Let $K=\mathbb{Q}(\sqrt{2})$ and $\mathfrak{p} = 7 \mathbb{Z}$. I found that $7 \mathcal{O}_K = (2\sqrt{2}+1)(-2\sqrt{2}+1)$. The Artin symbol $\left[ \frac{K/\mathbb{Q}}{(2\sqrt{2}+1)} \right]$ is defined as the unique $\sigma$ in ${\rm Gal}(K/\mathbb{Q})$ such that $$ \sigma(\alpha) = \alpha^{\text{Norm} (7\mathbb{Z})} = \alpha^7 \bmod (2\sqrt{2}+1). $$ Here, $\bmod (2\sqrt{2}+1)$ means that the difference belongs to the ideal $(2\sqrt{2}+1) = (2\sqrt{2}+1)\mathcal{O}_K = (2\sqrt{2}+1)\mathbb{Z}[\sqrt{2}]$.
Via Sage, I found that $\sigma$ is the identity and if I choose $\alpha = \frac{2}{5} \sqrt{2}$ I get $$\alpha^{\text{Norm} (7\mathbb{Z})} - \alpha = \alpha^7 - \alpha = -\frac{30226}{78125}\sqrt{2} \neq 0 \mod (2\sqrt{2}+1).$$
So, what is the point that I am missing?