Ring below Dedekind domain is Dedekind domain. Question
Let $R$ be a Dedekind domain with quotient field $K$ and $L$ a subfield of $K$ such that $R$ is integral over $R\cap L$. Show that $R'=R\cap L$ is a Dedekind domain.
Attempt
I have shown that $R'$ is integrally closed domain whose every non-zero prime ideal is maximal. Can anyone assist me on showing $R'$ to be Noetherian.
Any help is appreciated.
 A: Let $A$ be an integral domain with quotient field $K$, let $L$ be a subfield of $K$, and set $B = A \cap L$.  This note is adapted from elements of S. Oda's exposition in the paper On a Subring of an Integral Domain Obtained by Intersecting a Field.  In that paper, $B = A \cap L$ is termed a subring of $A$ with reduced quotient field.
Observe that we can assume $L$ is the quotient field of $B$.  (Indeed, if the quotient field of $B$ were $L'$, then it is easily checked that $L' \subseteq L$ and hence $B = A \cap L'$.)
For any fractional ideal $I$ of a domain $A$ with quotient field $K$, let $I^{-1} = (A :_K I)$ and $I_v = (I^{-1})^{-1}$.  The divisorial fractional ideals are the ideals such that $I_v = I$.  Recall the following facts about the $v$-operation:
(i) $A_v = A$ and $(qI)_v = qI_v$ for any $q \in K$
(ii) $I \subseteq I_v$ and $I \subseteq J \implies I_v \subseteq J_v$, and
(iii) $(I_v)_v = I_v$.
More generally these are the defining properties of so-called $*$-operations.
It is an easy exercise to show that $I_v = \bigcap_{q \in K, I \subseteq qA}qA$, i.e. the divisorial closure of an ideal is the intersection of the principal fractional ideals containing it.

Lemma: Let $B = A \cap L$, so that $L$ is the quotient field of $A \cap L$.  Let $v$ denote the $v$-operation in $A$.  For every divisorial ideal $I$ of $B$, it holds $I = (IA)_v \cap L$.

Proof: Let $q \in L$.  First it is trivial to argue that $qA \cap L = qB$.  Now suppose $I$ is divisorial, so that $I = \bigcap_{q \in L, I \subseteq qB} qB$.
Then by the characterization of divisorial ideals as intersections of principal fractional ideals containing them, we have $I \subseteq L \cap (IA)_v = L \cap \bigcap_{q \in K, IA \subseteq qA} qA \subseteq L \cap \bigcap_{q \in L, I \subseteq qB} qB = I$.  $\square$
A domain is called Mori if it satisfies the ascending chain condition on divisorial ideals.  This will be the lens through which we observe the descent of finiteness conditions from $A$ to $A \cap L$.

Examples of properties that always descend from $A$ to $A \cap L$.

*

*Mori

*Ascending Chain condition on principal ideals

*Valuation domain

*Integrally closed

*Completely integrally closed

*Krull

*DVR


Proof:
The descent of the chain conditions are immediate from the previous lemma:  a chain of divisorial (resp. principal) ideals in $B$ is contracted from a chain of divisorial (resp. principal) ideals in $A$, which stabilizes by assumption, hence the original chain stabilizes in $B$.
If $A$ is a valuation domain, $a, b \in B$, then WLOG $a \mid b$ in $A$ so that $ac = b$ for some $c \in A$.  But then $c = b/a \in L$, so $c \in A \cap L = B$ and $a \mid b$ in $B$.
Now if $A$ is integrally closed then we can write $A = \bigcap_\alpha V_\alpha$ for some collection of valuation overrings of $A$.  Then $B = A \cap L = \bigcap_\alpha (V_\alpha \cap L)$ is an intersection of valuation overrings of $B$, using the previous observation.
If $q \in L$ is almost integral over $B$, then we have $qI \subseteq I$ for some ideal $I$ of $B$.  Then also $qIA \subseteq IA$, i.e. $q$ is almost integral over $A$ as an element of $K$, and by assumption $q \in A$.  So $q \in A \cap L = B$.  (Alternatively you could use this approach to argue descent of integrally closed, because being integral is equivalent to $qI \subseteq I$ for an f.g. ideal ideal).
It is well-known that a domain is Krull iff it completely integrally closed and Mori.  For a reference, see section 2.4 of these course notes by S. Gabelli.
Being a Discrete Valuation Ring, i.e. a Noetherian valuation ring, is equivalent to being a Mori valuation ring.  See 2.15 in the linked notes.
$\square$
This list is by no means meant to be exhaustive.  But these observations are already enough to understand how the property of being Dedekind descends.
Because Dedekind domains are exactly the $1$-dimensional Krull domains, we have the following

Proposition: Let $A$ be a Dedekind domain.  Then $B = A \cap L$ is Dedekind iff it is $1$-dimensional.

In particular this works with your assumption that $A$ is integral over $A \cap L$.
When $A$ is integrally closed and $A \cap L \subseteq A$ is integral, local properties of $A$ also readily descend.  In this case we can apply the following:

Lemma [12.7 in R. Gilmer's Multiplicative Ideal Theory]: Let $B$ an integrally closed domain with field of fractions $L$, and $A \subseteq B$ an integral extension of domains.  For each prime $\mathfrak{q}$ of $A$ contracting to $\mathfrak{p}$ in $B$, it holds $B_\mathfrak{p} = A_\mathfrak{q} \cap L$.

In particular if $A$ is integrally closed then we have observed that $A \cap L$ is integrally closed.  So if $A \cap L \subseteq A$ is integral, we can apply the lemma along with our previous descent observations and get results like:

Lemma [Examples of local properties that descend when $A \cap L \subseteq A$ is integral]: Let $A \cap L \subseteq A$ be integral. If $A$ is Prüfer (i.e. locally a valuation domain) then $A \cap L$ is Prüfer.  If $A$ is locally a DVR then $A \cap L$ is locally a DVR.

