Mixing
Universal properties of immersion of schemes
$begingroup$
In this answer of Martin Brandenburg he refers to the universal properties of the open and closed immersion. As this looks kind of useful I tried to find a reference but I couldn't find any. So my question is
Is there a way to see an open immersion and a closed immersion (and perhaps general immersion) of schemes as a solution to a universal problem?
algebraic-geometry ringed-spaces
asked Nov 26 '17 at 19:56
yamete kudasaiyamete kudasai
1,160818
$endgroup$
add a comment |
$begingroup$
In this answer of Martin Brandenburg he refers to the universal properties of the open and closed immersion. As this looks kind of useful I tried to find a reference but I couldn't find any. So my question is
Is there a way to see an open immersion and a closed immersion (and perhaps general immersion) of schemes as a solution to a universal problem?
algebraic-geometry ringed-spaces
asked Nov 26 '17 at 19:56
yamete kudasaiyamete kudasai
1,160818
$endgroup$
$begingroup$
A my idea: it's the same "universal property" of the inclusion of a (non empty) subset in a set.
$endgroup$
– Armando j18eos
Nov 27 '17 at 15:21
$begingroup$
@Armandoj18eos Are you thinking in something like this? Given $Zsubset Y$ and $f:Xrightarrow Y$ with $f(X)subset Z$ then there exist a unique factorization of $f$ through the inclusion $i:Zrightarrow Y$. If so, I think this is true for open immersion but not in general for closed immersions. Perhaps some modification of this works in the case of closed immersions.
$endgroup$
– yamete kudasai
Nov 27 '17 at 18:07
$begingroup$
Yes, I am! For the closed immersion, the work is more delicated, but I think that this reasoning works. Are you agree?
$endgroup$
– Armando j18eos
Nov 27 '17 at 20:12
$begingroup$
It is not true that $f$ will factor through the closed subscheme $Z$ if $f$ has image inside $Z$. For example, the projection $k[X]to k[X]/(X^2)$ will not factor through $k[X]/(X)=k$.
$endgroup$
– user363120
Nov 27 '17 at 22:03
add a comment |
$begingroup$
In this answer of Martin Brandenburg he refers to the universal properties of the open and closed immersion. As this looks kind of useful I tried to find a reference but I couldn't find any. So my question is
Is there a way to see an open immersion and a closed immersion (and perhaps general immersion) of schemes as a solution to a universal problem?
algebraic-geometry ringed-spaces
asked Nov 26 '17 at 19:56
yamete kudasaiyamete kudasai
1,160818
$endgroup$
In this answer of Martin Brandenburg he refers to the universal properties of the open and closed immersion. As this looks kind of useful I tried to find a reference but I couldn't find any. So my question is
Is there a way to see an open immersion and a closed immersion (and perhaps general immersion) of schemes as a solution to a universal problem?
algebraic-geometry ringed-spaces
algebraic-geometry ringed-spaces
asked Nov 26 '17 at 19:56
yamete kudasaiyamete kudasai
1,160818
asked Nov 26 '17 at 19:56
yamete kudasaiyamete kudasai
1,160818
edited Jan 29 at 20:37
yamete kudasai
asked Nov 26 '17 at 19:56
yamete kudasaiyamete kudasai
1,160818
asked Nov 26 '17 at 19:56
yamete kudasaiyamete kudasai
1,160818
asked Nov 26 '17 at 19:56
yamete kudasaiyamete kudasai
1,160818
1,160818
$begingroup$
A my idea: it's the same "universal property" of the inclusion of a (non empty) subset in a set.
$endgroup$
– Armando j18eos
Nov 27 '17 at 15:21
$begingroup$
@Armandoj18eos Are you thinking in something like this? Given $Zsubset Y$ and $f:Xrightarrow Y$ with $f(X)subset Z$ then there exist a unique factorization of $f$ through the inclusion $i:Zrightarrow Y$. If so, I think this is true for open immersion but not in general for closed immersions. Perhaps some modification of this works in the case of closed immersions.
$endgroup$
– yamete kudasai
Nov 27 '17 at 18:07
$begingroup$
Yes, I am! For the closed immersion, the work is more delicated, but I think that this reasoning works. Are you agree?
