# the minimum polynomial of a unit

Let $A$ be a dvr of characteristic zero. Let $B/A$ be a finite integral extension of $A$.

Suppose that there exists a unit $x$ in $B$ such that $B=A[x]$. What can we say about the minimal polynomial of $x$ over $A$? Is it Eisenstein, do its coefficients satisfy some congruence properties, etc?

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Let $A$ be a domain with field of fractions $F$, integrally closed in $F$, and let $\alpha$ be algebraic over $F$.
Then $\alpha$ is integral over $A$ if and only if its monic irreducible polynomial over $F$ has coefficients in $A$.
Moreover, if $\alpha\neq 0$ and $f(x)$ is the monic irreducible polynomial of $\alpha$, then the monic irreducible polynomial of $\frac{1}{\alpha}$ is $x^nf(\frac{1}{x})$, where $n=\deg(f)$.
In particular, $\alpha$ and $\frac{1}{\alpha}$ are both integral over $A$ if and only if the monic irreducible polynomial of $\alpha$ has coefficients in $A$ and its constant term is a unit of $A$.
Yeah, because the constant term is a unit. So can we also say something about the derivative of $f$? (Here $f$ denotes the monic irreducible polynomial of $\alpha$.) More precisely, can we say something about the valuation of $f^\prime(\alpha)$? (This gives the valuation of the different of $B/A$. But can we make it more explicit?) – Hannah Jan 24 '12 at 21:15
@Hannah: I don't see how you can say much in general, since there are no restrictions on the "middle terms". Note that if $\alpha$ satisfies any monic polynomial with coefficients in $A$ and unit constant term, then $\alpha$ is a unit in the integral closure; so if you take any polynomial that has unit constant term, its roots are always units in the corresponding integral closure. Other than the fact that the coefficient of $x^k$ must be a multiple of $k+1$, I don't see much that can be said in general. – Arturo Magidin Jan 24 '12 at 21:20