# Let $D'$ the subring of $D$-integral elements of $E$. Why $D'$ is Dedekind?

I am studying the theorem:

Let $D$ be a Dedekind domain, $F$ its field of fractions, $E$ a finite dimensional extension field of $E$, $D'$ the subring of $D$-integral elements of $E$. Then $D'$ is Dedekind.

I am working with the book "Basic Algebra 2". At first it is taken in consider the case when $E$ is separable over $F$. Then it is assumed that $\operatorname{char}F=p$ and $E$ is purely inseparable over $F$. Then, for all $u$ in $E$ we have that $u^{q}$ is in $F$, where $q$ is a power of $p$. Let $F'$ be the algebraic closure of $F$ containing $E$ and let $F^{1/q}(D^{1/q})$ be the subfield of $F'$ of elements $v$ such that $v^q$ is in $F(D)$.

What is the set $F(D)$? Why $v\mapsto$$v^q$ is an isomorphism of $F^{1/q}$ onto $F$ mapping $D^{1/q}$ onto $D$. What is the set $D^{1/q}$?

How can I arrive in the conclusion that D' is Dedekind?

Would you help me, please? Thank you in advance.

• That's strange. I would expect $F(D)$ to be defined as the smallest subfield of $F'$ containing both $F$ and $D$, which is just $F$. – D_S Jul 4 '17 at 5:22

## 1 Answer

I'm going to follow the usual notation and write $\overline{F}$ for an algebraic closure of $F$ containing $E$, rather than $F'$.

You have probably misread what Jacobson wrote. He is writing "Let $F^{1/q}$ (respectively $D^{1/q}$) be the subfield (resp. subring) of elements $v$ of $\overline{F}$ such that $v^{1/q} \in F$ (resp. $D$)." He omits the word "respectively" so that it looks like $F(D)$ is a single object rather than two different objects.

In general, if $\phi: A \rightarrow B$ is an isomorphism between two rings, and if $B_0$ is a subring of $B$, then the preimage $\phi^{-1}(B_0)$ is a subring of $A$, and $\phi$ restricts to an isomorphism of rings $\phi^{-1}(B_0) \rightarrow B_0$.

The map $x \mapsto x^q$ is a ring homomorphism $\phi: \overline{F} \rightarrow \overline{F}$. It is automatically injective, as is any ring homomorphism from a field to a nonzero ring, and it is surjective, because $\overline{F}$ is algebraically closed. So it is an isomorphism.

By definition, $F^{1/q}$ is the preimage of $F$ under the map $\phi$. So we immediately see that $\phi$ restricts to an isomorphism $F^{1/q} \rightarrow F$. In the same way, $D^{1/q}$ is the preimage of $D$ under $\phi$, so we get an isomorphism $D^{1/q} \rightarrow D$.

You should be able to understand the proof now. Let me know if you have more questions.