Confusion about the definition of reduced scheme
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I'm following Eisenbud-Harris Geometry of Schemes, and I am a little bit confused about how they define the notion of a reduced scheme. In the affine case, if $X = text{Spec} , R$, then setting $X_{mathrm{red}} = text{Spec}, R_{mathrm{red}}$ where $R_{mathrm{red}}$ is the ring $R$ modulo its nil radical, we say that $X$ is reduced if $X = X_{mathrm{red}}$.
For an arbitrary scheme $X$, they define a quasi-coherent sheaf called the nilradical, that assigns to each open subset $U$ of $X$, the nilradical of $mathcal{O}_X(U)$, being $mathcal{O}_X$ the structure sheaf of $X$, and then say that the scheme $X$ is reduced if its associated closed subscheme $X_{mathrm{red}}$ is equal to $X$.
What does the statement "its associated closed subscheme $X_{mathrm{red}}$" mean?
The definition Eisenbud and Harris give of a closed subscheme $Y$ of a scheme $X$ is a closed topological subspace $|Y| subset |X|$ with a sheaf $mathcal{O}_Y$ that is the quotient sheaf of the structure sheaf of $X$ by a quasi-coherent sheaf of ideals $mathcal{J}$ such that the intersection of $Y$ with any affine open subset $U subset X$ is the closed subschema associated to the ideal $mathcal{J}(U)$.
Since we have a quasi-coherent sheaf $mathcal{N}$, we can quotient the structure sheaf $mathcal{O}_X$ of $X$ by $mathcal{N}$. Is then
$$X_{mathrm{red}} = (X, mathcal{O}_X/mathcal{N}), ?$$
The problem is that the definition of a closed subscheme $Y$ of a scheme $X$ says that the topological space $|Y|$ has to be a closed subspace of the topological space $|X|$, but in this case it would be the same space and not a proper subspace.
abstract-algebra algebraic-geometry sheaf-theory schemes
asked Nov 16 at 14:22
user313212
287520
add a comment |
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0
down vote
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I'm following Eisenbud-Harris Geometry of Schemes, and I am a little bit confused about how they define the notion of a reduced scheme. In the affine case, if $X = text{Spec} , R$, then setting $X_{mathrm{red}} = text{Spec}, R_{mathrm{red}}$ where $R_{mathrm{red}}$ is the ring $R$ modulo its nil radical, we say that $X$ is reduced if $X = X_{mathrm{red}}$.
For an arbitrary scheme $X$, they define a quasi-coherent sheaf called the nilradical, that assigns to each open subset $U$ of $X$, the nilradical of $mathcal{O}_X(U)$, being $mathcal{O}_X$ the structure sheaf of $X$, and then say that the scheme $X$ is reduced if its associated closed subscheme $X_{mathrm{red}}$ is equal to $X$.
What does the statement "its associated closed subscheme $X_{mathrm{red}}$" mean?
The definition Eisenbud and Harris give of a closed subscheme $Y$ of a scheme $X$ is a closed topological subspace $|Y| subset |X|$ with a sheaf $mathcal{O}_Y$ that is the quotient sheaf of the structure sheaf of $X$ by a quasi-coherent sheaf of ideals $mathcal{J}$ such that the intersection of $Y$ with any affine open subset $U subset X$ is the closed subschema associated to the ideal $mathcal{J}(U)$.
Since we have a quasi-coherent sheaf $mathcal{N}$, we can quotient the structure sheaf $mathcal{O}_X$ of $X$ by $mathcal{N}$. Is then
$$X_{mathrm{red}} = (X, mathcal{O}_X/mathcal{N}), ?$$
The problem is that the definition of a closed subscheme $Y$ of a scheme $X$ says that the topological space $|Y|$ has to be a closed subspace of the topological space $|X|$, but in this case it would be the same space and not a proper subspace.
abstract-algebra algebraic-geometry sheaf-theory schemes
asked Nov 16 at 14:22
user313212
287520
2
Indeed, in this case the two schemes have the same underlying topological space. This is not a problem, however.
– Sasha
Nov 16 at 15:40
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm following Eisenbud-Harris Geometry of Schemes, and I am a little bit confused about how they define the notion of a reduced scheme. In the affine case, if $X = text{Spec} , R$, then setting $X_{mathrm{red}} = text{Spec}, R_{mathrm{red}}$ where $R_{mathrm{red}}$ is the ring $R$ modulo its nil radical, we say that $X$ is reduced if $X = X_{mathrm{red}}$.
