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# 数学代写|数论代写Number Theory代考|MATH3240

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## 数学代写|数论代写Number Theory代考|Units of Norm 1

Let $m$ be a positive squarefree integer. Theorem 11.2.1 tells us that there exist positive integers $x$ and $y$ such that $x^2-m y^2=1$. Hence $\lambda=x+y \sqrt{m}$ is a unit of $O_K$, where $K=\mathbb{Q}(\sqrt{m})$, such that $\lambda>1$ and $N(\lambda)=1$. Since $\lambda^n \rightarrow \infty$ as $n \rightarrow \infty, O_K$ has infinitely many units of norm 1 , namely $\left{\lambda^n \mid n \in \mathbb{Z}\right}$. All of these units are of the form $u+v \sqrt{m}$, where $u$ and $v$ are integers such that $u^2-m v^2=1$. However, when $m \equiv 1(\bmod 4)$, there may be units in $O_K$ of the form $(u+v \sqrt{m}) / 2$, where $u$ and $v$ are both odd integers. For example $(3+\sqrt{5}) / 2$ is a unit of norm 1 in $O_{\mathbb{Q}(\sqrt{5})}$. In contrast, $O_{\mathbb{Q}(\sqrt{17})}$ does not contain any units of the form $(u+v \sqrt{17}) / 2$, where $u$ and $v$ are both odd integers, since $u^2-17 v^2= \pm 4$ cannot hold modulo 8 for odd integers $u$ and $v$.

Let $\lambda=x+y \sqrt{m}$ be a unit of $O_K(K=\mathbb{Q}(\sqrt{m}))$ of norm 1 with $x$ and $y$ both integers or possibly in the case $m \equiv 1(\bmod 4)$ both halves of odd integers. We now show how the signs of $x$ and $y$ determine to which of the four intervals $(-\infty,-1),(-1,0),(0,1)$, or $(1, \infty) \lambda$ belongs.

Theorem 11.3.1 Let $m$ be a positive squarefree integer. Let $x$ and $y$ both be integers or both halves of odd integers such that $x^2-m y^2=1$. Then
\begin{aligned} x+y \sqrt{m}>1 & \Longleftrightarrow x>0, y>0, \ 00, y<0, \ -10, \ x+y \sqrt{m}<-1 & \Longleftrightarrow x<0, y<0 . \end{aligned}

## 数学代写|数论代写Number Theory代考|Units of Norm −1

Let $m$ be a positive squarefree integer. We have already observed that the ring $O_{\mathbb{Q}(\sqrt{m})}$ of integers of the real quadratic field $\mathbb{Q}(\sqrt{m})$ may or may not contain units of norm -1 . Indeed $O_{\mathbb{Q}(\sqrt{2})}$ has units such as $1+\sqrt{2}$ of norm -1 whereas $O_{\mathbb{Q}(\sqrt{3})}$ does not contain any units of norm -1 . We suppose that $O_{\mathbb{Q}(\sqrt{m})}$ contains units of norm -1 and show that there exists a unique unit $\sigma>1$ in $O_{\mathbb{Q}(\sqrt{m})}$ of norm -1 such that all units in $O_{\mathbb{Q}(\sqrt{m})}$ of norm -1 are given by $\pm \sigma^{2 k+1}(k=0, \pm 1, \pm 2, \ldots)$ and all units in $O_{\mathbb{Q}(\sqrt{m})}$ of norm 1 are given by $\pm \sigma^{2 k}(k=0, \pm 1, \pm 2, \ldots)$.

Theorem 11.4.1 Let $m$ be a positive squarefree integer. Suppose that $O_{\mathbb{Q}(\sqrt{m})}$ contains units of norm -1 . Then there exists a unique unit $\sigma>1$ of norm -1 in $O_{\mathbb{Q}(\sqrt{m})}$ such that every unit in $O_{\mathbb{Q}(\sqrt{m})}$ is of the form $\pm \sigma^n$ for some integer $n$.

Proof: Let $\rho$ be a unit in $O_{\mathbb{Q}(\sqrt{m})}$ of norm -1 . Let $\rho^{\prime}$ denote its conjugate. Then
$$\rho \rho^{\prime}=N(\rho)=-1$$
so that
$$\rho^2 \rho^{\prime 2}=1$$
Thus $\rho^2$ is a unit of $O_{\mathbb{Q}(\sqrt{m})}$ of norm 1. Hence, by Theorem 11.3.2(b), we have
$$\rho^2= \pm \epsilon^n$$
for some integer $n$, where $\epsilon$ is the fundamental unit of $O_{\mathbb{Q}(\sqrt{m})}$ of norm 1. Clearly $\rho^2>0$ and $\epsilon^n>0$ so that
$$\rho^2=\epsilon^n .$$
If $n$ is even, say $n=2 k$, then
$$\rho^2=\epsilon^{2 k}$$
so that
$$\rho= \pm \epsilon^k$$
Hence
$$N(\rho)=N\left( \pm \epsilon^k\right)=N(\epsilon)^k=1,$$
contradicting $N(\rho)=-1$. Thus $n$ must be odd, say $n=2 l+1$, and so
$$\rho^2=\epsilon^{2 l+1} .$$
Hence
$$\epsilon=\left(\rho \epsilon^{-l}\right)^2 .$$

## 数学代写|数论代写Number Theory代考|Units of Norm 1

\begin{aligned} x+y \sqrt{m}>1 & \Longleftrightarrow x>0, y>0, \ 00, y<0, \ -10, \ x+y \sqrt{m}<-1 & \Longleftrightarrow x<0, y<0 . \end{aligned}

## 数学代写|数论代写Number Theory代考|Units of Norm −1

$$\rho \rho^{\prime}=N(\rho)=-1$$

$$\rho^2 \rho^{\prime 2}=1$$

$$\rho^2= \pm \epsilon^n$$

$$\rho^2=\epsilon^n .$$

$$\rho^2=\epsilon^{2 k}$$

$$\rho= \pm \epsilon^k$$

$$N(\rho)=N\left( \pm \epsilon^k\right)=N(\epsilon)^k=1,$$

$$\rho^2=\epsilon^{2 l+1} .$$

$$\epsilon=\left(\rho \epsilon^{-l}\right)^2 .$$

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