representations of G, we are naturally led to remark that it is defined as a subrepresentation of the permutation representation of G acting on the roots

$F=\bigoplus_{1\leq i\leq d}{\mathbf{C}\cdot [\alpha_i]},\text{ with } \sigma([\alpha_i])=[\sigma(\alpha_i)],$

which, as a representation, doesn’t depend on the particular polynomial (and its roots), but only on the group G, as a transitive permutation group acting on d objects.

So, for any particular transitive group G acting on d objects, one can construct the representation F, decompose it into sums of irreducibles πi (this is a problem of group theory), and then it follows that R can only be isomorphic to a direct sum of those representations occuring in R, with multiplicities at most equal to that in F: if

$F\simeq n_1\pi_1\oplus\cdots \oplus n_k\pi_k,\ \text{ with } n_i\geq 1,$

then R can only be of the form

$R\simeq m_1\pi_1\oplus\cdots \oplus m_k\pi_k,\ \text{ with }\ 0\leq m_i\leq n_i.$

Then one can look at each possibility in turn, and try to see which ones can actually occur as relations of roots of a polynomial. This is clearly easiest if F has few summands, particularly if there is no multiplicity (i.e., if each ni is 1), for then the subspace corresponding to a summand that occurs in R must be equal to that in F, which is itself uniquely determined.

[If there is multiplicity, say n1=2, then in the π1-isotypic component F1 in F, there are infinitely many subspaces isomorphic to π1, with multiplicity one, for instance

$V_{\lambda}=\{(v,\lambda v)\,\mid\,v\in V\}$

for each (fixed) scalar λ, where V is a fixed subspace of F1, G stable and isomorphic to π1 under this action. Each of these subspaces can lead to different relations.]

In the case of the symmetric group, the decomposition of the natural permutation representation is well-known:

$F=\mathbb{1}\oplus \pi$

where 1 is the subspace generated by

$[\alpha_1]+\cdots +[\alpha_d]$

while the complement subspace

$\pi=\{(\lambda_1,\ldots,\lambda_d)\,\mid\, \sum_{i}{\lambda_i}=0\}$

is irreducible. Then, 1 is in R precisely when the sum of the roots is zero (and we already know that this possibility can occur), and π can not be (contained in) a relation module, because in particular

$(1,-1,0,\ldots,0)\in \pi$

which leads (as before, but now in full generality) to the absurd relation

$\alpha_1=\alpha_2.$

In my paper on relations between zeros of zeta functions over finite fields, I applied this method to understand the possible additive relations (and the multiplicative ones, using the same technique with the module of multiplicative relations instead of R) when the Galois group is the Weyl group of Sp(2g). Here also the situation was simple, with three summands in F, each with multiplicity 1.

In fact, experimenting with Magma, I can see that most transitive permutation groups of small degree seem to have the property of not having multiplicity in the permutation representation. The first counter-example is the symmetric group S3, acting on itself by multiplication (there is no multiplicity in the natural action on three letters); indeed, if G acts on itself by multiplication, the permutation representation is isomorphic to the regular representation, and each irreducible occurs with multiplicity equal to its dimension, so such a transitive action always has multiplicity if the group is non-abelian. But most transitive groups are given with an action on a smaller set, and one finds that among the 173 non-trivial transitive groups of degree at most 11, only 5 exhibit some multiplicity. (Note that it is not clear at all if this impression can be quantified precisely).

This type of problem has been investigated a lot by K. Girstmair, and this paper of his (in Acta Arithmetica, 1999) is probably the most complete introduction to the subject.

### Kowalski

I am a professor of mathematics at ETH Zürich since 2008.

## 4 thoughts on “Relations betweens roots of polynomial equations”

1. The impression that transitive groups often have multiplicity-free decompositions of the representation into irreducibles does not look right.

While indeed for very small degrees, like up to 11, this could look possible, the combinatorial explosion happens quite soon afterwards.
The centraliser algebra of the permutation representation of G, that decomposes with multiplicities n_i as F above, has dimension d=sum_i n_i^2,
and d is equal to the number of the orbits of the stabiliser H of a point. When the degree |G|/|H| of the representation is small, there is often not enough room to fit in these n_i^2’s…

And this number d is at least
d>=1+(|G|/|H|-1)/|H|, that one can see by assuming that H acts regularly on all its orbits except the fixed point. So when |H| is much smaller than |G| we have d bigger than the number of different irreducibles of G, forcing some multiplicities be bigger than 1.

I would be curious to look at different relations between the roots that arise due to bigger than 1 multiplicities. Should not they somehow be “equivalent” under the centraliser algebra?

PS. there is a typo in the definition of R – the n_i’s should be m_i’s.

2. Thanks for the insight! So this seems to be one more case where small numbers lead to mistakes…

I think Girstmair’s paper (or one he refers to) gives examples of the more complicated possibilities for relations coming from higher multiplicity, but I don’t remember for sure.

(And thanks for the correction).

3. I heard that Girstmair himself was surprised by the following question:
“Prove that if a_1+a_2+a_3=0 for some roots of irreducible polynomial R(x) over Q, and polynomial is not of the form P(x^2), then 3|d (the degree of the polynomial)”.
It seems that the problem is still open. Though Dubickas and Smyth (in AMM problem) gave an example of degree 20 polynomial where this holds (though it is of the form P(x^2)).

4. A question like the one you mention is clearly of a different type than those presented in the post, asking to find groups having such and such a property.

One can even think that there might exist such a question “Does relation R hold for some polynomial over Q?”, where some group theory arguments then show that this polynomial would need to have a particular Galois group for which the inverse Galois problem is not solved!
(I don’t know any example of this scenario, however; in fact, I don’t know what would be the/a “smallest” group for which the inverse Galois group is not solved).