$endgroup$
– Armando j18eos
Nov 27 '17 at 20:12
$begingroup$
It is not true that $f$ will factor through the closed subscheme $Z$ if $f$ has image inside $Z$. For example, the projection $k[X]to k[X]/(X^2)$ will not factor through $k[X]/(X)=k$.
$endgroup$
– user363120
Nov 27 '17 at 22:03
add a comment |
$begingroup$
A my idea: it's the same "universal property" of the inclusion of a (non empty) subset in a set.
$endgroup$
– Armando j18eos
Nov 27 '17 at 15:21
$begingroup$
@Armandoj18eos Are you thinking in something like this? Given $Zsubset Y$ and $f:Xrightarrow Y$ with $f(X)subset Z$ then there exist a unique factorization of $f$ through the inclusion $i:Zrightarrow Y$. If so, I think this is true for open immersion but not in general for closed immersions. Perhaps some modification of this works in the case of closed immersions.
$endgroup$
– yamete kudasai
Nov 27 '17 at 18:07
$begingroup$
Yes, I am! For the closed immersion, the work is more delicated, but I think that this reasoning works. Are you agree?
$endgroup$
– Armando j18eos
Nov 27 '17 at 20:12
$begingroup$
It is not true that $f$ will factor through the closed subscheme $Z$ if $f$ has image inside $Z$. For example, the projection $k[X]to k[X]/(X^2)$ will not factor through $k[X]/(X)=k$.
$endgroup$
– user363120
Nov 27 '17 at 22:03
$begingroup$
A my idea: it's the same "universal property" of the inclusion of a (non empty) subset in a set.
$endgroup$
– Armando j18eos
Nov 27 '17 at 15:21
$begingroup$
A my idea: it's the same "universal property" of the inclusion of a (non empty) subset in a set.
$endgroup$
– Armando j18eos
Nov 27 '17 at 15:21
$begingroup$
@Armandoj18eos Are you thinking in something like this? Given $Zsubset Y$ and $f:Xrightarrow Y$ with $f(X)subset Z$ then there exist a unique factorization of $f$ through the inclusion $i:Zrightarrow Y$. If so, I think this is true for open immersion but not in general for closed immersions. Perhaps some modification of this works in the case of closed immersions.
$endgroup$
– yamete kudasai
Nov 27 '17 at 18:07
$begingroup$
@Armandoj18eos Are you thinking in something like this? Given $Zsubset Y$ and $f:Xrightarrow Y$ with $f(X)subset Z$ then there exist a unique factorization of $f$ through the inclusion $i:Zrightarrow Y$. If so, I think this is true for open immersion but not in general for closed immersions. Perhaps some modification of this works in the case of closed immersions.
$endgroup$
– yamete kudasai
Nov 27 '17 at 18:07
$begingroup$
Yes, I am! For the closed immersion, the work is more delicated, but I think that this reasoning works. Are you agree?
$endgroup$
– Armando j18eos
Nov 27 '17 at 20:12
$begingroup$
Yes, I am! For the closed immersion, the work is more delicated, but I think that this reasoning works. Are you agree?
$endgroup$
– Armando j18eos
Nov 27 '17 at 20:12
$begingroup$
It is not true that $f$ will factor through the closed subscheme $Z$ if $f$ has image inside $Z$. For example, the projection $k[X]to k[X]/(X^2)$ will not factor through $k[X]/(X)=k$.
$endgroup$
– user363120
Nov 27 '17 at 22:03
$begingroup$
It is not true that $f$ will factor through the closed subscheme $Z$ if $f$ has image inside $Z$. For example, the projection $k[X]to k[X]/(X^2)$ will not factor through $k[X]/(X)=k$.
$endgroup$
– user363120
Nov 27 '17 at 22:03
add a comment |
1 Answer
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oldest
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$begingroup$
Now I can answer this.
To simplify notation I will talk about subschemes rather than general immersions, although this are essentially the same.
Universal property of open subschemes: Let $X$ be a scheme and $Usubseteq X$ an open subscheme. A morphism of schemes $f:Yrightarrow X$ has a unique factorization $f:Yxrightarrow{f'} Uxrightarrow{i}X$ iff $f(X)subseteq U$ as sets.
