An algebro-geometric instance of the inverse function theorem?

Question

Let $f:X \to Y$ be a morphism of schemes of finite type over a field $k$. If $f$ is a closed immersion and the Zariski tangent space of $X$ at a $k$-point $x$ is isomorphic to the Zariski tangent space of $Y$ at $f(x)$, does there exist an open neighborhood $U$ to which the restriction of $f$ is an open immersion of $U$ in $Y$?

Answer 1

If you assume that $X$ is regular at $x$, this is true. For notation fix $\dim_x X = \dim \mathcal O_{X,x}$ and $T_x X = (\mathfrak m_x / \mathfrak m_x^2)^*$, and similarly for $Y$. Now first observe that $Y$ has to be regular in $x$ as well, because the surjection $\mathcal O_{Y,x} \twoheadrightarrow \mathcal O_{X,x}$ and the hypothesis $T_x X = T_x Y$ give us $$\dim_x Y \geq \dim_x X = \dim T_x X = \dim T_x Y \geq \dim_x Y.$$ So $\dim_x Y = \dim T_x Y$, which means that $\mathcal O_{Y,x}$ is a regular local ring. In particular it is an integral domain.

Next we want to show that $\varphi: \mathcal O_{Y,x} \twoheadrightarrow \mathcal O_{X,x}$ is an isomorphism. So assume there is $0 \neq g \in \mathcal O_{Y,x}$ with $\varphi(g) = 0$. Then the ideal $(g)$ has height at least $1$, because it contains the prime ideal $(0)$, as $\mathcal O_{Y,x}$ is an integral domain. But this means $$\dim \mathcal O_{X,x} \leq \dim \mathcal O_{Y,x} / (g) \leq \dim \mathcal O_{Y,x} - 1,$$ because any chain of prime ideals in $\mathcal O_{Y,x}/(g)$ gives a chain in $\mathcal O_{Y,x}$ which can be made longer by appending $(0)$ at the beginning. (Actually we know $\operatorname{ht}(g) = 1$ by Krull's Hauptidealsatz so that we have equality, but this is not important here). This is a contradiction to $\dim \mathcal O_{Y,x} = \dim \mathcal O_{X,x}$, so $\varphi$ is injective, hence an isomorphism.

Not let $U = \operatorname{Spec} A$ be an open affine neighbourhood of $x$ in $Y$. The closed immersion $U \cap X \to U$ is then given by some quotient $$ A \to A/I.$$ The support of $I$ (as a sheaf) is a closed subscheme of $U$, which does not contain $x$. Hence there is an open neighbourhood $V \subset U$ of $x$ with $V \cap X = V$ scheme-theoretically.

Answer 2

No. Take a singular nodal plane curve $X$ and embed it into $Y=\mathbb{A}^2$. At the singularity both tangent spaces are 2-dimensional. You need something like requiring $f$ to be smooth for this to work.