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Suppose $B$ is integral over $\mathbb{Z}$ such that the field $K$ of fractions of $B$ is finite over $\mathbb{Q}$ of degree $n$. Let $A$ denote the integral closure of $B$ in $K$. Denotes $r_1,r_2$ respectively the number of real embeddings and the number of imaginary embeddings. And let $r=r_1+r_2-1$. By theorem the unit group $A^\times\cong \mathbb{Z}^r\times G$ where $G$ is a finite cyclic group which consists of the roots of unity contained in $K$.

My question is:

Is the unit group $B^{\times}$ of rank $r$ too?

Let us consider a special case that $K=\mathbb{Q}[\sqrt{d}]$ for some square-free positive integer $d$. Then the answer is "Yes" in this case.

Now the rank of $A^{\times}$ is one. And $B\supset \mathbb{Z}[t\sqrt{d}]$ for some positive integer $t$. We know that the equation $X^2-Y^2d=\pm 1$ has infinitely many positive solutions $(x_n,y_n)$. And there are also infinitely many $y_n$ such that $t|y_n$. (A very nice partial "proof" can be found in this answer given by Old John). Hence $B$ has a unit of infinite order! That is to say in this case the rank of $B^{\times}$ must be 1 too.

Thanks.


Edit: I have checked the proof of the theorem and it works for me. However, we are now going to give another proof.

Facts. $A^{\times}\cap B=B^{\times}$ and for an $a\in A$ there is a positive natural number $s$ such that $sa\in B$.

It suffices to show that

For any $u\in A^{\times}$ there is a positive natural number $k$ such that $u^k\in B$.

Proof. Let $x^n+a_{n-1}x^{n-1}+\cdots +a_0$ be the characteristic polynomial of $u$. Then all $a_i$'s are integers and $a_0=\pm 1$. And we obtain: $$ u(1,u,\ldots u^{n-1})=(1,u,\ldots,u^{n-1})M, det(M)=\pm 1 $$ $$ u^k(1,u,\ldots,u^{n-1})=(1,u,\ldots,u^{n-1})M^k, k=1,2,\ldots $$ Say the first column of $M^k$ is $(b_0,b_1,\ldots,b_{n-1})$, then $u^k=b_0+b_1u+\cdots+b_{n-1}u^{n-1}$. There is a positive natural number $s$ such that $su\in B$, hence $u^k\in B$ if $s^{n-1}|b_i$ for $i=1,2,\ldots,n-1$.

But the matrix $M$ is invertible in $GL_{n}(\mathbb{Z}/s^{n-1})$, since $GL_{n}(\mathbb{Z}/s^{n-1})$ is a finite set, $M$ has finite order. It follows that there is a $k$ such that $M^k=1 \pmod{ s^{n-1}}$. With this $k$, $u^k\in B$.

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I am sorry. It seems that it can be directly proved using the proof of the Dirichlet unit theorem, just replace $B$ to $A$. I will check that. –  wxu Mar 15 '13 at 16:41
1  
After you check it, come back and answer your own question! –  David Speyer Mar 15 '13 at 17:05

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