Suppose $V$ is a quasi-affine set such that $k[V]\simeq k$, where $k[V]$ is the coordinate ring, and $k$ a field such that $k=\bar{k}$. Does this force $V$ to just be a point?
I'm curious because I know that if $V$ consists of $m$ points, then $k[V]\simeq k^m$ as $k$-algebras. So if $V$ is a singleton, then $k[V]\simeq k$, and my question is just the converse.
Does something like this make any sense? If $k[V]\simeq k$, then $k\simeq k[\mathbb{A}^n]/I(V)$, so since $k$ is a field, $I(V)$ is maximal, so it corresponds to a point and then $V$ is just a point? Or is it more complicated than that?
Yes, if $k[V]=k$ then $V$ is a single point.
Indeed, if $V\subset X$ is some open subset of the affine set $X$, the global sections $k[X]$ separate the points of $X$ (meaning that given two points of $X$ some regular function takes different values at them) and thus by restricting these regular functions to $V$ we obtain regular functions on $V$ that separate the points of $V$.
But if $k[V]=k$ these functions on $V$ cannot separate two points, so that $V$ contains only one point!