We know (eg, see Vakil's Foundations of Algebraic Geometry, 11.1.B) that a scheme of finite dimension $n$ admits an open covering by affines of dimension $\leq n$ with equality holding at least once.
I was wondering, if $X$ is furthermore irreducible and of finite type, then can we further assume that the covering by open affines above can be chosen to consist of
- (scheme morphisms whose underlying base-space maps are) homeomorphisms onto their images in $X$,
- with domains (Specs) all of dimension $n$?
You don't need $X$ to be irreducible; pure of dimension $n$ will do (all components have dimension $n$). Let's also assume reducedness for now and that $X$ is a finite-type $k$-scheme. Then the irreducible components of $X$, denoted $X_i$, are integral schemes and for any local ring $\mathcal{O}_{X_i,P}$, we have $\dim \mathcal{O}_{X_i,P}=\mathrm{trdeg}_k K(X_i)=n.$ In particular, the same is true for any affine open subscheme of $X_i$. So, cover each $X_i$ by affines $U_{ij}=\mathrm{Spec} A_{ij}$. This gives the cover you are after.