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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?

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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.

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