# The intersection of $O_K$ with $K^\ast$

Let $K/\mathbf{Q}$ be a number field with ring of integers $O_K$.

Is $O_K\cap K^\ast = O_K^\ast$?

I can't show that the inverse of an element in $O_K\cap K^\ast$ lies in $O_K^\ast$...

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$O_K \cap K^\times = O_K \cap (K\setminus 0) = O_K \setminus 0$. It might be helpful to think of $K = \mathbf{Q}$, then $O_K = \mathbf{Z}$ but $\mathbf{Z}^\times = \pm 1$. –  tkr Oct 18 '11 at 22:14
Can't possibly be true, since $K^* = K-\{0\}$. The intersection is just all nonzero elements of $\mathcal{O}_K$, and those are never all units (since $\mathcal{O}_K$ always includes $\mathbb{Z}$, but no other rationals). –  Arturo Magidin Oct 18 '11 at 22:17

If $K=\mathbb{Q}$ then the group of units is $K^\times=\mathbb{Q}-\{0\}$ and the ring of integers $\mathcal{O}_K=\mathbb{Z}$. From here we find that $\mathcal{O}_K\cap K^\times=\mathbb{Z}-\{0\}$ is only a multiplicative monoid whereas $\mathcal{O}_K{}^\times=\mathbb{Z}^\times=\{-1,1\}$.

The same applies to e.g. Gaussian integers, where $\mathcal{O}_K\cap K^\times=\mathbb{Z}[i]-\{0\}$ but $\mathcal{O}_K{}^\times=\{\pm1,\pm i\}$.

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Others have answered your question. I'll note that figuring out just exactly what is in $O_K^*$ can be tricky. If $K={\bf Q}(\sqrt{-1})$, then $O_K^*=\lbrace1,-1,i,-i\rbrace$. If $K={\bf Q}(\sqrt{-2})$, then $O_K^*=\lbrace1,-1\rbrace$. If $K={\bf Q}(\sqrt{-3})$, then $O_K^*=\lbrace1,-1,\rho,-\rho,\rho^{-1},-\rho^{-1}\rbrace$, where $\rho$ is a primitive 6th root of 1.

You may at some point come across Dirichlet's Unit Theorem, which sheds some light on $O_K^*$.

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Perhaps what you intended to show is the following:

Let $K$ be a number field, let $\mathcal{O}(K)$ be its ring of integers. Let $B\subseteq\mathcal{O}(K)$ be the collection $$B = \{r\in\mathcal{O}(K)\mid r\neq 0\text{ and } r^{-1}\text{ is integral over \mathcal{O}(K)}\}.$$ That is, all elements of $\mathcal{O}(K)$ whose inverse is integral. Then $B=\mathcal{O}(K)^*$.

This is true. Clearly, $\mathcal{O}(K)^*$ is contained in $B$. Conversely, if $r\in\mathcal{O}(K)$ has an integral inverse, then the inverse is integral over $\mathcal{O}(K)$, which is integral over $\mathbb{Z}$, so $r^{-1}$ is integral over $\mathbb{Z}$; since $r^{-1}\in K$, then $r^{-1}\in \mathcal{O}(K)$, so $r^{-1}\in\mathcal{O}(K)$, hence $r\in\mathcal{O}(K)^*$.

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