Mixing
Extending a morphism to a projective scheme
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
Let $C$ be a reduced affine Noetherian scheme of pure dimension 1 (all its irreducible components have dimension 1) and $p in C$ a regular closed point. Suppose we have a morphism $C backslash p to mathbbP^n$ that lies in a closed subscheme $Y subseteq mathbbP^n$. If this morphism extends to $C to mathbbP^n$, is it true that this extension must factor through $Y$?
Here's my proof. Since $p$ is a closed point, $p in C' subset C$ some irreducible component of $C$, necessarily of dimension 1. Note $C'$ is an irreducible Noetherian topological space of dimension 1 and $pi^-1(Y) cap C'$ is a closed subset of $C'$ containing $C' backslash p$. However, any proper closed subset of an irreducible Noetherian topological space of dimension 1 must have dimension 0 and thus be finite. But $C'$ is infinite as it has dimension 1 so $pi^-1(Y) cap C' = C'$ and hence $pi^-1(Y) ni p$. Now, since $C$ is reduced the schematic image is the closure of the image, but the image is contained in $Y$, and $Y$ is closed in $mathbbP^n$ and we conclude the schematic image of $C$ is contained in $Y$.
proof-verification algebraic-geometry curves dimension-theory
asked Jul 26 at 19:13
ggg
435313
add a comment |Â
up vote
1
down vote
favorite
Let $C$ be a reduced affine Noetherian scheme of pure dimension 1 (all its irreducible components have dimension 1) and $p in C$ a regular closed point. Suppose we have a morphism $C backslash p to mathbbP^n$ that lies in a closed subscheme $Y subseteq mathbbP^n$. If this morphism extends to $C to mathbbP^n$, is it true that this extension must factor through $Y$?
Here's my proof. Since $p$ is a closed point, $p in C' subset C$ some irreducible component of $C$, necessarily of dimension 1. Note $C'$ is an irreducible Noetherian topological space of dimension 1 and $pi^-1(Y) cap C'$ is a closed subset of $C'$ containing $C' backslash p$. However, any proper closed subset of an irreducible Noetherian topological space of dimension 1 must have dimension 0 and thus be finite. But $C'$ is infinite as it has dimension 1 so $pi^-1(Y) cap C' = C'$ and hence $pi^-1(Y) ni p$. Now, since $C$ is reduced the schematic image is the closure of the image, but the image is contained in $Y$, and $Y$ is closed in $mathbbP^n$ and we conclude the schematic image of $C$ is contained in $Y$.
proof-verification algebraic-geometry curves dimension-theory
asked Jul 26 at 19:13
ggg
435313
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $C$ be a reduced affine Noetherian scheme of pure dimension 1 (all its irreducible components have dimension 1) and $p in C$ a regular closed point. Suppose we have a morphism $C backslash p to mathbbP^n$ that lies in a closed subscheme $Y subseteq mathbbP^n$. If this morphism extends to $C to mathbbP^n$, is it true that this extension must factor through $Y$?
Here's my proof. Since $p$ is a closed point, $p in C' subset C$ some irreducible component of $C$, necessarily of dimension 1. Note $C'$ is an irreducible Noetherian topological space of dimension 1 and $pi^-1(Y) cap C'$ is a closed subset of $C'$ containing $C' backslash p$. However, any proper closed subset of an irreducible Noetherian topological space of dimension 1 must have dimension 0 and thus be finite. But $C'$ is infinite as it has dimension 1 so $pi^-1(Y) cap C' = C'$ and hence $pi^-1(Y) ni p$. Now, since $C$ is reduced the schematic image is the closure of the image, but the image is contained in $Y$, and $Y$ is closed in $mathbbP^n$ and we conclude the schematic image of $C$ is contained in $Y$.
proof-verification algebraic-geometry curves dimension-theory
asked Jul 26 at 19:13
ggg
435313
Let $C$ be a reduced affine Noetherian scheme of pure dimension 1 (all its irreducible components have dimension 1) and $p in C$ a regular closed point. Suppose we have a morphism $C backslash p to mathbbP^n$ that lies in a closed subscheme $Y subseteq mathbbP^n$. If this morphism extends to $C to mathbbP^n$, is it true that this extension must factor through $Y$?
