## Compute the norm of a given element in an algebraic number field

Compute the norm of $1 + \sqrt{7}$ in $\mathbb{Q}(i,\sqrt{7})$.

Recall that the norm of $1+\sqrt{7}$ over $\mathbb{Q}(i,\sqrt{7})$ is the product of its conjugates there.

Let $\theta = i + \sqrt{7}$. We claim that $\mathbb{Q}(\theta) = \mathbb{Q}(i,\sqrt{7})$. Indeed, $\sqrt{7} = (20\theta - \theta^3)/16$ and $i = (\theta^3 - 4\theta)/16$. Moreover, we have $1 + \sqrt{7} = 1 + \frac{5}{4}\theta - \frac{1}{16}\theta^3 = q(\theta)$.

Note that $i+\sqrt{7}$ is a root of $p(x) = x^4 - 12x^2 + 64$, and that the other roots of this polynomial are $\pm i \pm \sqrt{7}$. We claim that $p$ is in fact the minimal polynomial of $\theta$ over $\mathbb{Q}$. To see this, it suffices to show that $\mathbb{Q}(\theta)$ does not have degree 2 over $\mathbb{Q}$. To that end, suppose $i \in \mathbb{Q}(\sqrt{7})$; that is, suppose $a+b\sqrt{7}$ exists such that $(a+b\sqrt{7})^2 = -1$. Comparing coefficients, we have $a^2 + 7b^2 = -1$, a contradiction since the left hand side is positive while the right is negative. So $\mathbb{Q}(i,\sqrt{7})$ has degree 2 over $\mathbb{Q}$, and thus $\mathbb{Q}(\theta)$ has degree 4 over $\mathbb{Q}$. In particular, $\pm i \pm \sqrt{7}$ are the conjugates of $\theta$.

Thus we see that the conjugates of $1+\sqrt{7}$ for $\mathbb{Q}(\theta)$ are $1+\sqrt{7}$ (twice) and $1 - \sqrt{7}$ (twice). Thus the norm of $1+\sqrt{7}$ is \$latex N(1+\sqrt{7}) = (1+\sqrt{7})^2(1-\sqrt{7})^2 = 36