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edited for correctness
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Keith Millar
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The above lemma is essentially a direct consequence of Chinese remainder theorem; for any $I$ and $\nu$, there is $p\in K$ such that $pI$ is coprimecritical to $\nu$the proof, in which case $pI$ becomes principalbut standard in $\mathcal{O}_{K_\nu}$, so you can just divide bynumber theory as the image ofring $p$$\mathcal{O}_{K_\nu}$ is a PID.

The above lemma is essentially a direct consequence of Chinese remainder theorem; for any $I$ and $\nu$, there is $p\in K$ such that $pI$ is coprime to $\nu$, in which case $pI$ becomes principal in $\mathcal{O}_{K_\nu}$, so you can just divide by the image of $p$.

The above lemma is critical to the proof, but standard in number theory as the ring $\mathcal{O}_{K_\nu}$ is a PID.

fixed explanations of the real points
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Keith Millar
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Notably, this dictionary also makes it possible to view that, for example, $\mathcal{O}_{\mathbb{R}}^\times=[-1,1]^\times$$\mathcal{O}_{\mathbb{C}}^\times=\mathbb{T}$, so $x(\mathbb{R})$ is the subset of complex numbers $u$ in the unit disccircle such that $\frac{x}{u}\in\mathbb{R}$. However, any algebraic number $x$ then has $\frac{x}{|x|}=u$ in the unit disccircle, so then $u$ is a real point. This therefore becomes trivial in the calculation of the group.

Notably, this dictionary also makes it possible to view that, for example, $\mathcal{O}_{\mathbb{R}}^\times=[-1,1]^\times$, so $x(\mathbb{R})$ is the subset of complex numbers $u$ in the unit disc such that $\frac{x}{u}\in\mathbb{R}$. However, any algebraic number $x$ then has $\frac{x}{|x|}=u$ in the unit disc, so then $u$ is a real point. This therefore becomes trivial in the calculation of the group.

Notably, this dictionary also makes it possible to view that, for example, $\mathcal{O}_{\mathbb{C}}^\times=\mathbb{T}$, so $x(\mathbb{R})$ is the subset of complex numbers $u$ in the unit circle such that $\frac{x}{u}\in\mathbb{R}$. However, any algebraic number $x$ then has $\frac{x}{|x|}=u$ in the unit circle, so then $u$ is a real point. This therefore becomes trivial in the calculation of the group.

minor correction again
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Keith Millar
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  1. $\mathcal{O}_{-}^\times$ is like an abelian variety defined over $\mathbb{Q}$.
  2. Principal homogeneous spaces for $\mathcal{O}_{-}^\times$ are (nonzero) algebraic numbers $x$.
  3. For a principal homogeneous space $x$ of $\mathcal{O}_{-}^\times$, $x(K)$ is the subset of $\mathcal{O}_\bar{K}^\times$ consisting of $u$ such that $\frac{x}{u}\in K$. Members of $x(K)$ are $K$-rational points.
  4. $x(K_\nu)$ is then the subset of $\mathcal{O}_{\bar{K}_\nu}^\times$ consisting of $u$ such that $\frac{x_\nu}{u}\in K_\nu$, where $x_\nu$ is the image of $x$ along some chosen embedding $\bar{K}\rightarrow\bar{K}_\nu$.
  1. $\mathcal{O}_{-}^\times$ is like an abelian variety defined over $\mathbb{Q}$.
  2. Principal homogeneous spaces for $\mathcal{O}_{-}^\times$ are algebraic numbers $x$.
  3. For a principal homogeneous space $x$ of $\mathcal{O}_{-}^\times$, $x(K)$ is the subset of $\mathcal{O}_\bar{K}^\times$ consisting of $u$ such that $\frac{x}{u}\in K$. Members of $x(K)$ are $K$-rational points.
  4. $x(K_\nu)$ is then the subset of $\mathcal{O}_{\bar{K}_\nu}^\times$ consisting of $u$ such that $\frac{x_\nu}{u}\in K_\nu$, where $x_\nu$ is the image of $x$ along some chosen embedding $\bar{K}\rightarrow\bar{K}_\nu$.
  1. $\mathcal{O}_{-}^\times$ is like an abelian variety defined over $\mathbb{Q}$.
  2. Principal homogeneous spaces for $\mathcal{O}_{-}^\times$ are (nonzero) algebraic numbers $x$.
  3. For a principal homogeneous space $x$ of $\mathcal{O}_{-}^\times$, $x(K)$ is the subset of $\mathcal{O}_\bar{K}^\times$ consisting of $u$ such that $\frac{x}{u}\in K$. Members of $x(K)$ are $K$-rational points.
  4. $x(K_\nu)$ is then the subset of $\mathcal{O}_{\bar{K}_\nu}^\times$ consisting of $u$ such that $\frac{x_\nu}{u}\in K_\nu$, where $x_\nu$ is the image of $x$ along some chosen embedding $\bar{K}\rightarrow\bar{K}_\nu$.
clarified real point structure
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Keith Millar
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removed decidedly unecessary information
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Keith Millar
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Names of references
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LSpice
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added clarification on the meaning of Sha
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Keith Millar
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Keith Millar
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