# Formula for sum of three roots in base field

Let $\mathbb K$ be a perfect field and let $P$ be an irreducible, monic polynomial with coefficients in $\mathbb K$ : $P=X^d+\sum_{k=0}^{d-1}a_kX^{k}$. Let $\alpha_1,\alpha_2,\ldots,\alpha_d$ be the (necessarily distinct) roots of $P$. Then we have :

(*) If $g=\alpha_1+\alpha_2$ satisfies $g\in{\mathbb K}$, then $g=-\frac{2a_{d-1}}{d}$.

To see this, note that $P(g-X)$ and $P$ have roots $\alpha_1$ and $\alpha_2$ in common, so by irreducibility we have $P(g-X)=(-1)^d P$. So $\sigma : x \mapsto g-x$ gives a permutation of the roots of $P$. If $d$ is odd, then $\sigma$ has a fixed point, so $\frac{g}{2}$ is a root of $P$ contradicting irreducibility. So $d$ is even, and we obtain (*) by looking at the coefficients in degree $d-1$.

I wonder if (*) can be generalized to three variables instead of two : if $h=\alpha_1+\alpha_2+\alpha_3$ satisfies $h\in{\mathbb K}$, can we express $h$ in terms of the coefficients of $P$ ? Resultants tell us that $h$ will satisfy a relation of degree $d(d-1)(d-2)$, but this relation probably simplifies considerably with the additional assumption $h\in{\mathbb K}$.

The first non-trivial case is $d=6$.

• why are the $\alpha_k$ distinct ? – mercio Nov 6 '15 at 19:14
• @mercio If any two $\alpha_k$ are distinct, $gcd(P,P')$ is nontrivial, so $P$ cannot be irreducible. – Ewan Delanoy Nov 6 '15 at 19:44
• what if $K$ is not perfect and $P$ is not separable and $P'$ is zero ? – mercio Nov 6 '15 at 19:50
• @mercio corrected, thanks. – Ewan Delanoy Nov 6 '15 at 19:51

Since $P$ is irreducible, its Galois group $G$ is transitive and sends every $\alpha_i$ to every $\alpha_j$ an equal number of times for each pair.
Then, assuming the characteristic of $K$ doesn't divide $d$ nor $|G|$, $|G|h = \sum_{\sigma \in G} \sigma(h) = \sum_{\sigma \in G} \sigma(\alpha_1 + \alpha_2 + \alpha_3) = \frac{3|G|}d\sum_{k=1}^d \alpha_k = - \frac{3|G|a_{d-1}}d$
Therefore $h = - \frac{3a_{d-1}}d$.
• now I wonder if the result is still true when the characteristic divides $|G|$ but not $d$. – mercio Nov 6 '15 at 19:52