Just to comment to @Karl_Peter if $X/k$ is any finite type scheme and $X(k)\ne\varnothing$ then $X$ is connected if and only if it's geometrically connected (e.g. see this: math.stackexchange.com/a/3168956/16497)

@dorebell It is true that if $X$ and $Y$ are finite type over a field $k$ then surjective on closed points implies surjective. Indeed, the image is constructible by Chevalley's theorem. So, it's complement is constructible. But, closed points are very dense, so if the complement was non-empty it would contain a closed point.

group structures on affine space (e.g. Witt schemes). Am I missing something silly?

@Angelo Hey Angelo--sorry to dig up such an old question. You said that this is not true for non-perfect, and I don't see why. The automorphism sheaf of $\mathbb{A}^1$ is still an extension of the additive group and multiplicative group, and computing their cohomology is insensitive to doing so on the flat or etale site, so it shouldn't matter whether we're dealing with the separable or algebraic closure. I know that it's true there are non-trivial twists of the additive group over non-perfect fields, but this is not a twisting problem but the fact that one can show there are non-standard

Oh, wait, I'm being dumb. The Albanese variety IS a birational invariant. I was stupid and thinking the fact that Picard group was not a birational invariant meant that the Albanese variety is not, but this is obviously false--it's the component group of the Picard scheme which is not a birational invariant, the Albanese is fine. I'll leave my silly comment above there in case anyone else makes the same mistake. I guess it also explains a fractional, fractional part of the result you cited. In fact, I guess that the birational invariance of the Albanese is why $H^1$ is usually OK. Cheers!

for crystalline cohomology (that I know--I suppose this is part of what your cited result says). Anyways, anything you could say would be very helpful. Thanks!

I guess what's confusing to me is the following. What I said above is not super unintuitive, I suppose, since I've really just said that if $X$ and $Y$ are birational, then their Albanese varieties are birational, and so intuitively their associated $p$-divisible groups (which are classified by the Dieudonne module/first cristalline cohomology) agree. I don't know why this should be the case. Namely, for etale cohomology one expects first cohomology with, say, constant coefficients to not change under birational equivalence since $\pi_1$ doesn't change. But, there's no analagous thing

isomorphic. This is surprising since, of course, the Albanese variety itself is far from being a birational invariant. Have I made a mistake somewhere and, if not, can you say anything about this?

Perhaps I am being silly, but your discussion of the crystalline case seems to imply the following, which I find surprising. Namely, if $X/k$ is smooth proper then $H^1_\text{cris}(X/K_0)$ always has slopes bounded by $1$. In fact, it's the Dieudonne isocrystals associated with $A[p^\infty]$ where $A=\text{Alb}(X)$. So from what you said above it seems to imply, unless I've misread something, that if $X$ and $Y$ are birational that the Dieudonne isocrystals of their Albanese varieties are isomorphic and thus, consequently, that the $p$-divisible groups of their Albanese varieties are

meaning the smallest subcategories closed under the six operations and containing an 'important subcategory' (vector bundles for Coh, and lisse Z_\ell sheaves for constructible)?

@user40276 I see--I suppose that makes sense, but I have no idea how you deduced that from the OPs question or their above comment. Anyways, is there actually a 'six functors formalism' in the abstract? Namely, as far as I know there's no abstract definition of a 'six functors formalism' to make precise the statement 'Qcoh is the largest subcategory of Ab having a 'six functors formalism' '. Is there one I am unaware of, or were you just giving an intuition about what sort of thing the OP wants? Isn't it true that in both Qcoh and constructible they are the 'six operations hull'

in the etale setting. I apologize if I have misinterpreted your question. Dorebell seems to think you're asking for the analogous objects in etale cohomology for coherents in coherent(=Zariski) cohomology. If that's true then, yes, I agree that for most applications of etale cohomology one is usually interested in the cohomology of a lisse $\mathbb{Z}_\ell$-sheaf (or perhaps a constructible $\mathbb{Z}_\ell$-sheaf).

Are you asking about coherent sheaves for the etale topology? One of the first thing one shows (using faithfully flat descent) is that, essentially, $\mathrm{QCoh}(X_{\acute{e}\text{t}})\cong \mathrm{QCoh}(X_{\mathrm{Zar}}$ in the obvious way (namely if $\mathcal{F}$ is quasi-coherent on $X_{\acute{e}\text{t}}$ then its pullback to the Zariski site is a quasi-coherent) and that this preserves global sections. In particular, the etale cohomology is the same as the Zariski cohomology of coherent sheaves. In other words NOTHING new is gained by thinking about quasi-coherent sheaves