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Tangent space of a scheme and subschemes of length two

Mathematics Asked by Federico on October 19, 2020

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.

One Answer

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

This is a crucial point. Remember, that a closed subscheme $Z subset X$ is an equivalence class of closed immersions $Z hookrightarrow X$, where two such morphisms $f, g$ are equivalent if and only if there is an isomphism $phi: Z to Z$, making the following diagram commute: $$begin{align*}Z & xrightarrow{f} X\ phi downarrow & nearrow g \ Zend{align*}$$

So both things can happen:

  1. You can have two different morphisms $f, g: Z hookrightarrow X$, which define the same subscheme $Z subset X$.
  2. You can have two different subschemes $Z_1, Z_2 subset X$, which are abstractly isomorphic as schemes $Z_1 cong Z_2$, but not as subschemes of $X$, even if they have the same points in $X$.

So much for the general theory, now to your question.

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

$DeclareMathOperator{Spec}{Spec}$I don't belive the claim of Huybrechts is true as written.

  1. You already mentioned the first problem I see: The zero-vector, which is the morphism $Spec(k[epsilon]) to Spec(k) to X$, does not define a subscheme of length two.

  2. Two different morphisms $f,g: Spec(k[epsilon]) to X$(i.e. two different tangent vectors), define the same subscheme in $X$, if there is an isomorphism of $k[epsilon]$ commuting with $f$ and $g$. But any such isomorphism is given by a map $k[epsilon] to k[epsilon], epsilon mapsto c cdot epsilon$ for any $c in k setminus {0}$. So two tangent vectors define the same subscheme if and only if they differ by a scalar.

Those two points show, that the length two subschemes concentrated at $x$ are rather the projectification of the tangent vectors, i.e. the space $mathbb{P}(T_x)$ (or $mathbb{P}(T_x^*)$, depending on your notation).

Answered by red_trumpet on October 19, 2020

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