Here's my proof. Since $p$ is a closed point, $p in C' subset C$ some irreducible component of $C$, necessarily of dimension 1. Note $C'$ is an irreducible Noetherian topological space of dimension 1 and $pi^-1(Y) cap C'$ is a closed subset of $C'$ containing $C' backslash p$. However, any proper closed subset of an irreducible Noetherian topological space of dimension 1 must have dimension 0 and thus be finite. But $C'$ is infinite as it has dimension 1 so $pi^-1(Y) cap C' = C'$ and hence $pi^-1(Y) ni p$. Now, since $C$ is reduced the schematic image is the closure of the image, but the image is contained in $Y$, and $Y$ is closed in $mathbbP^n$ and we conclude the schematic image of $C$ is contained in $Y$.
proof-verification algebraic-geometry curves dimension-theory
asked Jul 26 at 19:13
ggg
435313
asked Jul 26 at 19:13
ggg
435313
asked Jul 26 at 19:13
ggg
435313
asked Jul 26 at 19:13
ggg
435313
435313
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Yes, this is fine - it may be quicker to note that morphisms of schemes are continuous and that $f(overlineX)subset overlinef(X)$ for a continuous map of topological spaces $f:Sto T$ and $Xsubset S$.
For other extension problems about morphisms of schemes, you may wish to look up the notions of separated and proper.
answered Jul 26 at 20:37
KReiser
7,47011230
You're right, that is faster, but don't we need something to ensure that $overlineC backslash p = C$ (e.g. $C$ is connected)? I don't mean to be overly technical but most of my wariness about this is that I don't understand the topology of a pure dimension 1 Noetherian scheme if it's not irreducible.
– ggg
Jul 26 at 21:06
1
The assumption that $C$ has no irreducible components of dimension $0$ is enough to guarantee that $overlineCsetminusp=C$. All you need is that $Csetminusp$ is dense in $C$, which is clear - $Csetminusp$ is an open set which intersects each irreducible component nontrivially (here's where we use no dimension $0$ components), and every nontrivial open subset of an irreducible set is dense.
– KReiser
Jul 26 at 21:12
Ah yes, so $C backslash p $ being dense is already forced by the assumptions. Thank you!
– ggg
Jul 26 at 22:06
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Yes, this is fine - it may be quicker to note that morphisms of schemes are continuous and that $f(overlineX)subset overlinef(X)$ for a continuous map of topological spaces $f:Sto T$ and $Xsubset S$.
For other extension problems about morphisms of schemes, you may wish to look up the notions of separated and proper.
answered Jul 26 at 20:37
KReiser
7,47011230
You're right, that is faster, but don't we need something to ensure that $overlineC backslash p = C$ (e.g. $C$ is connected)? I don't mean to be overly technical but most of my wariness about this is that I don't understand the topology of a pure dimension 1 Noetherian scheme if it's not irreducible.
– ggg
Jul 26 at 21:06
1
The assumption that $C$ has no irreducible components of dimension $0$ is enough to guarantee that $overlineCsetminusp=C$. All you need is that $Csetminusp$ is dense in $C$, which is clear - $Csetminusp$ is an open set which intersects each irreducible component nontrivially (here's where we use no dimension $0$ components), and every nontrivial open subset of an irreducible set is dense.
– KReiser
Jul 26 at 21:12
Ah yes, so $C backslash p $ being dense is already forced by the assumptions. Thank you!
– ggg
Jul 26 at 22:06
add a comment |Â
up vote
2
down vote
accepted
Yes, this is fine - it may be quicker to note that morphisms of schemes are continuous and that $f(overlineX)subset overlinef(X)$ for a continuous map of topological spaces $f:Sto T$ and $Xsubset S$.
For other extension problems about morphisms of schemes, you may wish to look up the notions of separated and proper.
answered Jul 26 at 20:37
KReiser
7,47011230
You're right, that is faster, but don't we need something to ensure that $overlineC backslash p = C$ (e.g. $C$ is connected)? I don't mean to be overly technical but most of my wariness about this is that I don't understand the topology of a pure dimension 1 Noetherian scheme if it's not irreducible.
– ggg
Jul 26 at 21:06
1
The assumption that $C$ has no irreducible components of dimension $0$ is enough to guarantee that $overlineCsetminusp=C$. All you need is that $Csetminusp$ is dense in $C$, which is clear - $Csetminusp$ is an open set which intersects each irreducible component nontrivially (here's where we use no dimension $0$ components), and every nontrivial open subset of an irreducible set is dense.
– KReiser
Jul 26 at 21:12
Ah yes, so $C backslash p $ being dense is already forced by the assumptions. Thank you!
