If $Y$ has irreducible components $Y_1, cdots, Y_n$, then the $overline{Y_i}$ are the irreducible components...
Let $X$ be a noetherian topological space, $Y$ a subspace having irreducible components $Y_1, cdots, Y_n$. Prove that the $bar{Y_i}$ are the irreducible components of $bar{Y}$.
I think as the irreducible components are the maximal irreducible subsets, it is enough to prove
When a set $Z subseteq X$ is irreducible, so is $bar{Z}$.
Is this true? If it is, is it still true when the noetherian condition is dropped?
general-topology
edited Sep 19 '11 at 3:57
Arturo Magidin
260k32584904
asked Sep 19 '11 at 3:41
ShinyaSakai
2,96512450
add a comment |
Let $X$ be a noetherian topological space, $Y$ a subspace having irreducible components $Y_1, cdots, Y_n$. Prove that the $bar{Y_i}$ are the irreducible components of $bar{Y}$.
I think as the irreducible components are the maximal irreducible subsets, it is enough to prove
When a set $Z subseteq X$ is irreducible, so is $bar{Z}$.
Is this true? If it is, is it still true when the noetherian condition is dropped?
general-topology
edited Sep 19 '11 at 3:57
Arturo Magidin
260k32584904
asked Sep 19 '11 at 3:41
ShinyaSakai
2,96512450
Double check your typing. It makes wonders. Specially in titles, which is the very first thing people see of your question!
– Mariano Suárez-Álvarez
Sep 19 '11 at 3:47
@Mariano Suárez-Alvarez♦: Thank you very much for reminding me. I don't know what careless mistake I have made in my title because it might have been kindly corrected by Arturo Magidin. I will be more careful next time. Thanks again.
– ShinyaSakai
Sep 19 '11 at 12:30
add a comment |
Let $X$ be a noetherian topological space, $Y$ a subspace having irreducible components $Y_1, cdots, Y_n$. Prove that the $bar{Y_i}$ are the irreducible components of $bar{Y}$.
I think as the irreducible components are the maximal irreducible subsets, it is enough to prove
When a set $Z subseteq X$ is irreducible, so is $bar{Z}$.
Is this true? If it is, is it still true when the noetherian condition is dropped?
general-topology
edited Sep 19 '11 at 3:57
Arturo Magidin
260k32584904
asked Sep 19 '11 at 3:41
ShinyaSakai
2,96512450
Let $X$ be a noetherian topological space, $Y$ a subspace having irreducible components $Y_1, cdots, Y_n$. Prove that the $bar{Y_i}$ are the irreducible components of $bar{Y}$.
I think as the irreducible components are the maximal irreducible subsets, it is enough to prove
When a set $Z subseteq X$ is irreducible, so is $bar{Z}$.
Is this true? If it is, is it still true when the noetherian condition is dropped?
general-topology
general-topology
edited Sep 19 '11 at 3:57
Arturo Magidin
260k32584904
asked Sep 19 '11 at 3:41
ShinyaSakai
2,96512450
edited Sep 19 '11 at 3:57
Arturo Magidin
260k32584904
asked Sep 19 '11 at 3:41
ShinyaSakai
2,96512450
edited Sep 19 '11 at 3:57
Arturo Magidin
260k32584904
edited Sep 19 '11 at 3:57
Arturo Magidin
260k32584904
edited Sep 19 '11 at 3:57
Arturo Magidin
260k32584904
260k32584904
asked Sep 19 '11 at 3:41
ShinyaSakai
2,96512450
asked Sep 19 '11 at 3:41
ShinyaSakai
2,96512450
asked Sep 19 '11 at 3:41
ShinyaSakai
2,96512450
2,96512450
Double check your typing. It makes wonders. Specially in titles, which is the very first thing people see of your question!
– Mariano Suárez-Álvarez
Sep 19 '11 at 3:47
@Mariano Suárez-Alvarez♦: Thank you very much for reminding me. I don't know what careless mistake I have made in my title because it might have been kindly corrected by Arturo Magidin. I will be more careful next time. Thanks again.
– ShinyaSakai
Sep 19 '11 at 12:30
add a comment |
Double check your typing. It makes wonders. Specially in titles, which is the very first thing people see of your question!
– Mariano Suárez-Álvarez
Sep 19 '11 at 3:47
@Mariano Suárez-Alvarez♦: Thank you very much for reminding me. I don't know what careless mistake I have made in my title because it might have been kindly corrected by Arturo Magidin. I will be more careful next time. Thanks again.
– ShinyaSakai
Sep 19 '11 at 12:30
Double check your typing. It makes wonders. Specially in titles, which is the very first thing people see of your question!
– Mariano Suárez-Álvarez
Sep 19 '11 at 3:47
Double check your typing. It makes wonders. Specially in titles, which is the very first thing people see of your question!
– Mariano Suárez-Álvarez
Sep 19 '11 at 3:47
@Mariano Suárez-Alvarez♦: Thank you very much for reminding me. I don't know what careless mistake I have made in my title because it might have been kindly corrected by Arturo Magidin. I will be more careful next time. Thanks again.
