I found in Huybrecht's book Fourier Mukai transforms in Algebraic Geometry the following statement

A tangent vector $v$ at $x \in X$ is the data of a length two subscheme $Z$ concentrated at x

Here $X$ is a $k$-scheme and $x$ is a closed point of $X$. I tried to prove this equivalence, but I am not sure whether I did it correctly or not. I know that tangent vectors are in bijection with the set of $k$-scheme homomorphisms $Hom_{k} \left( \text{Spec} \; k[\epsilon], X \right)$, where I set $k[\epsilon] = k[\epsilon] \left/ (\epsilon^2) \right.$. Now my idea was to do the following:

To every $\phi \in Hom_{k} \left( \text{Spec} \; k[\epsilon], X \right)$ I attach its schematic image $Z_{\phi}$. This is a subscheme of $X$ supported at $x$ and the stalk at $x$ is given by $\mathcal{O}_{X,x} \left/ \text{ker} \phi_{x} \simeq k[\epsilon] \right.$ for every morphism which is not the trivial one (i.e. the one factorizing for the inclusion of the closed point, which gives a subscheme of length 1). Therefore, $Z_{\phi}$ is the required subscheme.

For every subscheme $i : Z \rightarrow X$ of length two let us consider $\mathcal{O}_{Z,x} \simeq \mathcal{O}_{X,x} \left/ \mathcal{I}_{Z,x} \right.$. As this is a length 2 module over $\mathcal{O}_{X,x}$, and $\mathcal{O}_{X,x}$ is a local ring, we have a short exact sequence $$ 0 \rightarrow k(x) \rightarrow \mathcal{O}_{X,x} \left/ \mathcal{I}_{Z,x} \right. \rightarrow k(x) \rightarrow 0 $$ of $k(x)$-vector spaces. Therefore, $\mathcal{O}_{X,x} \left/ \mathcal{I}_{Z,x} \simeq k(x) \oplus k(x) \right.$ as a vector space, and the multiplication on the left can easily be transferred on the right because $\left( m_{x} \left/ \mathcal{I}_{Z,x} \right. \right)^2 = 0$, namely we have $(a,b) \cdot (c,d) = (ac, ad+bc)$. We now define $\mathcal{O}_{X,x} \rightarrow k[\epsilon]$ as the composition of the projection $\mathcal{O}_{X,x} \rightarrow \mathcal{O}_{Z,x}$, the isomorphism $\mathcal{O}_{X,x} \left/ \mathcal{I}_{Z,x} \simeq k(x) \oplus k(x) \right.$ and the map $(a,b) \mapsto a + \epsilon b$.

Is this correct? I fear it is not, because it seems like all the subschemes have the same structure sheaf, namely $k[\epsilon]$, with the same multiplication structure.

  • $\begingroup$ You can have distinct but isomorphic closed subschemes! $\endgroup$ – loch Jan 22 at 10:47
  • $\begingroup$ So is my answer correct? Therefore, all the subschemes are isomorphic? But then where is the information of the tangent vector? $\endgroup$ – Federico Jan 22 at 10:55
  • $\begingroup$ @loch Sorry, I forgot to tag you. $\endgroup$ – Federico Jan 22 at 17:47

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