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For $A=\mathbb{Z}[x]/(f)$ with quotient field $K$ and ring of integers $B$, does $U(B)/U(A)$ have a name?

For instance $u = \tfrac{1+\sqrt{5}}{2}$ is a unit in $\mathbb{Q}[\sqrt{5}]$, but neither $u$ nor $u^2$ has integer coefficients in the basis $\{ 1, \sqrt{5} \}$. Of course $u^3$ has integer coefficients (spooky if you haven't tried it!) and in fact $u^n$ has integer coefficients iff $0 \equiv n \mod 3$.

For quadratic fields with basis $\{ 1, \sqrt{n} \}$ for $n$ square-free, one almost always has $U(A) = U(B)$. If not, then $[ U(B) : U(A) ] = 3$.

That's crazy, and it should have a name. For instance, I'd like to find out if the following is true, but I don't even know what to look for:

Is $U(B)/U(A)$ always finite? [ where $B$ is the ring of integers of an order $A$ in a number ring ]

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In general, if $A \subset B$ is an extension of rings, I would call $B^{\times}/A^{\times}$ the relative unit group. (I am probably not the only one, but I couldn't say how widespread this is.) I feel reasonably confident that there is no specialized terminology in the case of nonmaximal orders.

To answer your non-terminological question: yes, if $A \subset B$ are orders in the same number field, the group $B^{\times}/A^{\times}$ is finite. This follows from the fact that one can prove the Dirichlet Unit Theorem equally well for a nonmaximal order $A$ in a number field $K$: $A^{\times}$ is still finitely generated with torsion subgroup equal to the roots of unity in $A$ and free rank equal to $r+s-1$, where $r$ is the number of real places and $s$ is the number of complex places of $K$. Thus we have finitely generated abelian groups $A^{\times} \subset B^{\times}$ with the same free rank, so $B^{\times}/A^{\times}$ is finite.

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    Sounds good to me. The unit theorem at least tells me an answer exists, and the rest of your answer tells me I'm not simply missing the name of some well-known Kash or $M$$a$gma comman$d$ that calculates this using class field theory or something.2011-02-15