The surjective image of a proper scheme is proper
$begingroup$
Proper is supposed to generalized compactness so something like this should be true.
Let $X$ be a proper $S$-scheme and let $f:Xrightarrow Y$ be a surjective morphism of $S$-schemes. Is it true that $Y$ also a proper $S$-scheme?
I think that I can prove that $g:Y rightarrow S$ is universally closed. In fact, given a $S$-scheme $Z$ we have that the composition
$$f_Z:Xtimes_S Zrightarrow Ytimes_S Z$$
is surjective as this property is invariant under base change. So if $W$ is a closed subset of $Ytimes_S Z$ we have that
$$g_Z(W)=(g_Zcirc f_Z)(f_Z^{-1}(W))$$
is closed as $g_Zcirc f_Z=Xtimes_S Zrightarrow Z$ is a base chage of $Xrightarrow S$.
Also, as is show in here, Y should be separated. (Edit: This was not true as KReiser example shows)
Now, what about finite type?
Remark: Because of this it is enough to prove locally of finite type.
Edit: Corrected the proof of universally closed.
algebraic-geometry schemes
asked Dec 13 '18 at 16:53
Walter SimonWalter Simon
16910
$endgroup$
add a comment |
$begingroup$
Proper is supposed to generalized compactness so something like this should be true.
Let $X$ be a proper $S$-scheme and let $f:Xrightarrow Y$ be a surjective morphism of $S$-schemes. Is it true that $Y$ also a proper $S$-scheme?
I think that I can prove that $g:Y rightarrow S$ is universally closed. In fact, given a $S$-scheme $Z$ we have that the composition
$$f_Z:Xtimes_S Zrightarrow Ytimes_S Z$$
is surjective as this property is invariant under base change. So if $W$ is a closed subset of $Ytimes_S Z$ we have that
$$g_Z(W)=(g_Zcirc f_Z)(f_Z^{-1}(W))$$
is closed as $g_Zcirc f_Z=Xtimes_S Zrightarrow Z$ is a base chage of $Xrightarrow S$.
Also, as is show in here, Y should be separated. (Edit: This was not true as KReiser example shows)
Now, what about finite type?
Remark: Because of this it is enough to prove locally of finite type.
Edit: Corrected the proof of universally closed.
algebraic-geometry schemes
asked Dec 13 '18 at 16:53
Walter SimonWalter Simon
16910
$endgroup$
add a comment |
$begingroup$
Proper is supposed to generalized compactness so something like this should be true.
Let $X$ be a proper $S$-scheme and let $f:Xrightarrow Y$ be a surjective morphism of $S$-schemes. Is it true that $Y$ also a proper $S$-scheme?
I think that I can prove that $g:Y rightarrow S$ is universally closed. In fact, given a $S$-scheme $Z$ we have that the composition
$$f_Z:Xtimes_S Zrightarrow Ytimes_S Z$$
is surjective as this property is invariant under base change. So if $W$ is a closed subset of $Ytimes_S Z$ we have that
$$g_Z(W)=(g_Zcirc f_Z)(f_Z^{-1}(W))$$
is closed as $g_Zcirc f_Z=Xtimes_S Zrightarrow Z$ is a base chage of $Xrightarrow S$.
Also, as is show in here, Y should be separated. (Edit: This was not true as KReiser example shows)
Now, what about finite type?
Remark: Because of this it is enough to prove locally of finite type.
Edit: Corrected the proof of universally closed.
algebraic-geometry schemes
asked Dec 13 '18 at 16:53
Walter SimonWalter Simon
16910
$endgroup$
Proper is supposed to generalized compactness so something like this should be true.
Let $X$ be a proper $S$-scheme and let $f:Xrightarrow Y$ be a surjective morphism of $S$-schemes. Is it true that $Y$ also a proper $S$-scheme?
I think that I can prove that $g:Y rightarrow S$ is universally closed. In fact, given a $S$-scheme $Z$ we have that the composition
$$f_Z:Xtimes_S Zrightarrow Ytimes_S Z$$
is surjective as this property is invariant under base change. So if $W$ is a closed subset of $Ytimes_S Z$ we have that
$$g_Z(W)=(g_Zcirc f_Z)(f_Z^{-1}(W))$$
is closed as $g_Zcirc f_Z=Xtimes_S Zrightarrow Z$ is a base chage of $Xrightarrow S$.