Proof: Only $Leftarrow$ is non trivial. If $f(X)subseteq U$ then $f=icirc f'$ as maps between sets. Now as $f_*=i_*circ f'_*$ and $i^{-1}circ i_*=text{Id}$, we can apply the functor $i^{-1}$ to the morphism
$f^sharp:mathcal{O}_Xrightarrow f_*mathcal{O}_Y$
to get a morphism $mathcal{O}_Urightarrow f'_*mathcal{O}_Y$. If we define $f'^sharp$ as this morphism we can see the factorization $f=icirc f'$ at the level of schemes. $square$
Universal property of closed subschemes: Let $Zsubseteq X$ be defined by the quasicoherent ideal $mathcal{J}subset mathcal{O}_X$, i.e, $Z=text{Supp}(mathcal{O}_X/mathcal{J})$ and $mathcal{O}_Z=i^{-1}(mathcal{O}_X/mathcal{J})$.
Then a morphism $f:Yrightarrow X$ has a unique factorization $f:Yxrightarrow{f'} Zxrightarrow{i}X$ iff $f$ vanish over $mathcal{J}$, i.e, $f^sharp(mathcal{J}(U))=0 forall Usubseteq X$.
Proof: $Leftarrow$ is the most interesting part.
First we prove that $f$ can be factored as a map of sets, for this we need to prove that $f(Y)subseteq Z$ or equivalently $(mathcal{O}_X/mathcal{J})_{f(y)}neq 0 forall yin Y$. This follows because as $f$ vanish at $mathcal{J}$ the morphism $f^sharp$ can be factored to a morphism $mathcal{O}_X/mathcal{J}rightarrow f_*mathcal{O}_Y$ and using the $(f^{-1},f_*)$ adjunction we get a morphism $f^{-1}(mathcal{O}_X/mathcal{J})rightarrow mathcal{O}_Y$. Now looking at the stalk at $yin Y$ we get a ring morphism $(mathcal{O}_X/mathcal{J})_{f(y)}rightarrow mathcal{O}_{Y,y}$ and as $mathcal{O}_{Y,y}neq 0$ we should have $(mathcal{O}_X/mathcal{J})_{f(y)}neq 0$.
Now we show that $f=icirc f'$ can be extended to the level of schemes. As we already have $f'$ at the level of sets is enough to define $f'^sharp$. For this, again using the identities $f_*=i_*circ f'_*$ and $i^{-1}circ i_*=text{Id}$ we can apply $i^{-1}$ to the morphism $mathcal{O}_X/mathcal{J}rightarrow f_*mathcal{O}_Y$ to get a morphism $i^{-1}(mathcal{O}_X/mathcal{J})rightarrow f'_*mathcal{O}_Y$ that we can use as $f'^sharp$. $square$
Using a combination of this two properties is possible to give a universal property for locally closed subschemes but I am not sure how useful it is.
Notes:
The condition $f^sharp(mathcal{J}(U))=0 ; forall Usubseteq X$ can be restated as $f^*(mathcal{J})=0$
For a locally ringed space $(X,mathcal{O}_X)$ and a subset $i:Yhookrightarrow X$ the locally ringed space $(Y,i^{-1}mathcal{O}_X)$ has the following property: A morphism $f:Zrightarrow X$ factorize through $Y$ iff $f(Z)subset Y$ as sets. This generalize the universal property for open immersions and gives a universal property for the inclusions $text{Spec } mathcal{O}_{X,x}hookrightarrow X$ and $text{Spec }k(x)hookrightarrow X$.
answered May 7 '18 at 18:16
yamete kudasaiyamete kudasai
1,160818
$endgroup$
add a comment |
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$begingroup$
Now I can answer this.
To simplify notation I will talk about subschemes rather than general immersions, although this are essentially the same.
Universal property of open subschemes: Let $X$ be a scheme and $Usubseteq X$ an open subscheme. A morphism of schemes $f:Yrightarrow X$ has a unique factorization $f:Yxrightarrow{f'} Uxrightarrow{i}X$ iff $f(X)subseteq U$ as sets.
