Rational points on singular curves and their normalization Let $X$ be a curve over a field $k$. Assume that $X$ is geometrically connected, geometrically reduced and stable.
Let $Y\to X$ be the normalization. Is $Y(k) = X(k)$?
 A: To elaborate on Andrew's answer, there are two things that can occur:
The map $Y(k) \to X(k)$ can fail to be injective, or can fail to be surjective.
More precisely, if $x \in X(k)$ is a node, then the fibre of $Y$ over $x$ will
be a degree two reduced zero-dimensional variety, which will contain either two $k$-points
(in which case the map $Y(k) \to X(k)$ is not injective; this is an in Andrew's example), or no $k$-points, in which case $Y(k) \to X(k)$ will not be surjective.
The non-surjective case can't occur if $k$ is algebraically closed, since then the fibre over a node is just two copies of Spec $k$, but here is a case with $k = \mathbb R$ where it occurs:
Take $X$ to be the curve $y^2 = x^3 - x^2$. (To see that the normalization has no $\mathbb R$-valued points lying over the node, note that these points
correspond to the tangent directions to the two branches through the node.  These tangent directions have the equation $x^2 + y^2 = 0$, whose linear factors are defined over $\mathbb C$, but not over $\mathbb R$.)
A: The answer is no. Consider the curve $C: y^2=x^2+x^3,$ and let $\pi:\widetilde C \to C$ be the blowup of $C$ at $0.$ Then the fibre $\pi^{-1}(0)=\{p,q\}$ consists of two points, while  $\pi$ is an isomorphism everywhere else. Since $\widetilde C$ is a desinglularization of $C,$ it is isomorphic to the normalization of $C.$