Also, as is show in here, Y should be separated. (Edit: This was not true as KReiser example shows)
Now, what about finite type?
Remark: Because of this it is enough to prove locally of finite type.
Edit: Corrected the proof of universally closed.
algebraic-geometry schemes
algebraic-geometry schemes
asked Dec 13 '18 at 16:53
Walter SimonWalter Simon
16910
asked Dec 13 '18 at 16:53
Walter SimonWalter Simon
16910
edited Dec 14 '18 at 14:15
Walter Simon
asked Dec 13 '18 at 16:53
Walter SimonWalter Simon
16910
asked Dec 13 '18 at 16:53
Walter SimonWalter Simon
16910
asked Dec 13 '18 at 16:53
Walter SimonWalter Simon
16910
16910
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Unfortunately this is not true for two reasons. First of all, recall that proper is equivalent to universally closed, separated, and of finite type: if $Y$ isn't separated, it can't be proper. For example, $Y$ being a projective line with a doubled point and $X$ being two copies of the projective line with $f$ being the natural identification map provides a counterexample to your statement.
Secondly, it may be the case that a surjective map is not of finite type: let $k$ be a field, $R=k[x_1,x_2,cdots]$, $I=(x_1,cdots)$, and consider $operatorname{Spec} R/I^2 to operatorname{Spec} k$. This map is a homeomorphism after any base change (and thus it's universally closed), but is not locally of finite type.
So one really does need the target of a surjective morphism to be separated and of finite type in order to conclude that the image of a proper scheme is again proper.
answered Dec 14 '18 at 7:01
KReiserKReiser
9,73221435
$endgroup$
1
$begingroup$
Oh, now I have a contradiction in my head. Using the result proved in this question I think I can deduce that the codomain is separable. But I am also convinced that your example of two copies of $mathbb{P}^1$ gluing in one $mathbb{P}^1$ with a double point is a counterexample. Do you have any idea of what is happening here?
$endgroup$
– Walter Simon
Dec 14 '18 at 11:08
1
$begingroup$
Now I got it. I was assuming that $X$ proper implies $Xrightarrow Y$ proper, but actually we need $Y$ to be separated for this (see here). Also now I realized that it is intuitive to ask for $Y$ to be separated because properness generalize compactness + Hausdorff so we need to avoid the possibility that the surjective map is a weird projection that can mess with the Hausdorff part. Thank you for your answer!
$endgroup$
– Walter Simon
Dec 14 '18 at 13:21
$begingroup$
I have just realised that there is a surjective morphism from the (obviously proper) scheme $text{Spec }k$ to $text{Spec }R/I^2$ so your second example is exactly what we needed.
$endgroup$
– Walter Simon
Dec 16 '18 at 10:48
add a comment |
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$begingroup$
Unfortunately this is not true for two reasons. First of all, recall that proper is equivalent to universally closed, separated, and of finite type: if $Y$ isn't separated, it can't be proper. For example, $Y$ being a projective line with a doubled point and $X$ being two copies of the projective line with $f$ being the natural identification map provides a counterexample to your statement.
Secondly, it may be the case that a surjective map is not of finite type: let $k$ be a field, $R=k[x_1,x_2,cdots]$, $I=(x_1,cdots)$, and consider $operatorname{Spec} R/I^2 to operatorname{Spec} k$. This map is a homeomorphism after any base change (and thus it's universally closed), but is not locally of finite type.
So one really does need the target of a surjective morphism to be separated and of finite type in order to conclude that the image of a proper scheme is again proper.
answered Dec 14 '18 at 7:01
KReiserKReiser
9,73221435
$endgroup$
1
$begingroup$
Oh, now I have a contradiction in my head. Using the result proved in this question I think I can deduce that the codomain is separable. But I am also convinced that your example of two copies of $mathbb{P}^1$ gluing in one $mathbb{P}^1$ with a double point is a counterexample. Do you have any idea of what is happening here?
$endgroup$
– Walter Simon
Dec 14 '18 at 11:08
1
$begingroup$
Now I got it. I was assuming that $X$ proper implies $Xrightarrow Y$ proper, but actually we need $Y$ to be separated for this (see here). Also now I realized that it is intuitive to ask for $Y$ to be separated because properness generalize compactness + Hausdorff so we need to avoid the possibility that the surjective map is a weird projection that can mess with the Hausdorff part. Thank you for your answer!
