Regular Functions are Morphisms to $\mathbb A^1_{\mathbb Z}$

Question

My question refers to an answer given to a former thread of mine:

Linear Morphism between Schemes

Brandon wrote:

A regular function is the same thing as a map to $\mathbb A^1_{\mathbb Z}$, so $\eta(x)$ really does live on $\mathbb A^1_{\mathbb Z}$. It's good to think about why this should be.

If we start with a scheme $U$ and we call it's global sections $\mathcal{O}_U(U)$ "regular functions".

But I don't see how these regular functions give rise for morphisms to $\mathbb A^1_{\mathbb Z}$?

A naive approach would be to take a $η \in \mathcal{O}_U(U)$ which provides following map: $η:U→∐_{x∈U}\kappa(x),x↦κ(x)=\mathcal{O}_{U,x}/p_x$ where $p_x$ is the prime ideal of an affine neighbourhood of $x \in U$.

But how does it map to $\mathbb A^1_{\mathbb Z}$? Does anybody has an idea what Brandon had here in mind?

Answer 1

The affine line $\mathbb{A}^1_{\mathbb{Z}}=\operatorname{Spec}\mathbb{Z}[T]$ represents the functor $X\mapsto \mathcal{O}_X(X)$ on the category of schemes: $$ \operatorname{Hom}_{\operatorname{Sch}}(X,\operatorname{Spec}\mathbb{Z}[T])=\operatorname{Hom}_{\operatorname{Rings}}(\mathbb{Z}[T],\mathcal{O}_X(X))=\mathcal{O}_X(X), $$ where the last identification is via evaluation at $T$.

Answer 2

We have a functor $\text{Spec}\colon \mathsf{Ring}^{\text{op}}\to \mathsf{Sch}$ whose left adjoint is the global sections functor $\Gamma\colon \mathsf{Sch}\to \mathsf{Ring}^{\text{op}}$ (the global sections functor, where $\Gamma(X) = \mathcal{O}_X(X)$). Now $\mathbb{A}^1_\mathbb{Z} = \text{Spec}(\mathbb{Z}[x])$, and $\mathbb{Z}[x]$ is the free ring on one generator, so a map $f\colon \mathbb{Z}[x]\to R$ is determined by the element $f(x)\in R$. So:

\begin{align*} \text{Hom}_{\mathsf{Sch}}(X,\text{Spec}(\mathbb{Z}[x])) &\cong \text{Hom}_{\mathsf{Ring}^{\text{op}}}(\Gamma(X),\mathbb{Z}[x])\\ &\cong \text{Hom}_{\mathsf{Ring}}(\mathbb{Z}[x],\Gamma(X))\\ &\cong \Gamma(X). \end{align*}

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Ok, that's pretty abstract. More concretely, suppose you have a global section $a\in \Gamma(X)$. This determines a map $\mathbb{Z}[x]\to \Gamma(X) = \mathcal{O}_X(X)$ which is "evaluate the polynomial on $x = a$". For any affine open $\text{Spec}(A)\subseteq X$, this map restricts to a ring map $\mathbb{Z}[x]\to \mathcal{O}_X(A) = A$. So we get a map $\text{Spec}(A)\to \mathbb{A}^1_\mathbb{Z} = \text{Spec}(\mathbb{Z}[x])$ given pointwise by taking preimages of prime ideals under the map determined by $x\mapsto a$. This is a coherent family of maps from an affine open cover of $X$ to $\mathbb{A}^1_\mathbb{Z}$, which glue to a map $X\to \mathbb{A}^1_\mathbb{Z}$.

Answer 3

Here's the general fact you need.

If $X$ is a scheme and $A$ is a ring, then morphisms $X\to\operatorname{Spec}(A)$ are in a 1-1 correspondence with ring homomorphisms $A\to\mathcal O_X(X)$.

Explicitly, one direction of this correspondence is given by taking a morphism $f:X\to\operatorname{Spec}(A)$, and looking at the sheaf (of rings) morphism $\mathcal O_{\operatorname{Spec}(A)}\to f_*\mathcal O_X$ (recall this is part of the data of a morphism of schemes). Then looking at how this morphism acts on global sections of $\operatorname{Spec}(A)$ we have a ring homomorphism $$A=\mathcal O_{\operatorname{Spec}(A)}(\operatorname{Spec}(A))\to f_*\mathcal O_X(\operatorname{Spec}(A))=\mathcal O_X(X).$$

Back to the point, if you take $X=U$ and $A=\mathbb Z[x]$, we see morphisms $U\to\mathbb A_{\mathbb Z}^1$ are in 1-1 correspondence with ring homomorphisms $\Bbb Z[x]\to\mathcal O_X(X)$, but in general if $R$ is a ring then ring homomorphisms $\Bbb Z[x]\to R$ are in bijection with elements of $R$ simply by choosing where $x$ goes, so this gives us what we want.