"Consider the irreducible polynomial $g(x)=x^3+6x^2+3x+13 \in \mathbb{Q}[x]$
Let $\alpha$ be any zero of $g$ and $ L =\mathbb{Q}(\alpha) $. Compute $N(\alpha-1)$ and $ \operatorname{Tr}(\alpha-1)$."
What I have so far:
We consider the map $A_{\alpha - 1} : L \to L$. This is given by $A_{\alpha - 1}(x) = (\alpha - 1)x$, where $x\in \mathbb{Q}(\alpha)$
Now we consider how this acts on our basis $\{1,\alpha,\alpha^2\}$
We get $1\to-1+\alpha$, $\alpha\to0-\alpha+\alpha^2 $ and $\alpha^2\to-13-3\alpha-7\alpha^2$, giving us the transformation matrix of:
$\begin{bmatrix}-1 & 0 & -13\\1 & -1 & -3\\0 & 1 & -7\end{bmatrix}$
So $N(\alpha-1) = -1(7+3) - 13(1) = -23$ and $Tr(\alpha-1) = -1-1-7 =-9$
However, if I transform the polynomial by $x = x -1$, we get
$h(x)= g(x-1)= x^3+3x^2-6x+15$
which I believe suggests we have a norm of $15$ and a trace of $-6$
Where have I gone wrong? Is it just an arithmetic mistake somewhere?