proof of finite inseparable extensions of Dedekind domains are Dedekind

Let $R$ be a Dedekind domain with field of fractions $K$ and $L/K$ be a field extension. We suppose that $K$ has characteristic (http://planetmath.org/characteristic) $p>0$ and that there is a $q=p^{r}$ such that $x^{q}\\in K$ for all $x\\in L$ . In particular, this is satisfied if it is a purely inseparable and finite extension.

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

We cannot apply the same method of proof as for the proof of finite separable extensions of Dedekind domains are Dedekind, because here $A$ does not have to be finitely generated as an $R$ -module.

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 $\\mathfrak{a}^{q-1}\\mathfrak{b}^{-1}_{A}$ .