How do you prove that $\displaystyle 1+\sqrt{2}$ is a fundamental unit in $\displaystyle Q(\sqrt{2})$?
I'm just hitting dead ends with all my attempts.
Suppose that 1 < x is a unit with norm +1 in the ring of integers. Then x' (conjugate, got by changing the sign on the square root) is 1/x and the trace x+x' = x+1/x is an integer. You have a unit of trace 2, so if x is smaller then x+x' = x+1/x is an integer less than 2. But x>1 so x+x' > 1, a contradiction.
Quite correct. But if x is a unit of norm -1 then x^2 is a unit of norm (-1)^2=+1. So you need to firstly find the fundamental totally positive unit (smallest unit > 1 with norm +1) and then check its square root to see whether that too is a unit in the field.But the norm of a unit can be -1 not just 1
In this case it's easy to check that $\displaystyle 3+2\sqrt2$ is a fundamental totally positive unit. The square root $\displaystyle 1+\sqrt2$ is also in the field $\displaystyle {\mathbf Q}\left(\sqrt2\right)$ and hence is the fundamental unit, and has norm -1.
BTW, when I said 'trace 2' in my previous post, I should have said 'trace 6', referring to $\displaystyle 3+2\sqrt2$: that was a slip.