– ShinyaSakai
Sep 19 '11 at 12:30
@Mariano Suárez-Alvarez♦: Thank you very much for reminding me. I don't know what careless mistake I have made in my title because it might have been kindly corrected by Arturo Magidin. I will be more careful next time. Thanks again.
– ShinyaSakai
Sep 19 '11 at 12:30
add a comment |
2 Answers
2
active
oldest
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This is true in general; you do not need $X$ to be a noetherian space.
Continuing user3296's argument: the only thing left to check is that $C cap Z$ and $D cap Z$ are proper in $Z$. You do not need them closed in $X$ -- irreducibility is not a relative property: a space $Z$ is reducible if it is the union of two proper closed -in-itself subsets.
To see that for example $C cap Z subsetneq Z$, you have to convince yourself that $C cap Z = Z$ implies $C = overline{C} = overline{C cap Z} = overline Z$. The first equality is true because $C$ is closed in the closed set $overline Z$ (convince yourself that this implies that $C$ is closed). And the second because, on one hand, $C = overline C$ is contained in $overline Z = overline{C cap Z}$, and on the other hand, $C$ contains $C cap Z$ so that $overline C$ contains $overline{C cap Z}$.
answered Oct 23 '11 at 7:00
sibilant
467212
I thought of the "closedness" part of the definition for irreducibility in a wrong way. Thank you very much!
– ShinyaSakai
Oct 23 '11 at 18:08
add a comment |
As to your second gray box:
Suppose $overline{Z}$ is reducible, say $overline{Z} = C cup D$ with $C$ and $D$ proper closed subsets of $overline{Z}$. Then $C cap Z$ and $D cap Z$ are closed in $Z$, and $Z = (C cap Z) cup (D cap Z)$. There is one last thing to check -- what is it, and how do you check it?
answered Sep 19 '11 at 4:51
Daniel McLaury
15.5k32977
Thanks very much for your answer. I think I should check that both of them are closed and proper. But I have no idea...
– ShinyaSakai
Sep 19 '11 at 12:37
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
This is true in general; you do not need $X$ to be a noetherian space.
Continuing user3296's argument: the only thing left to check is that $C cap Z$ and $D cap Z$ are proper in $Z$. You do not need them closed in $X$ -- irreducibility is not a relative property: a space $Z$ is reducible if it is the union of two proper closed -in-itself subsets.
To see that for example $C cap Z subsetneq Z$, you have to convince yourself that $C cap Z = Z$ implies $C = overline{C} = overline{C cap Z} = overline Z$. The first equality is true because $C$ is closed in the closed set $overline Z$ (convince yourself that this implies that $C$ is closed). And the second because, on one hand, $C = overline C$ is contained in $overline Z = overline{C cap Z}$, and on the other hand, $C$ contains $C cap Z$ so that $overline C$ contains $overline{C cap Z}$.
answered Oct 23 '11 at 7:00
sibilant
467212
I thought of the "closedness" part of the definition for irreducibility in a wrong way. Thank you very much!
– ShinyaSakai
Oct 23 '11 at 18:08
add a comment |
This is true in general; you do not need $X$ to be a noetherian space.
Continuing user3296's argument: the only thing left to check is that $C cap Z$ and $D cap Z$ are proper in $Z$. You do not need them closed in $X$ -- irreducibility is not a relative property: a space $Z$ is reducible if it is the union of two proper closed -in-itself subsets.
To see that for example $C cap Z subsetneq Z$, you have to convince yourself that $C cap Z = Z$ implies $C = overline{C} = overline{C cap Z} = overline Z$. The first equality is true because $C$ is closed in the closed set $overline Z$ (convince yourself that this implies that $C$ is closed). And the second because, on one hand, $C = overline C$ is contained in $overline Z = overline{C cap Z}$, and on the other hand, $C$ contains $C cap Z$ so that $overline C$ contains $overline{C cap Z}$.
answered Oct 23 '11 at 7:00
sibilant
467212
I thought of the "closedness" part of the definition for irreducibility in a wrong way. Thank you very much!
– ShinyaSakai
Oct 23 '11 at 18:08
add a comment |
This is true in general; you do not need $X$ to be a noetherian space.
Continuing user3296's argument: the only thing left to check is that $C cap Z$ and $D cap Z$ are proper in $Z$. You do not need them closed in $X$ -- irreducibility is not a relative property: a space $Z$ is reducible if it is the union of two proper closed -in-itself subsets.
To see that for example $C cap Z subsetneq Z$, you have to convince yourself that $C cap Z = Z$ implies $C = overline{C} = overline{C cap Z} = overline Z$. The first equality is true because $C$ is closed in the closed set $overline Z$ (convince yourself that this implies that $C$ is closed). And the second because, on one hand, $C = overline C$ is contained in $overline Z = overline{C cap Z}$, and on the other hand, $C$ contains $C cap Z$ so that $overline C$ contains $overline{C cap Z}$.
answered Oct 23 '11 at 7:00
sibilant
467212
This is true in general; you do not need $X$ to be a noetherian space.
