etale morphisms and equidimensionality

Let $(R,m)\rightarrow (R',m')$ be a local ring map (of noetherian rings) which is étale local, namely it essentially of finite type, it is flat, $mR'=m'$ and the field extension $R/m\subset R'/m'$ is finite and separable.

Which properties on $R$ ensure that $R'$ is equidimensional? For example, if $R$ is a domain, is it true that $R'$ is equidimensional?

Evidently if $$R$$ is not itself equidimensional there is no hope, so assume $$R$$ is a domain. Note that for every minimal prime $$\mathfrak p\subset R’$$, $$\mathfrak p\cap R=(0)$$ since $$R\to R’$$ is faithfully flat, hence satisfies going down. A sufficient condition to ensure that $$R’$$ be equidimensional is that $$R$$ be universally catenary, because then we have the formula $$\dim R’/\mathfrak p=\dim R+\operatorname{trdeg}_R(R’/\mathfrak p)-\operatorname{trdeg}_{k(m)}k(m’).$$ Here $$\operatorname{trdeg}_R(R’/\mathfrak p)$$ means transcendence degree of the function fields. It is zero in our case since the function field of $$R’/\mathfrak p$$ is a finite separable extension of that of $$R$$; $$\operatorname{trdeg}_{k(m)}k(m’)=0$$ for the same reason.