Theorem (Lang-Nishimura). Let $X \,-\!\!\rightarrow Y$ be a rational map between $k$-varieties, where $Y$ is proper. If $X$ has a smooth $k$-point, then $Y$ has a $k$-point.
Does the theorem of Lang-Nishimura still hold if any of the following changes are made?
- The assumption that $Y$ is proper is dropped.
- The given $k$-point on $X$ is not assumed to be smooth.
Thanks in advance!
The theorem of Lang-Nishimura can fail if either of the aforementioned changes is made.
Let $k$ be a finite field. Let $X = \mathbb{A}_k^1$, which is irreducible. Let $Y$ be the open subvariety $X - X(k)$ of $X$. The identity $Y \to Y$ represents a rational map $X \,-\!\!\rightarrow Y$. Moreover, $X$ has a smooth $k$-point (in fact, all of $X$ is smooth over $k$), but $Y(k) = \emptyset$.
Let $k = \mathbb{R}$. Let $X$ be the affine curve $x^2 + y^2 = 0$ in $\mathbb{A}_\mathbb{R}^2$, which has exactly one $\mathbb{R}$-point, namely $P := (0, 0)$. Note that $X$ is not smooth at $P$. Since the polynomial $x^2 + y^2$ is irreducible over $\mathbb{R}$, the variety $X$ is irreducible. Let $Y$ be the subscheme of $\mathbb{A}_\mathbb{R}^1$ defined by $t^2 + 1 = 0$. Since $Y$ is finite over $\mathbb{R}$, it is proper. Then $(x, y) \mapsto y/x$ defines a rational map $X \,-\!\!\rightarrow Y$ defined everywhere except $P$. But $Y(\mathbb{R}) = \emptyset$.
One can also find a counterexample in which $X$ and $Y$ are geometrically integral. For example, let $X$ be the projective closure in $\mathbb{P}^2_\mathbb{R}$ of the affine curve $y^2 = -(x^4 + 1)$ over $\mathbb{R}$, and let $Y$ be its smooth projective model. One can show that $X$ has a nonsmooth $\mathbb{R}$-point at infinity, and that $Y(\mathbb{R})$ is empty.