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

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