Let $X, Y$ be irreducible projective schemes over $\mathbb{C}$ and $X \subset Y$. Let $x \in X$ be a closed point. Assume that for any positive integer $n$ and any morphism from $\mathrm{Spec} (\mathbb{C}[t]/(t^n))$ to $Y$ such that its composition with the natural morphism from $\mathrm{Spec}(\mathbb{C})$ to $\mathrm{Spec}(\mathbb{C}[t]/(t^n))$ corresponds to the closed point $x$, we have that this morphism factors through $X$. Does this imply that there exists an open neighbourhood $U$ of $x$ in $Y$ such that $U$ is contained in $X$?


Suppose that $X$ has smaller dimension at $x$ than $Y$. Embed $Y\subseteq\mathbb{P}^n$ using a square of a very ample line bundle. For a general linear subspace $L$ in $\mathbb{P}^n$ through $x$ of the right codimension, $X\cap L$ will be finite and $Y\cap L$ will be a curve. Normalizing that curve and completing at some preimage of $x$ gives you a map $Spec(k[[t]])\to Y$, mapping the closed point to $x$, whose image is not contained in $X$. But then some $k[[t]]/(t^n)$ does not map to $X$.

The above shows that the answer is yes if $Y$ is reduced. See answer_bot's comment below for a counterexample if $Y$ is non-reduced.

  • 3
    $\begingroup$ This argument only works if $Y$ is reduced. A counter example to the question is $\mathbf{C}[x, y]/(x^2, y^2) \to \mathbf{C}[x, y]/(x^2, xy, y^2)$. $\endgroup$ – answer_bot Apr 24 '14 at 20:05
  • $\begingroup$ answer_bot: of course, thanks for spotting that! $\endgroup$ – Piotr Achinger Apr 24 '14 at 21:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.