Proof: Only $Leftarrow$ is non trivial. If $f(X)subseteq U$ then $f=icirc f'$ as maps between sets. Now as $f_*=i_*circ f'_*$ and $i^{-1}circ i_*=text{Id}$, we can apply the functor $i^{-1}$ to the morphism
$f^sharp:mathcal{O}_Xrightarrow f_*mathcal{O}_Y$
to get a morphism $mathcal{O}_Urightarrow f'_*mathcal{O}_Y$. If we define $f'^sharp$ as this morphism we can see the factorization $f=icirc f'$ at the level of schemes. $square$
Universal property of closed subschemes: Let $Zsubseteq X$ be defined by the quasicoherent ideal $mathcal{J}subset mathcal{O}_X$, i.e, $Z=text{Supp}(mathcal{O}_X/mathcal{J})$ and $mathcal{O}_Z=i^{-1}(mathcal{O}_X/mathcal{J})$.
Then a morphism $f:Yrightarrow X$ has a unique factorization $f:Yxrightarrow{f'} Zxrightarrow{i}X$ iff $f$ vanish over $mathcal{J}$, i.e, $f^sharp(mathcal{J}(U))=0 forall Usubseteq X$.
Proof: $Leftarrow$ is the most interesting part.
First we prove that $f$ can be factored as a map of sets, for this we need to prove that $f(Y)subseteq Z$ or equivalently $(mathcal{O}_X/mathcal{J})_{f(y)}neq 0 forall yin Y$. This follows because as $f$ vanish at $mathcal{J}$ the morphism $f^sharp$ can be factored to a morphism $mathcal{O}_X/mathcal{J}rightarrow f_*mathcal{O}_Y$ and using the $(f^{-1},f_*)$ adjunction we get a morphism $f^{-1}(mathcal{O}_X/mathcal{J})rightarrow mathcal{O}_Y$. Now looking at the stalk at $yin Y$ we get a ring morphism $(mathcal{O}_X/mathcal{J})_{f(y)}rightarrow mathcal{O}_{Y,y}$ and as $mathcal{O}_{Y,y}neq 0$ we should have $(mathcal{O}_X/mathcal{J})_{f(y)}neq 0$.
Now we show that $f=icirc f'$ can be extended to the level of schemes. As we already have $f'$ at the level of sets is enough to define $f'^sharp$. For this, again using the identities $f_*=i_*circ f'_*$ and $i^{-1}circ i_*=text{Id}$ we can apply $i^{-1}$ to the morphism $mathcal{O}_X/mathcal{J}rightarrow f_*mathcal{O}_Y$ to get a morphism $i^{-1}(mathcal{O}_X/mathcal{J})rightarrow f'_*mathcal{O}_Y$ that we can use as $f'^sharp$. $square$
Using a combination of this two properties is possible to give a universal property for locally closed subschemes but I am not sure how useful it is.
Notes:
The condition $f^sharp(mathcal{J}(U))=0 ; forall Usubseteq X$ can be restated as $f^*(mathcal{J})=0$
For a locally ringed space $(X,mathcal{O}_X)$ and a subset $i:Yhookrightarrow X$ the locally ringed space $(Y,i^{-1}mathcal{O}_X)$ has the following property: A morphism $f:Zrightarrow X$ factorize through $Y$ iff $f(Z)subset Y$ as sets. This generalize the universal property for open immersions and gives a universal property for the inclusions $text{Spec } mathcal{O}_{X,x}hookrightarrow X$ and $text{Spec }k(x)hookrightarrow X$.
answered May 7 '18 at 18:16
yamete kudasaiyamete kudasai
1,160818
$endgroup$
add a comment |
$begingroup$
Now I can answer this.
To simplify notation I will talk about subschemes rather than general immersions, although this are essentially the same.
Universal property of open subschemes: Let $X$ be a scheme and $Usubseteq X$ an open subscheme. A morphism of schemes $f:Yrightarrow X$ has a unique factorization $f:Yxrightarrow{f'} Uxrightarrow{i}X$ iff $f(X)subseteq U$ as sets.
