Let $X$ be a proper ring variety over a field $k$. Note first that $X$ is geometrically connected since it has a $k$-rational point ($0$ and $1$ are both rational points). Let $Y$ be the base-change of $X$ to an algebraic closure $\overline{k}$. Note that $Y_{red}$ is a subring of $Y$ (the ring operations $Y\times Y\to Y$ restrict to morphisms $Y_{red}\times Y_{red}\to Y_{red}$ since $Y_{red}\times Y_{red}$ is reduced). Since $Y$ is connected, so is $Y_{red}$, and it follows that $Y_{red}$ is irreducible (since its additive group structure makes it homogeneous). If we show that $0=1$ in $Y_{red}$, that will imply $0=1$ in $Y$, and so $0=1$ in $X$ as well so $X$ is the trivial ring variety.
The upshot of all of this is that $Y_{red}$ is a proper ring variety over $\overline{k}$ and it suffices to show that $Y_{red}$ is trivial. In other words, we may replace $X$ with $Y_{red}$ and assume that $k$ is algebraically closed.
Now consider the morphism $f:X\times X\to X\times X$ given by $f(x,y)=(xy,y)$. Since we are assuming $k$ is algebraically closed, $X\times X$ is irreducible. Note that the fiber of $f$ over $(1,1)$ is just $\{(1,1)\}$. It follows that the generic nonempty fiber of $f$ is $0$-dimensional and so the image of $f$ has the same dimension as $X\times X$. Since $X\times X$ is proper the image of $f$ is closed and since $X\times X$ is irreducible the image is dense, and so $f$ is surjective. In particular, there exist $x,y\in X$ such that $f(x,y)=(1,0)$. This implies $y=0$ and so $1=xy=0$. Thus $0=1$ in $X$ and $X$ is the trivial ring variety.
(This argument does not use the full ring structure of $X$, but merely uses the fact that $X$ has a multiplication operation with an element $1$ such that $x1=x$ for all $x$ and an element $0$ such that $x0=0$ for all $x$.)
More generally, let $R_i, i \in I$ be a finite collection of commutative rings and let $S$ be a commutative ring. Check that $\text{Hom}(\prod R_i, S)$ can naturally be identified with the data of
- a decomposition $S = \prod S_i$ of $S$ into a product over the same index set, and
- a tuple $f_i : R_i \to S_i$ of ring homomorphisms.
(There are several ways to think about this. A hint is to examine the images of the primitive idempotents $e_i = (0, \dots 1, \dots 0)$, where the $1$ occurs in the $i^{th}$ place.)
When each $r_i = \mathbb{Z}$ the ring homomorphisms $f_i$ are unique so the data is the data of a decomposition of $S$ into a product over $I$. Geometrically this is the same thing as a decomposition of $\text{Spec } S$ into $I$ disjoint components, which is in turn the same thing as a continuous (equivalently, locally constant) function $\text{Spec } S \to I$.
The case $|I| = 2$ is particularly easy to think about because we only need to track one nontrivial idempotent. $\mathbb{Z} \times \mathbb{Z} \cong \mathbb{Z}[e]/(e^2 - e)$ is the free commutative ring on an idempotent, so
$$\text{Hom}(\mathbb{Z} \times \mathbb{Z}, S) \cong \{ e \in S : e^2 = e \}$$
is precisely the set of idempotents in $S$, which you can check is in natural bijection with the set of ways to decompose $\text{Spec } S$ into two disjoint components.
Geometrically, the general claim is that a morphism $\text{Spec } S \to \bigsqcup \text{Spec } R_i$ disconnects $\text{Spec } S$ into $I$ components, the pullbacks of the morphism along the inclusions of each $\text{Spec } R_i$; more formally, the slice category over $\bigsqcup \text{Spec } R_i$ is naturally isomorphic to the product of the slice categories over each $\text{Spec } R_i$, which is one of the axioms defining an extensive category. This means intuitively that coproducts behave "disjointly," the way coproducts of sets or spaces do (but not the way coproducts of, say, groups do).
It's very false that $\text{Hom}(\mathbb{Z}^I, S) = S^I$ (although it's true that $\mathbb{Z}^I \otimes S \cong S^I$). The LHS is a set and the RHS is a ring, and the LHS is covariant in $I$ while the RHS is contravariant.
Another way to think about this result is to first check that $\text{Mod}(\prod R_i) \cong \prod \text{Mod}(R_i)$; that is, the category of modules over a finite product breaks up into a finite product of module categories, and the equivalence is given very explicitly via the primitive idempotents above. This is even an equivalence of symmetric monoidal categories, and so induces an equivalence from the category of commutative $\prod R_i$-algebras to the product of categories of commutative $R_i$-algebras, and the equivalence is the identification above.
Edit: Here's a cute corollary. Consider morphisms $\text{Hom}(R^I, R^J)$ where $I, J$ are finite sets and $R$ is a ring with no nontrivial endomorphisms (e.g. $\mathbb{Z}$, any localization of $\mathbb{Z}$, any prime field, but also weirder examples like $\mathbb{R}$). We get that such maps correspond exactly to morphisms $J \to I$ of finite sets, hence:
Claim: Let $R$ be a ring with no nontrivial endomorphisms. Then $\text{FinSet} \ni I \mapsto \text{Spec } R^I \in \text{Aff}$ is a fully faithful embedding.
Best Answer
Yes.
The opposites confuse me about which of "left" and "right" I'm supposed to say. It should be left with the correct choice of ops.
$\alpha$ deserves to be called "affinization." It's the universal map from a scheme or generalized scheme into an affine scheme; that is, it's the left adjoint of the inclusion of affine schemes into schemes / generalized schemes. As a simple example, the affinization of projective space is a point.