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In Conrad's Grothendieck Duality and Base Change, p. 6, it is said:

When $f:X\to Y$ is a smooth map of schemes, then $\omega_{X/Y}$ denotes the top exterior power of the locally free finite rank sheaf $\Omega_{X/Y}^1$ on $X$.

My question is: for $k$ a field, what is “the top power of $\Omega_{X/Y}^1$” when $f$ equals $\coprod_{n\geq 0}\mathbb{A}_k^n\to\operatorname{Spec}k$?

Elías Guisado Villalgordo
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  • Funnily enough, Hartshorne's Residues and Duality doesn't give a precise definition either. In III.1, p. 140, they only define $\omega_{X/Y}$ for a smooth morphism $f$ of pure relative dimension $n$ as $\omega_{X/Y}=\Omega^n_{X/Y}$. In the page immediatly after, he proceeds to invoke $\omega_{X/Y}$ for a smooth morphism that does not necessarily have pure relative dimension. – Elías Guisado Villalgordo Oct 30 '23 at 19:32
  • The imprecisions in Hartshorne's Residues and Duality is why Brian Conrad wrote his book, so I don't think it's that surprising. – KReiser Oct 30 '23 at 21:37
  • I sometimes have the feeling that algebraic geometry is just authors through the decades passing the buck to the next generation to come. – Elías Guisado Villalgordo Oct 30 '23 at 22:46
  • I mean, the alternative is everyone getting everything correct on the first go-round. I don't know about you, but I have rarely if ever achieved that. (Also, there are some fields that have had worse times.) Anyways, this particular case is not the worst example :) – KReiser Oct 30 '23 at 23:09

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On $\Bbb A^n_k$, it is $\bigwedge^n \Omega_{\Bbb A^n_k/k}^1$. If you have a smooth morphism, $\Omega_{X/Y}^1$ is locally free, and so $X$ is a disjoint union of open sets where $\Omega_{X/Y}^1$ is of constant rank. Take the appropriate exterior power on each of these sets. If you schemes are nice, you can reduce this to the appropriate exterior power on each connected component, and that's a morally instructive picture.

KReiser
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  • Do you mean this definition works in all instances from Conrad's book where notation $\omega_{X/Y}$ is used for a smooth morphism? Also, connected components may not be open, right? – Elías Guisado Villalgordo Oct 30 '23 at 16:49
  • (Okay, connected components are open since a locally Noetherian topological space is locally connected.) – Elías Guisado Villalgordo Oct 30 '23 at 16:56
  • Sure. Dualizing complexes should be local on the source. – KReiser Oct 30 '23 at 17:00
  • I just realized $X$ is an arbitrary scheme in Conrad's definition and does not need to be locally Noetherian. I think the correct definition would go like this: Since $f$ is smooth, the function $r_f:x\in X\mapsto \operatorname{rank}{\mathcal{O}{X,x}}\Omega_{X/Y,x}^1$ is $\mathbb{N}$-valued and locally constant. Thus $X=\coprod U_n$ is an open cover, where $U_n=r_f^{-1}(n)$. Hence, we define $\omega_{X/Y}|{U_n}=\Omega^n{U_n/Y}$. – Elías Guisado Villalgordo Oct 30 '23 at 17:11
  • Yes, that's what I was trying to tell you. (Except for me I rarely think about non-locally-connected schemes, so the locally constant bit translated to connected components.) – KReiser Oct 30 '23 at 17:31