Proof: Only $Leftarrow$ is non trivial. If $f(X)subseteq U$ then $f=icirc f'$ as maps between sets. Now as $f_*=i_*circ f'_*$ and $i^{-1}circ i_*=text{Id}$, we can apply the functor $i^{-1}$ to the morphism
$f^sharp:mathcal{O}_Xrightarrow f_*mathcal{O}_Y$
to get a morphism $mathcal{O}_Urightarrow f'_*mathcal{O}_Y$. If we define $f'^sharp$ as this morphism we can see the factorization $f=icirc f'$ at the level of schemes. $square$
Universal property of closed subschemes: Let $Zsubseteq X$ be defined by the quasicoherent ideal $mathcal{J}subset mathcal{O}_X$, i.e, $Z=text{Supp}(mathcal{O}_X/mathcal{J})$ and $mathcal{O}_Z=i^{-1}(mathcal{O}_X/mathcal{J})$.
Then a morphism $f:Yrightarrow X$ has a unique factorization $f:Yxrightarrow{f'} Zxrightarrow{i}X$ iff $f$ vanish over $mathcal{J}$, i.e, $f^sharp(mathcal{J}(U))=0 forall Usubseteq X$.
Proof: $Leftarrow$ is the most interesting part.
First we prove that $f$ can be factored as a map of sets, for this we need to prove that $f(Y)subseteq Z$ or equivalently $(mathcal{O}_X/mathcal{J})_{f(y)}neq 0 forall yin Y$. This follows because as $f$ vanish at $mathcal{J}$ the morphism $f^sharp$ can be factored to a morphism $mathcal{O}_X/mathcal{J}rightarrow f_*mathcal{O}_Y$ and using the $(f^{-1},f_*)$ adjunction we get a morphism $f^{-1}(mathcal{O}_X/mathcal{J})rightarrow mathcal{O}_Y$. Now looking at the stalk at $yin Y$ we get a ring morphism $(mathcal{O}_X/mathcal{J})_{f(y)}rightarrow mathcal{O}_{Y,y}$ and as $mathcal{O}_{Y,y}neq 0$ we should have $(mathcal{O}_X/mathcal{J})_{f(y)}neq 0$.
Now we show that $f=icirc f'$ can be extended to the level of schemes. As we already have $f'$ at the level of sets is enough to define $f'^sharp$. For this, again using the identities $f_*=i_*circ f'_*$ and $i^{-1}circ i_*=text{Id}$ we can apply $i^{-1}$ to the morphism $mathcal{O}_X/mathcal{J}rightarrow f_*mathcal{O}_Y$ to get a morphism $i^{-1}(mathcal{O}_X/mathcal{J})rightarrow f'_*mathcal{O}_Y$ that we can use as $f'^sharp$. $square$
Using a combination of this two properties is possible to give a universal property for locally closed subschemes but I am not sure how useful it is.
Notes:
The condition $f^sharp(mathcal{J}(U))=0 ; forall Usubseteq X$ can be restated as $f^*(mathcal{J})=0$
For a locally ringed space $(X,mathcal{O}_X)$ and a subset $i:Yhookrightarrow X$ the locally ringed space $(Y,i^{-1}mathcal{O}_X)$ has the following property: A morphism $f:Zrightarrow X$ factorize through $Y$ iff $f(Z)subset Y$ as sets. This generalize the universal property for open immersions and gives a universal property for the inclusions $text{Spec } mathcal{O}_{X,x}hookrightarrow X$ and $text{Spec }k(x)hookrightarrow X$.
answered May 7 '18 at 18:16
yamete kudasaiyamete kudasai
1,160818
$endgroup$
add a comment |
$begingroup$
Now I can answer this.
To simplify notation I will talk about subschemes rather than general immersions, although this are essentially the same.
Universal property of open subschemes: Let $X$ be a scheme and $Usubseteq X$ an open subscheme. A morphism of schemes $f:Yrightarrow X$ has a unique factorization $f:Yxrightarrow{f'} Uxrightarrow{i}X$ iff $f(X)subseteq U$ as sets.
Proof: Only $Leftarrow$ is non trivial. If $f(X)subseteq U$ then $f=icirc f'$ as maps between sets. Now as $f_*=i_*circ f'_*$ and $i^{-1}circ i_*=text{Id}$, we can apply the functor $i^{-1}$ to the morphism
$f^sharp:mathcal{O}_Xrightarrow f_*mathcal{O}_Y$
to get a morphism $mathcal{O}_Urightarrow f'_*mathcal{O}_Y$. If we define $f'^sharp$ as this morphism we can see the factorization $f=icirc f'$ at the level of schemes. $square$
Universal property of closed subschemes: Let $Zsubseteq X$ be defined by the quasicoherent ideal $mathcal{J}subset mathcal{O}_X$, i.e, $Z=text{Supp}(mathcal{O}_X/mathcal{J})$ and $mathcal{O}_Z=i^{-1}(mathcal{O}_X/mathcal{J})$.