$endgroup$
– Walter Simon
Dec 14 '18 at 13:21
$begingroup$
I have just realised that there is a surjective morphism from the (obviously proper) scheme $text{Spec }k$ to $text{Spec }R/I^2$ so your second example is exactly what we needed.
$endgroup$
– Walter Simon
Dec 16 '18 at 10:48
add a comment |
$begingroup$
Unfortunately this is not true for two reasons. First of all, recall that proper is equivalent to universally closed, separated, and of finite type: if $Y$ isn't separated, it can't be proper. For example, $Y$ being a projective line with a doubled point and $X$ being two copies of the projective line with $f$ being the natural identification map provides a counterexample to your statement.
Secondly, it may be the case that a surjective map is not of finite type: let $k$ be a field, $R=k[x_1,x_2,cdots]$, $I=(x_1,cdots)$, and consider $operatorname{Spec} R/I^2 to operatorname{Spec} k$. This map is a homeomorphism after any base change (and thus it's universally closed), but is not locally of finite type.
So one really does need the target of a surjective morphism to be separated and of finite type in order to conclude that the image of a proper scheme is again proper.
answered Dec 14 '18 at 7:01
KReiserKReiser
9,73221435
$endgroup$
1
$begingroup$
Oh, now I have a contradiction in my head. Using the result proved in this question I think I can deduce that the codomain is separable. But I am also convinced that your example of two copies of $mathbb{P}^1$ gluing in one $mathbb{P}^1$ with a double point is a counterexample. Do you have any idea of what is happening here?
$endgroup$
– Walter Simon
Dec 14 '18 at 11:08
1
$begingroup$
Now I got it. I was assuming that $X$ proper implies $Xrightarrow Y$ proper, but actually we need $Y$ to be separated for this (see here). Also now I realized that it is intuitive to ask for $Y$ to be separated because properness generalize compactness + Hausdorff so we need to avoid the possibility that the surjective map is a weird projection that can mess with the Hausdorff part. Thank you for your answer!
$endgroup$
– Walter Simon
Dec 14 '18 at 13:21
$begingroup$
I have just realised that there is a surjective morphism from the (obviously proper) scheme $text{Spec }k$ to $text{Spec }R/I^2$ so your second example is exactly what we needed.
$endgroup$
– Walter Simon
Dec 16 '18 at 10:48
add a comment |
$begingroup$
Unfortunately this is not true for two reasons. First of all, recall that proper is equivalent to universally closed, separated, and of finite type: if $Y$ isn't separated, it can't be proper. For example, $Y$ being a projective line with a doubled point and $X$ being two copies of the projective line with $f$ being the natural identification map provides a counterexample to your statement.
Secondly, it may be the case that a surjective map is not of finite type: let $k$ be a field, $R=k[x_1,x_2,cdots]$, $I=(x_1,cdots)$, and consider $operatorname{Spec} R/I^2 to operatorname{Spec} k$. This map is a homeomorphism after any base change (and thus it's universally closed), but is not locally of finite type.
So one really does need the target of a surjective morphism to be separated and of finite type in order to conclude that the image of a proper scheme is again proper.
answered Dec 14 '18 at 7:01
KReiserKReiser
9,73221435
$endgroup$
Unfortunately this is not true for two reasons. First of all, recall that proper is equivalent to universally closed, separated, and of finite type: if $Y$ isn't separated, it can't be proper. For example, $Y$ being a projective line with a doubled point and $X$ being two copies of the projective line with $f$ being the natural identification map provides a counterexample to your statement.
Secondly, it may be the case that a surjective map is not of finite type: let $k$ be a field, $R=k[x_1,x_2,cdots]$, $I=(x_1,cdots)$, and consider $operatorname{Spec} R/I^2 to operatorname{Spec} k$. This map is a homeomorphism after any base change (and thus it's universally closed), but is not locally of finite type.
So one really does need the target of a surjective morphism to be separated and of finite type in order to conclude that the image of a proper scheme is again proper.
answered Dec 14 '18 at 7:01
KReiserKReiser
9,73221435
answered Dec 14 '18 at 7:01
KReiserKReiser
9,73221435
answered Dec 14 '18 at 7:01
KReiserKReiser
9,73221435
answered Dec 14 '18 at 7:01
KReiserKReiser
9,73221435
9,73221435
1
$begingroup$
Oh, now I have a contradiction in my head. Using the result proved in this question I think I can deduce that the codomain is separable. But I am also convinced that your example of two copies of $mathbb{P}^1$ gluing in one $mathbb{P}^1$ with a double point is a counterexample. Do you have any idea of what is happening here?