– ggg
Jul 26 at 22:06
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Yes, this is fine - it may be quicker to note that morphisms of schemes are continuous and that $f(overlineX)subset overlinef(X)$ for a continuous map of topological spaces $f:Sto T$ and $Xsubset S$.
For other extension problems about morphisms of schemes, you may wish to look up the notions of separated and proper.
answered Jul 26 at 20:37
KReiser
7,47011230
Yes, this is fine - it may be quicker to note that morphisms of schemes are continuous and that $f(overlineX)subset overlinef(X)$ for a continuous map of topological spaces $f:Sto T$ and $Xsubset S$.
For other extension problems about morphisms of schemes, you may wish to look up the notions of separated and proper.
answered Jul 26 at 20:37
KReiser
7,47011230
answered Jul 26 at 20:37
KReiser
7,47011230
answered Jul 26 at 20:37
KReiser
7,47011230
answered Jul 26 at 20:37
KReiser
7,47011230
7,47011230
You're right, that is faster, but don't we need something to ensure that $overlineC backslash p = C$ (e.g. $C$ is connected)? I don't mean to be overly technical but most of my wariness about this is that I don't understand the topology of a pure dimension 1 Noetherian scheme if it's not irreducible.
– ggg
Jul 26 at 21:06
1
The assumption that $C$ has no irreducible components of dimension $0$ is enough to guarantee that $overlineCsetminusp=C$. All you need is that $Csetminusp$ is dense in $C$, which is clear - $Csetminusp$ is an open set which intersects each irreducible component nontrivially (here's where we use no dimension $0$ components), and every nontrivial open subset of an irreducible set is dense.
– KReiser
Jul 26 at 21:12
Ah yes, so $C backslash p $ being dense is already forced by the assumptions. Thank you!
– ggg
Jul 26 at 22:06
add a comment |Â
You're right, that is faster, but don't we need something to ensure that $overlineC backslash p = C$ (e.g. $C$ is connected)? I don't mean to be overly technical but most of my wariness about this is that I don't understand the topology of a pure dimension 1 Noetherian scheme if it's not irreducible.
– ggg
Jul 26 at 21:06
1
The assumption that $C$ has no irreducible components of dimension $0$ is enough to guarantee that $overlineCsetminusp=C$. All you need is that $Csetminusp$ is dense in $C$, which is clear - $Csetminusp$ is an open set which intersects each irreducible component nontrivially (here's where we use no dimension $0$ components), and every nontrivial open subset of an irreducible set is dense.
– KReiser
Jul 26 at 21:12
Ah yes, so $C backslash p $ being dense is already forced by the assumptions. Thank you!
– ggg
Jul 26 at 22:06
You're right, that is faster, but don't we need something to ensure that $overlineC backslash p = C$ (e.g. $C$ is connected)? I don't mean to be overly technical but most of my wariness about this is that I don't understand the topology of a pure dimension 1 Noetherian scheme if it's not irreducible.
– ggg
Jul 26 at 21:06
You're right, that is faster, but don't we need something to ensure that $overlineC backslash p = C$ (e.g. $C$ is connected)? I don't mean to be overly technical but most of my wariness about this is that I don't understand the topology of a pure dimension 1 Noetherian scheme if it's not irreducible.
– ggg
Jul 26 at 21:06
1
1
The assumption that $C$ has no irreducible components of dimension $0$ is enough to guarantee that $overlineCsetminusp=C$. All you need is that $Csetminusp$ is dense in $C$, which is clear - $Csetminusp$ is an open set which intersects each irreducible component nontrivially (here's where we use no dimension $0$ components), and every nontrivial open subset of an irreducible set is dense.
– KReiser
Jul 26 at 21:12
The assumption that $C$ has no irreducible components of dimension $0$ is enough to guarantee that $overlineCsetminusp=C$. All you need is that $Csetminusp$ is dense in $C$, which is clear - $Csetminusp$ is an open set which intersects each irreducible component nontrivially (here's where we use no dimension $0$ components), and every nontrivial open subset of an irreducible set is dense.
– KReiser
Jul 26 at 21:12
Ah yes, so $C backslash p $ being dense is already forced by the assumptions. Thank you!
– ggg
Jul 26 at 22:06
Ah yes, so $C backslash p $ being dense is already forced by the assumptions. Thank you!
– ggg
Jul 26 at 22:06
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2863716%2fextending-a-morphism-to-a-projective-scheme%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password