$
\newcommand{\A}{\Bbb A}
\newcommand{\Set}{\mathsf{Set}}
\newcommand{\Sch}{\mathsf{Sch}}
$
Define a ring variety to be a variety$^{[1]}$ $X$ over a field $k$, such that the functor of points
$$\Sch_k \to \Set,
\quad S \mapsto \mathrm{Hom}_k(S, X)$$
factors through the forgetful functor from the category of rings to $\sf Set$.
The exercise 1.4 here claims that the only ring variety which is proper over $k$ is the point $\mathrm{Spec}(k)$. My question is to understand why.
(Exercise 1 here assumes $k$ to be algebraically closed).
Notice that this not true without the completeness hypothesis.
Typically, $\A^n_k : S \mapsto O_S(S)^n = \Gamma(S, O_S)^n$ is an affine ring variety.
When $n=r^2$, we can also endow $\A^{r^2}_k : S \mapsto M_r(O_S(S))$ with a structure of non-commutative ring variety. But I don't know what happens if the variety is proper.
Related: (1).
$^{[1]}$
i.e. a separated and integral scheme of finite type.
Best Answer
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 $x$.)