Then a morphism $f:Yrightarrow X$ has a unique factorization $f:Yxrightarrow{f'} Zxrightarrow{i}X$ iff $f$ vanish over $mathcal{J}$, i.e, $f^sharp(mathcal{J}(U))=0 forall Usubseteq X$.
Proof: $Leftarrow$ is the most interesting part.
First we prove that $f$ can be factored as a map of sets, for this we need to prove that $f(Y)subseteq Z$ or equivalently $(mathcal{O}_X/mathcal{J})_{f(y)}neq 0 forall yin Y$. This follows because as $f$ vanish at $mathcal{J}$ the morphism $f^sharp$ can be factored to a morphism $mathcal{O}_X/mathcal{J}rightarrow f_*mathcal{O}_Y$ and using the $(f^{-1},f_*)$ adjunction we get a morphism $f^{-1}(mathcal{O}_X/mathcal{J})rightarrow mathcal{O}_Y$. Now looking at the stalk at $yin Y$ we get a ring morphism $(mathcal{O}_X/mathcal{J})_{f(y)}rightarrow mathcal{O}_{Y,y}$ and as $mathcal{O}_{Y,y}neq 0$ we should have $(mathcal{O}_X/mathcal{J})_{f(y)}neq 0$.
Now we show that $f=icirc f'$ can be extended to the level of schemes. As we already have $f'$ at the level of sets is enough to define $f'^sharp$. For this, again using the identities $f_*=i_*circ f'_*$ and $i^{-1}circ i_*=text{Id}$ we can apply $i^{-1}$ to the morphism $mathcal{O}_X/mathcal{J}rightarrow f_*mathcal{O}_Y$ to get a morphism $i^{-1}(mathcal{O}_X/mathcal{J})rightarrow f'_*mathcal{O}_Y$ that we can use as $f'^sharp$. $square$
Using a combination of this two properties is possible to give a universal property for locally closed subschemes but I am not sure how useful it is.
Notes:
The condition $f^sharp(mathcal{J}(U))=0 ; forall Usubseteq X$ can be restated as $f^*(mathcal{J})=0$
For a locally ringed space $(X,mathcal{O}_X)$ and a subset $i:Yhookrightarrow X$ the locally ringed space $(Y,i^{-1}mathcal{O}_X)$ has the following property: A morphism $f:Zrightarrow X$ factorize through $Y$ iff $f(Z)subset Y$ as sets. This generalize the universal property for open immersions and gives a universal property for the inclusions $text{Spec } mathcal{O}_{X,x}hookrightarrow X$ and $text{Spec }k(x)hookrightarrow X$.
answered May 7 '18 at 18:16
yamete kudasaiyamete kudasai
1,160818
$endgroup$
Now I can answer this.
To simplify notation I will talk about subschemes rather than general immersions, although this are essentially the same.
Universal property of open subschemes: Let $X$ be a scheme and $Usubseteq X$ an open subscheme. A morphism of schemes $f:Yrightarrow X$ has a unique factorization $f:Yxrightarrow{f'} Uxrightarrow{i}X$ iff $f(X)subseteq U$ as sets.
Proof: Only $Leftarrow$ is non trivial. If $f(X)subseteq U$ then $f=icirc f'$ as maps between sets. Now as $f_*=i_*circ f'_*$ and $i^{-1}circ i_*=text{Id}$, we can apply the functor $i^{-1}$ to the morphism
$f^sharp:mathcal{O}_Xrightarrow f_*mathcal{O}_Y$
to get a morphism $mathcal{O}_Urightarrow f'_*mathcal{O}_Y$. If we define $f'^sharp$ as this morphism we can see the factorization $f=icirc f'$ at the level of schemes. $square$
Universal property of closed subschemes: Let $Zsubseteq X$ be defined by the quasicoherent ideal $mathcal{J}subset mathcal{O}_X$, i.e, $Z=text{Supp}(mathcal{O}_X/mathcal{J})$ and $mathcal{O}_Z=i^{-1}(mathcal{O}_X/mathcal{J})$.
