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Math Help - Fundamental units in a number field

  1. #1
    DMT
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    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.
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  2. #2
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    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.
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  3. #3
    DMT
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    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.
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    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.
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