Let $\mathbb{K}$ be a finite-degree algebraic field extension of $\mathbb{Q}$, and let $\xi$ be algebraic over $\mathbb{K}$ of degree $d$ minimal polynomial: $$X^{d}-\sum_{k=0}^{d-1}\alpha_{k}X^{k}\in\mathbb{K}\left[X\right]$$ i.e., so that: $$\xi^{d}=\sum_{k=0}^{d-1}\alpha_{k}\xi^{k}$$ Letting $m$ be a positive integer, one can use the above identity to re-write $\xi^{m+d}$ as a linear combination of $1,\xi,\xi^{2},\ldots,\xi^{d-1}$. Namely, for each $m$, there are polynomials $$P_{m}\left(\alpha_{k},\alpha_{k-1},\ldots,\alpha_{k-m}\right)$$ so that $$\xi^{d+m}=\sum_{k=0}^{d-1}P_{m}\left(\alpha_{k},\alpha_{k-1},\ldots,\alpha_{k-m}\right)\xi^{k}$$ with the convention that $\alpha_{j}=0$ for all $j\leq-1$. Doing the computations by hand for the first few values of $m$ gives:
$P_{0}=\alpha_{k}$
$P_{1}=\alpha_{k-1}+\alpha_{k}\alpha_{d-1}$
$P_{2}=\alpha_{k-2}+\alpha_{k-1}\alpha_{d-1}+\alpha_{k}\left(\alpha_{d-1}^{2}+\alpha_{d-2}\right)$
$P_{3}=\alpha_{k-3}+\alpha_{k-2}\alpha_{d-1}+\alpha_{k-1}\left(\alpha_{d-1}^{2}+\alpha_{d-2}\right)+\alpha_{k}\left(\alpha_{d-1}^{3}+2\alpha_{d-1}\alpha_{d-2}+\alpha_{d-3}\right)$
Clearly, there are polynomials:
$Q_{0}=1$
$Q_{1}=\alpha_{d-1}$
$Q_{2}=\alpha_{d-1}^{2}+\alpha_{d-2}$
$Q_{3}=\alpha_{d-1}^{3}+2\alpha_{d-1}\alpha_{d-2}+\alpha_{d-3}$
So that: $$P_{m}=\alpha_{k-m}Q_{0}+\alpha_{k-\left(m-1\right)}Q_{1}+\alpha_{k-\left(m-2\right)}Q_{2}+\cdots+\alpha_{k}Q_{m}=\sum_{j=0}^{m}\alpha_{k-m+j}Q_{j}$$ at least for $m\leq d-1$. My question is: what is the general formula for $Q_{j}$? And what is the general formula for $P_{m}$, including when $m\geq d$?