Mixing
Tangent space of a scheme and subschemes of length two
$begingroup$
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.
algebraic-geometry schemes
asked Jan 21 at 21:55
FedericoFederico
918313
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add a comment |
$begingroup$
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.
algebraic-geometry schemes
asked Jan 21 at 21:55
FedericoFederico
918313
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
algebraic-geometry schemes
asked Jan 21 at 21:55
FedericoFederico
918313
$endgroup$
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.
algebraic-geometry schemes
algebraic-geometry schemes
asked Jan 21 at 21:55
FedericoFederico
918313
asked Jan 21 at 21:55
FedericoFederico
918313
edited Jan 21 at 22:02
Federico
asked Jan 21 at 21:55
FedericoFederico
918313
asked Jan 21 at 21:55
FedericoFederico
918313
asked Jan 21 at 21:55
FedericoFederico
918313
918313
$begingroup$
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
add a comment |
$begingroup$
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
$begingroup$
You can have distinct but isomorphic closed subschemes!
$endgroup$
– loch
Jan 22 at 10:47
$begingroup$
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
$begingroup$
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
$begingroup$
@loch Sorry, I forgot to tag you.
$endgroup$
– Federico
Jan 22 at 17:47
add a comment |
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$begingroup$
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