Suppose $F=\mathbb{Q(\alpha)}$ and that $\alpha$ is the root of the irreducible polynomial $p(x)=x^n+a_{n-1}x^{n-1}+\cdots + a_0$. I am told that the trace of $\alpha$ is $-a_{n-1}$, but I don't understand why.
I know that we have $\mathbb{Q(\alpha)}\cong \mathbb{Q}[x]/p(x)$, and that the norm is equal to the sum of the conjugates of $\alpha$, but how do I calculate this?
In the Galois closure (the splitting field of $p(x)$), we know $$ p(x)=\prod_{i=1}^n (x-\alpha_i) $$ where the $\alpha=\alpha_1,\dots,\alpha_n$ are the conjugates of $\alpha$. Now multiply out this product and look at the coefficient of the degree $n-1$ term: it's the sum of all the conjugates.
EDIT: More explicitly, if we multiply out this product, we have $$ p(x)=x^n-(\sum_{i=1}^n\alpha_i)x^{n-1}+\cdots +(-1)^n\prod_{i=1}^n\alpha_i=x^n+a_{n-1}x^{n-1}+\cdots +a_0. $$ Hence $Tr_{\mathbb{Q}(\alpha)/\mathbb{Q}}(\alpha)=\sum_{i=1}^n\alpha_i=-a_{n-1}$ and $N_{\mathbb{Q}(\alpha)/\mathbb{Q}}(\alpha)=\prod_{i=1}^n\alpha_i=(-1)^na_0$.