$endgroup$
– Walter Simon
Dec 14 '18 at 11:08
1
$begingroup$
Now I got it. I was assuming that $X$ proper implies $Xrightarrow Y$ proper, but actually we need $Y$ to be separated for this (see here). Also now I realized that it is intuitive to ask for $Y$ to be separated because properness generalize compactness + Hausdorff so we need to avoid the possibility that the surjective map is a weird projection that can mess with the Hausdorff part. Thank you for your answer!
$endgroup$
– Walter Simon
Dec 14 '18 at 13:21
$begingroup$
I have just realised that there is a surjective morphism from the (obviously proper) scheme $text{Spec }k$ to $text{Spec }R/I^2$ so your second example is exactly what we needed.
$endgroup$
– Walter Simon
Dec 16 '18 at 10:48
add a comment |
1
$begingroup$
Oh, now I have a contradiction in my head. Using the result proved in this question I think I can deduce that the codomain is separable. But I am also convinced that your example of two copies of $mathbb{P}^1$ gluing in one $mathbb{P}^1$ with a double point is a counterexample. Do you have any idea of what is happening here?
$endgroup$
– Walter Simon
Dec 14 '18 at 11:08
1
$begingroup$
Now I got it. I was assuming that $X$ proper implies $Xrightarrow Y$ proper, but actually we need $Y$ to be separated for this (see here). Also now I realized that it is intuitive to ask for $Y$ to be separated because properness generalize compactness + Hausdorff so we need to avoid the possibility that the surjective map is a weird projection that can mess with the Hausdorff part. Thank you for your answer!
$endgroup$
– Walter Simon
Dec 14 '18 at 13:21
$begingroup$
I have just realised that there is a surjective morphism from the (obviously proper) scheme $text{Spec }k$ to $text{Spec }R/I^2$ so your second example is exactly what we needed.
$endgroup$
– Walter Simon
Dec 16 '18 at 10:48
1
1
$begingroup$
Oh, now I have a contradiction in my head. Using the result proved in this question I think I can deduce that the codomain is separable. But I am also convinced that your example of two copies of $mathbb{P}^1$ gluing in one $mathbb{P}^1$ with a double point is a counterexample. Do you have any idea of what is happening here?
$endgroup$
– Walter Simon
Dec 14 '18 at 11:08
$begingroup$
Oh, now I have a contradiction in my head. Using the result proved in this question I think I can deduce that the codomain is separable. But I am also convinced that your example of two copies of $mathbb{P}^1$ gluing in one $mathbb{P}^1$ with a double point is a counterexample. Do you have any idea of what is happening here?
$endgroup$
– Walter Simon
Dec 14 '18 at 11:08
1
1
$begingroup$
Now I got it. I was assuming that $X$ proper implies $Xrightarrow Y$ proper, but actually we need $Y$ to be separated for this (see here). Also now I realized that it is intuitive to ask for $Y$ to be separated because properness generalize compactness + Hausdorff so we need to avoid the possibility that the surjective map is a weird projection that can mess with the Hausdorff part. Thank you for your answer!
$endgroup$
– Walter Simon
Dec 14 '18 at 13:21
$begingroup$
Now I got it. I was assuming that $X$ proper implies $Xrightarrow Y$ proper, but actually we need $Y$ to be separated for this (see here). Also now I realized that it is intuitive to ask for $Y$ to be separated because properness generalize compactness + Hausdorff so we need to avoid the possibility that the surjective map is a weird projection that can mess with the Hausdorff part. Thank you for your answer!
$endgroup$
– Walter Simon
Dec 14 '18 at 13:21
$begingroup$
I have just realised that there is a surjective morphism from the (obviously proper) scheme $text{Spec }k$ to $text{Spec }R/I^2$ so your second example is exactly what we needed.
$endgroup$
– Walter Simon
Dec 16 '18 at 10:48
$begingroup$
I have just realised that there is a surjective morphism from the (obviously proper) scheme $text{Spec }k$ to $text{Spec }R/I^2$ so your second example is exactly what we needed.
$endgroup$
– Walter Simon
Dec 16 '18 at 10:48
add a comment |
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