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Let $X$ be an $S$-scheme. That is there exists a morphism of schemes $X \to S$, in particular we can view $X$ as a family over $S$.

Also, suppose that $X$ is a $k$ scheme, i.e there exists a morphism $X \to \operatorname{Spec} k$?

How would you describe the structure sheaf on $X$ as a scheme over $k$ vs. the structure sheaf on $X$ as a scheme over $S$?

More generally, I am a bit confused as to how the ring of functions on a scheme $X$ is affected by a scheme parameterizing $X$.

Maybe there exists something like a relative structure sheaf $\mathcal{O}_{X/S}$ vs. $\mathcal{O}_X$...

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  • $\begingroup$ A morphism $X \to \operatorname{Spec} k$ for any commutative ring $k$ gives $\mathcal{O}_X$ the structure of a sheaf of $k$-algebras, rather than just a sheaf of rings. $\endgroup$
    – Joppy
    Commented Mar 18, 2018 at 0:36

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There is no such thing as "the structure sheaf on $X$ as a scheme over $S$" or "the structure sheaf on $X$ as a scheme over $k$". There's just the structure sheaf $\mathcal{O}_X$ of $X$ as a scheme, which does not depend on any base scheme.

The additional structure that a morphism $f:X\to S$ does give you on the structure sheaf is a morphism of sheaves of rings $f^{-1}\mathcal{O}_S\to\mathcal{O}_X$. In other words, it makes $\mathcal{O}_X$ not just a sheaf of rings but a sheaf of algebras over the sheaf of rings $f^{-1}\mathcal{O}_S$. (In the specific case that $S=\operatorname{Spec} k$ for a field $k$, the sheaf of rings $f^{-1}\mathcal{O}_S$ is just the constant sheaf $k$ on $X$, so this is equivalent to making $\mathcal{O}_X$ a sheaf of $k$-algebras.)

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No change.

Let $\mathcal{X}$ denote the relative scheme $X \xrightarrow{\pi} S$. I use a different letter so that we can distinguish between "schemes" and "schemes over $S$".

Let $\mathbb{A}^1 = \mathrm{Spec}(\mathbb{Z}[x])$ denote the "affine line". For a scheme $S$, let $\mathbb{A}^1_S = \mathbb{A}^1 \times S$ denote the "affine line over $S$".

Recall that the structure sheaf is conceived of as the collection of algebraic functions. In fact, we can make this rather literal: the structure sheaf is a representable presheaf! There is a natural (in $U$) isomorphism $\mathcal{O}_X(U) \cong \hom_{\mathbf{Sch}}(U, \mathbb{A}^1)$.

Given a scheme $\mathcal{X}$ over $S$ viewed as an $S$-indexed family of schemes, we can think of there being functions on $\mathcal{X}_s$. There are a few ways how we might aggregate these together; the one we want is:

  • A function on $\mathcal{X}$ is an $S$-indexed family of functions

This boils down to the same thing as "a function on $X$". If this isn't clear, compare with the case of sets: given a family $f_s$ of functions, we can construct a new function $g$ by $g(s,x) = f_s(x)$. Conversely, given $g$, we can reconstruct the family $f_s$.

An alternative is that we might ask for an $S$-indexed family of function spaces. I don't know if this higher-order notion has any reasonable incarnation in algebraic geometry.


Another reason why this is the right thing is that we want algebraic geometry over $S$ to mimic algebraic geometry over $\mathbb{Z}$. Central to this is the role of the affine line; we would like functions on $\mathcal{X}$ to be maps to the affine line over $S$! That is,

$${\mathcal{O}}_{\mathcal{X}} \cong \hom_{\mathbf{Sch}/S}(-, \mathbb{A}^1_S) $$

However, we have a general identity for any relative scheme $\mathcal{X}$ and ordinary scheme $Y$:

$$ \hom_{\mathbf{Sch}/S}(\mathcal{X}, Y \times S \to S) \cong \hom_{\mathbf{Sch}}(X, Y) $$

So, $\mathcal{O}_{\mathcal{X}}$ and $\mathcal{O}_X$ should be basically the same thing.

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