Then a morphism $f:Yrightarrow X$ has a unique factorization $f:Yxrightarrow{f'} Zxrightarrow{i}X$ iff $f$ vanish over $mathcal{J}$, i.e, $f^sharp(mathcal{J}(U))=0 forall Usubseteq X$.
Proof: $Leftarrow$ is the most interesting part.
First we prove that $f$ can be factored as a map of sets, for this we need to prove that $f(Y)subseteq Z$ or equivalently $(mathcal{O}_X/mathcal{J})_{f(y)}neq 0 forall yin Y$. This follows because as $f$ vanish at $mathcal{J}$ the morphism $f^sharp$ can be factored to a morphism $mathcal{O}_X/mathcal{J}rightarrow f_*mathcal{O}_Y$ and using the $(f^{-1},f_*)$ adjunction we get a morphism $f^{-1}(mathcal{O}_X/mathcal{J})rightarrow mathcal{O}_Y$. Now looking at the stalk at $yin Y$ we get a ring morphism $(mathcal{O}_X/mathcal{J})_{f(y)}rightarrow mathcal{O}_{Y,y}$ and as $mathcal{O}_{Y,y}neq 0$ we should have $(mathcal{O}_X/mathcal{J})_{f(y)}neq 0$.
Now we show that $f=icirc f'$ can be extended to the level of schemes. As we already have $f'$ at the level of sets is enough to define $f'^sharp$. For this, again using the identities $f_*=i_*circ f'_*$ and $i^{-1}circ i_*=text{Id}$ we can apply $i^{-1}$ to the morphism $mathcal{O}_X/mathcal{J}rightarrow f_*mathcal{O}_Y$ to get a morphism $i^{-1}(mathcal{O}_X/mathcal{J})rightarrow f'_*mathcal{O}_Y$ that we can use as $f'^sharp$. $square$
Using a combination of this two properties is possible to give a universal property for locally closed subschemes but I am not sure how useful it is.
Notes:
The condition $f^sharp(mathcal{J}(U))=0 ; forall Usubseteq X$ can be restated as $f^*(mathcal{J})=0$
For a locally ringed space $(X,mathcal{O}_X)$ and a subset $i:Yhookrightarrow X$ the locally ringed space $(Y,i^{-1}mathcal{O}_X)$ has the following property: A morphism $f:Zrightarrow X$ factorize through $Y$ iff $f(Z)subset Y$ as sets. This generalize the universal property for open immersions and gives a universal property for the inclusions $text{Spec } mathcal{O}_{X,x}hookrightarrow X$ and $text{Spec }k(x)hookrightarrow X$.
answered May 7 '18 at 18:16
yamete kudasaiyamete kudasai
1,160818
edited Jan 29 at 20:39
answered May 7 '18 at 18:16
yamete kudasaiyamete kudasai
1,160818
answered May 7 '18 at 18:16
yamete kudasaiyamete kudasai
1,160818
answered May 7 '18 at 18:16
yamete kudasaiyamete kudasai
1,160818
1,160818
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$begingroup$
A my idea: it's the same "universal property" of the inclusion of a (non empty) subset in a set.
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– Armando j18eos
Nov 27 '17 at 15:21
$begingroup$
@Armandoj18eos Are you thinking in something like this? Given $Zsubset Y$ and $f:Xrightarrow Y$ with $f(X)subset Z$ then there exist a unique factorization of $f$ through the inclusion $i:Zrightarrow Y$. If so, I think this is true for open immersion but not in general for closed immersions. Perhaps some modification of this works in the case of closed immersions.
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– yamete kudasai
Nov 27 '17 at 18:07
$begingroup$
Yes, I am! For the closed immersion, the work is more delicated, but I think that this reasoning works. Are you agree?
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– Armando j18eos
Nov 27 '17 at 20:12
$begingroup$
It is not true that $f$ will factor through the closed subscheme $Z$ if $f$ has image inside $Z$. For example, the projection $k[X]to k[X]/(X^2)$ will not factor through $k[X]/(X)=k$.
$endgroup$
– user363120
Nov 27 '17 at 22:03