Continuing user3296's argument: the only thing left to check is that $C cap Z$ and $D cap Z$ are proper in $Z$. You do not need them closed in $X$ -- irreducibility is not a relative property: a space $Z$ is reducible if it is the union of two proper closed -in-itself subsets.
To see that for example $C cap Z subsetneq Z$, you have to convince yourself that $C cap Z = Z$ implies $C = overline{C} = overline{C cap Z} = overline Z$. The first equality is true because $C$ is closed in the closed set $overline Z$ (convince yourself that this implies that $C$ is closed). And the second because, on one hand, $C = overline C$ is contained in $overline Z = overline{C cap Z}$, and on the other hand, $C$ contains $C cap Z$ so that $overline C$ contains $overline{C cap Z}$.
answered Oct 23 '11 at 7:00
sibilant
467212
answered Oct 23 '11 at 7:00
sibilant
467212
answered Oct 23 '11 at 7:00
sibilant
467212
answered Oct 23 '11 at 7:00
sibilant
467212
467212
I thought of the "closedness" part of the definition for irreducibility in a wrong way. Thank you very much!
– ShinyaSakai
Oct 23 '11 at 18:08
add a comment |
I thought of the "closedness" part of the definition for irreducibility in a wrong way. Thank you very much!
– ShinyaSakai
Oct 23 '11 at 18:08
I thought of the "closedness" part of the definition for irreducibility in a wrong way. Thank you very much!
– ShinyaSakai
Oct 23 '11 at 18:08
I thought of the "closedness" part of the definition for irreducibility in a wrong way. Thank you very much!
– ShinyaSakai
Oct 23 '11 at 18:08
add a comment |
As to your second gray box:
Suppose $overline{Z}$ is reducible, say $overline{Z} = C cup D$ with $C$ and $D$ proper closed subsets of $overline{Z}$. Then $C cap Z$ and $D cap Z$ are closed in $Z$, and $Z = (C cap Z) cup (D cap Z)$. There is one last thing to check -- what is it, and how do you check it?
answered Sep 19 '11 at 4:51
Daniel McLaury
15.5k32977
Thanks very much for your answer. I think I should check that both of them are closed and proper. But I have no idea...
– ShinyaSakai
Sep 19 '11 at 12:37
add a comment |
As to your second gray box:
Suppose $overline{Z}$ is reducible, say $overline{Z} = C cup D$ with $C$ and $D$ proper closed subsets of $overline{Z}$. Then $C cap Z$ and $D cap Z$ are closed in $Z$, and $Z = (C cap Z) cup (D cap Z)$. There is one last thing to check -- what is it, and how do you check it?
answered Sep 19 '11 at 4:51
Daniel McLaury
15.5k32977
Thanks very much for your answer. I think I should check that both of them are closed and proper. But I have no idea...
– ShinyaSakai
Sep 19 '11 at 12:37
add a comment |
As to your second gray box:
Suppose $overline{Z}$ is reducible, say $overline{Z} = C cup D$ with $C$ and $D$ proper closed subsets of $overline{Z}$. Then $C cap Z$ and $D cap Z$ are closed in $Z$, and $Z = (C cap Z) cup (D cap Z)$. There is one last thing to check -- what is it, and how do you check it?
answered Sep 19 '11 at 4:51
Daniel McLaury
15.5k32977
As to your second gray box:
Suppose $overline{Z}$ is reducible, say $overline{Z} = C cup D$ with $C$ and $D$ proper closed subsets of $overline{Z}$. Then $C cap Z$ and $D cap Z$ are closed in $Z$, and $Z = (C cap Z) cup (D cap Z)$. There is one last thing to check -- what is it, and how do you check it?
answered Sep 19 '11 at 4:51
Daniel McLaury
15.5k32977
answered Sep 19 '11 at 4:51
Daniel McLaury
15.5k32977
answered Sep 19 '11 at 4:51
Daniel McLaury
15.5k32977
answered Sep 19 '11 at 4:51
Daniel McLaury
15.5k32977
15.5k32977
Thanks very much for your answer. I think I should check that both of them are closed and proper. But I have no idea...
– ShinyaSakai
Sep 19 '11 at 12:37
add a comment |
Thanks very much for your answer. I think I should check that both of them are closed and proper. But I have no idea...
– ShinyaSakai
Sep 19 '11 at 12:37
Thanks very much for your answer. I think I should check that both of them are closed and proper. But I have no idea...
– ShinyaSakai
Sep 19 '11 at 12:37
Thanks very much for your answer. I think I should check that both of them are closed and proper. But I have no idea...
– ShinyaSakai
Sep 19 '11 at 12:37
add a comment |
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Double check your typing. It makes wonders. Specially in titles, which is the very first thing people see of your question!
– Mariano Suárez-Álvarez
Sep 19 '11 at 3:47
@Mariano Suárez-Alvarez♦: Thank you very much for reminding me. I don't know what careless mistake I have made in my title because it might have been kindly corrected by Arturo Magidin. I will be more careful next time. Thanks again.
– ShinyaSakai
Sep 19 '11 at 12:30