# Trace of an algebraic number

Let $\alpha$ have minimal polynomial $m(x) = x^n + a_{n-1} x^{n-1} + \dots + a_0.$ Show that \begin{equation*} \mathrm{Tr}(\alpha^k) = -k a_{n-k} - \sum_{i=1}^{k-1} a_{n-i} \mathrm{Tr}(\alpha^{k-i}) \end{equation*} I see that \begin{equation*} \mathrm{Tr}(\alpha^k) = \sum_{i=1}^{n} \sigma_i(\alpha^k) = \sum_{i=1}^{n} \sigma_i^k(\alpha)= \sigma_1^k(\alpha)+ \sigma_2^k(\alpha)+\cdots+ \sigma_n^k(\alpha). \end{equation*} which are related to symmetric polynomials and Newton's Identities. Newton's Identities. I am not able to put things together. How do I go about this. Thanks in advance.

• I’m not sure, but maybe if you try the companion matrix of $\alpha$ (and it made things really hard for this geezer when you used $\alpha$ for the root and $a_i$ for the coefficients of its minimal polynomial). Indeed, the companion matrix seems to be the key. A real mess, though. Feb 11, 2016 at 4:59

Let $X_1, \dotsc, X_n$ be the roots of $m$.

Denote by $e_j(X_1, \dotsc, X_n)$ the $j$-th (degree is $j$) elementary symmetric polynomial in $X_1, \dotsc, X_n$ and denote by $p_j$ the power-sum $p_j := \sum\limits_{i=1}^n X_i^j$.

Let us write down the claim in terms of the $X_i$:

$$\sum_{j=1}^n X_i^k = -k(-1)^ke_k(X_1, \dotsc, X_n) - \sum_{i=1}^{k-1}(-1)^ie_i(X_1, \dotsc, X_n)\sum_{j=1}^n X_i^{k-i}$$

Or in short terms, we have to show:

$$p_k=-k(-1)^ke_k - \sum_{i=1}^{k-1}(-1)^ie_ip_{k-i} ~~~~~~~~~~ (\star)$$

Now we use Newton's identities as stated in the Wikipedia article, you quoted:

$$ke_k=\sum_{i=1}^k (-1)^{i-1}e_{k-i}p_i$$

Write down the summand for $i=k$ explicitly and then change the order of summation and we get:

$$ke_k=(-1)^{k-1}p_k + \sum_{i=1}^{k-1} (-1)^{k-i-1}e_{i}p_{k-i}$$

Multiply with $(-1)^{k}$:

$$k(-1)^ke_k = -p_k - \sum_{i=1}^{k-1} (-1)^{i}e_{i}p_{k-i}$$

This is precisely $(\star)$.