For an arbitrary scheme $X$, they define a quasi-coherent sheaf called the nilradical, that assigns to each open subset $U$ of $X$, the nilradical of $mathcal{O}_X(U)$, being $mathcal{O}_X$ the structure sheaf of $X$, and then say that the scheme $X$ is reduced if its associated closed subscheme $X_{mathrm{red}}$ is equal to $X$.
What does the statement "its associated closed subscheme $X_{mathrm{red}}$" mean?
The definition Eisenbud and Harris give of a closed subscheme $Y$ of a scheme $X$ is a closed topological subspace $|Y| subset |X|$ with a sheaf $mathcal{O}_Y$ that is the quotient sheaf of the structure sheaf of $X$ by a quasi-coherent sheaf of ideals $mathcal{J}$ such that the intersection of $Y$ with any affine open subset $U subset X$ is the closed subschema associated to the ideal $mathcal{J}(U)$.
Since we have a quasi-coherent sheaf $mathcal{N}$, we can quotient the structure sheaf $mathcal{O}_X$ of $X$ by $mathcal{N}$. Is then
$$X_{mathrm{red}} = (X, mathcal{O}_X/mathcal{N}), ?$$
The problem is that the definition of a closed subscheme $Y$ of a scheme $X$ says that the topological space $|Y|$ has to be a closed subspace of the topological space $|X|$, but in this case it would be the same space and not a proper subspace.
abstract-algebra algebraic-geometry sheaf-theory schemes
asked Nov 16 at 14:22
user313212
287520
I'm following Eisenbud-Harris Geometry of Schemes, and I am a little bit confused about how they define the notion of a reduced scheme. In the affine case, if $X = text{Spec} , R$, then setting $X_{mathrm{red}} = text{Spec}, R_{mathrm{red}}$ where $R_{mathrm{red}}$ is the ring $R$ modulo its nil radical, we say that $X$ is reduced if $X = X_{mathrm{red}}$.
For an arbitrary scheme $X$, they define a quasi-coherent sheaf called the nilradical, that assigns to each open subset $U$ of $X$, the nilradical of $mathcal{O}_X(U)$, being $mathcal{O}_X$ the structure sheaf of $X$, and then say that the scheme $X$ is reduced if its associated closed subscheme $X_{mathrm{red}}$ is equal to $X$.
What does the statement "its associated closed subscheme $X_{mathrm{red}}$" mean?
The definition Eisenbud and Harris give of a closed subscheme $Y$ of a scheme $X$ is a closed topological subspace $|Y| subset |X|$ with a sheaf $mathcal{O}_Y$ that is the quotient sheaf of the structure sheaf of $X$ by a quasi-coherent sheaf of ideals $mathcal{J}$ such that the intersection of $Y$ with any affine open subset $U subset X$ is the closed subschema associated to the ideal $mathcal{J}(U)$.
Since we have a quasi-coherent sheaf $mathcal{N}$, we can quotient the structure sheaf $mathcal{O}_X$ of $X$ by $mathcal{N}$. Is then
$$X_{mathrm{red}} = (X, mathcal{O}_X/mathcal{N}), ?$$
The problem is that the definition of a closed subscheme $Y$ of a scheme $X$ says that the topological space $|Y|$ has to be a closed subspace of the topological space $|X|$, but in this case it would be the same space and not a proper subspace.
abstract-algebra algebraic-geometry sheaf-theory schemes
abstract-algebra algebraic-geometry sheaf-theory schemes
asked Nov 16 at 14:22
user313212
287520
asked Nov 16 at 14:22
user313212
287520
asked Nov 16 at 14:22
user313212
287520
asked Nov 16 at 14:22
user313212
287520
asked Nov 16 at 14:22
user313212
287520
287520
2
Indeed, in this case the two schemes have the same underlying topological space. This is not a problem, however.
– Sasha
Nov 16 at 15:40
add a comment |
2
Indeed, in this case the two schemes have the same underlying topological space. This is not a problem, however.
– Sasha
Nov 16 at 15:40
2
2
Indeed, in this case the two schemes have the same underlying topological space. This is not a problem, however.
– Sasha
Nov 16 at 15:40
Indeed, in this case the two schemes have the same underlying topological space. This is not a problem, however.
– Sasha
Nov 16 at 15:40
add a comment |
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Indeed, in this case the two schemes have the same underlying topological space. This is not a problem, however.
– Sasha
Nov 16 at 15:40