A global section "is" a map to $\mathbb{A}^1$

Question

Suppose $T$ is a scheme and $\mathbb{A}^1=\operatorname{Spec}\mathbb{Z}[x]$. I'd like to understand why the global sections of $T$, $\Gamma(\mathcal{O}_T)$, can be interpreted as maps to $\mathbb{A}^1$. Using II.2.4 in Hatshorne, we have $\operatorname{Hom}(T,\mathbb{A}^1)\cong\operatorname{Hom}(\mathbb{Z}[x],\Gamma(\mathcal{O}_T))$, but why is this isomorphic to $\Gamma(\mathcal{O}_T)$? I am having trouble constructing a map from the former to the later, let alone an isomorphism.

Answer 1

Ring maps out of $\Bbb Z[x]$ are entirely determined by where they send $x$ - recall $1\mapsto 1$, and then $x$ can go anywhere. This shows that $\operatorname{Hom}_{\text{Ring}}(\Bbb Z[x],A)=A$ for any commutative unital ring $A$.

_{(I'm surprised this isn't a duplicate anywhere - or at least no related questions and approach0 doesn't show anything in a first glance. If it turns out I missed something, please leave a comment and I'll take care of it.)}