Is still $A\otimes_{R} K$ a Dedekind ring if $A$ is? Let $A$ be a Dedekind ring containing a ring $R$, let $Q$ be the quotient field of $R$ and let $Q\subset K$ be a finite field extension.
Is the ring $A\otimes_{R} K$ a Dedekind ring? If not, are there some "minimal" assumptions I could make on $R$ or $K$ under which $A\otimes_{R} K$ will be a Dedekind ring? 
Many thanks!
 A: Let $R=\mathbf{F}_p[T]$, so $Q=\mathbf{F}_p(T)$. Now take $K=\mathbf{F}_p(T^{1/p})=Q[X]/(X^p-T)$. If $A$ is the integral closure of $R$ in $K$, then $A$ is a Dedekind domain finite over $R$. We have
$A\otimes_RK=(A\otimes_RQ)\otimes_QK=K\otimes_QK=K[X]/(X^p-T)=K[X]/((X-T^{1/p})^p)$,
which is a non-reduced local ring. 
In terms of hypotheses on the rings involved which ensure an affirmative answer, nothing general comes to mind immediately. It would be helpful if the OP clarified exactly what is meant by a Dedekind ring (for example, if such a ring must be $1$-dimensional or if e.g. fields are allowed). For me, a Dedekind domain is a $1$-dimensional Noetherian domain for which the localization at each nonzero prime ideal is a discrete valuation ring. I don't know if the term "Dedekind ring" is meant to be synonymous with "Dedekind domain" or if it is meant to specify a generalization of the latter concept, presumably in which the condition that the ring be a domain is dropped. 
One thing to be noted is that $A\otimes_RK$ is a $K$-algebra. If $A$ happens to be a finite $R$-module, as was the case in my example, then $A\otimes_RK$ will be a finite $K$-algebra. Any such ring is zero-dimensional and semi-local, and is in fact canonically isomorphic to the product of its localizations at maximal (equivalently prime) ideals, each of which is a finite Artin local $K$-algebra (never a discrete valuation ring).
Even if $A$ is not finite over $R$, $A\otimes_RK$ may not be Dedekind. For example, let $R=\mathbf{Z}$, $A=\mathbf{Z}_p$, and $K$ a number field (i.e. a finite extension of $\mathbf{Q}=\mathrm{Frac}(R)$). Then
$A\otimes_RK=\mathbf{Z}_p\otimes_\mathbf{Z}K
=(\mathbf{Z}_p\otimes_\mathbf{Z}\mathbf{Q})\otimes_\mathbf{Q}K
=\mathbf{Q}_p\otimes_\mathbf{Q}K=\prod_{v\mid p}K_v$,
where the product is over the places of $K$ lying over $p$ and $K_v$ is the $v$-adic completion of $K$. So again this is not Dedekind. 
If we assume that $R$ is a field and $A$ is of finite type over $R$, then we can say something positive. First, $A\otimes_RK$ will be of finite type over $K$, hence Noetherian. Noether normalization together with flatness of $R\to K$ implies that $1=\dim(A)=\dim(A\otimes_RK)$. Still, this is not enough. The ring $A\otimes_RK$ may not be a domain (it may even fail to be reduced if the field $R$ is imperfect and $K$ is an inseparable extension of $R$). However, if we assume that $A$ is geometrically integral over $R$ (this is equivalent to the requirement that $R$ is algebraically closed in $\mathrm{Frac}(A)$), then $A\otimes_RK$ will be a domain. Some nontrivial commutative algebra will then imply that $A\otimes_RK$ is regular, which is a reformulation of the condition that its localizations at nonzero prime ideals are discrete valuation rings. So we can conclude that $A\otimes_RK$ is a Dedekind domain. (None of this actually uses that $K$ is finite over $R$, and is valid for $K$ an arbitrary extension of $R$.)
