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everyone. I am having difficulties with this question.

Let $K = \mathbb{Q}(\sqrt{d})$ be a real quadratic field , and $ \mathcal{O}_K$ be the ring of integers of $K$. By making use of $u$, a fundamental unit (coming from a solution of the Pell equation), show that there is no such thing as the smallest possible positive integer in $\mathcal{O}_K$ (in the sense of the usual order of the real numbers).

Thank you in advance.

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    The title does not reflect the question.2011-04-28

3 Answers 3

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If $u$ is a non-trivial unit then either $|u|<1$ or $|1/u|<1$. So, there is a unit $u$ with $|u|<1$. Then the powers of $u$ are a sequence of units converging to $0$. (Here, $|\cdot|$ is the real absolute value, not the field norm.)

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Hint. Suppose $u$ is a fundamental unit of $K$. Then one of $u$, $u^{-1}$, $-u$, or $-u^{-1}$ is positive and smaller than $1$ (in the usual order of $\mathbb{R}$).

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If you are having trouble with this how about an example: $\mathbb Q(\sqrt{3})$.

In this field we have $2^2 - 3 \cdot 1^2 = 1$ and this is the smallest so let $u = 2 - 1 \cdot \sqrt{3}$ be a fundamental unit.

Now here are some calculations:

 u^1 = 0.2679491924311227064725536585...  u^2 = 0.0717967697244908258902146339...  u^3 = 0.0192378864668405970883048774...  u^4 = 0.0051547761428715624630048756...  u^5 = 0.0013812181046456527637146252...  u^6 = 0.0003700962757110485918536254...  u^7 = 0.0000991669981985416036998764... 

but they all have norm 1, so what's going on!

The answer is that norm measures not the size of the number in the "real" sense but in a higher dimensional lattice.

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    The answer is that one is equivocating on the meaning of "norm": there's the **field norm** $\mathbb{Q}(\sqrt{3})\to\mathbb{Q}$ given by the propduct of all conjugates, and then there's the *archimedean absolute value/norm* given by the usual absolute value/complex norm.2011-04-28