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.

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- May 26th 2006, 10:47 AMDMTFundamental units in a number field
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. - May 26th 2006, 01:53 PMrgep
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, 12:09 AMDMT
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, 03:55 AMrgepQuote:

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.