# Fundamental units in a number field

• May 26th 2006, 11:47 AM
DMT
Fundamental units in a number field
How do you prove that $1+\sqrt{2}$ is a fundamental unit in $Q(\sqrt{2})$?

I'm just hitting dead ends with all my attempts.
• May 26th 2006, 02:53 PM
rgep
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.
• May 28th 2006, 01:09 AM
DMT
But the norm of a unit can be -1 not just 1, which gives x-1/x for the trace, which is not guaranteed to be >1 if x>1.
• May 28th 2006, 04:55 AM
rgep
Quote:

But the norm of a unit can be -1 not just 1
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.

In this case it's easy to check that $3+2\sqrt2$ is a fundamental totally positive unit. The square root $1+\sqrt2$ is also in the field ${\mathbf Q}\left(\sqrt2\right)$ 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 $3+2\sqrt2$: that was a slip.