# proof of finite inseparable extensions of Dedekind domains are Dedekind

We show that the integral closure  $A$ of $R$ in $L$ is a Dedekind domain.

We use the characterization of Dedekind domains as integral domains  in which all nonzero ideals are invertible   (see proof that a domain is Dedekind if its ideals are invertible). Note that for any $x\in A$, $x^{q}$ is in $K$ and is integral over $R$ so, by integral closure, $x^{q}\in R$.

So, let $\mathfrak{a}$ be a nonzero ideal in $A$, and let $\mathfrak{b}$ be the ideal of $R$ generated by terms of the form $a^{q}$ for $a\in\mathfrak{a}$,

 $\mathfrak{b}=\left(a^{q}:a\in\mathfrak{a}\right)_{R}.$

Then, as $R$ is a Dedekind domain, there is a fractional ideal  $\mathfrak{b}^{-1}$ of $R$ such that $\mathfrak{b}\mathfrak{b}^{-1}=R$, and write $\mathfrak{b}^{-1}_{A}$ for the fractional ideal of $A$ generated by $\mathfrak{b}^{-1}$. Then,

 $1\in R=\mathfrak{b}\mathfrak{b}^{-1}\subseteq\mathfrak{a}^{q}\mathfrak{b}^{-1}% _{A}.$ (1)

On the other hand, if $a_{1},\ldots,a_{q}\in\mathfrak{a}$ and $b\in\mathfrak{b}^{-1}$ then

 $(a_{1}\cdots a_{q}b)^{q}=(a_{1}^{q}b)\cdots(a_{q}^{q}b)\in R,$

so $a_{1}\cdots a_{q}b$ is integral over $R$ and is in $A$. Therefore, $\mathfrak{a}^{q}\mathfrak{b}^{-1}_{A}\subseteq A$. Combining with (1) gives $\mathfrak{a}^{q}\mathfrak{b}^{-1}_{A}=A$, so $\mathfrak{a}$ is invertible with inverse     $\mathfrak{a}^{q-1}\mathfrak{b}^{-1}_{A}$.

Title proof of finite inseparable extensions of Dedekind domains are Dedekind ProofOfFiniteInseparableExtensionsOfDedekindDomainsAreDedekind 2013-03-22 18:35:42 2013-03-22 18:35:42 gel (22282) gel (22282) 5 gel (22282) Proof msc 13A15 msc 13F05