The nonsingular variety is a manifold and irreduciblity
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For the claim that a nonsingular variety is a smooth manifold, do we need to require the nonsingular variety to be irreducible? I am thinking that each irreducible component is a smooth manifold and different components might have different dimensions, which might lead to a problem of the dimension of the manifold.
algebraic-geometry smooth-manifolds
asked Dec 27 '18 at 12:33
66666666
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add a comment |
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For the claim that a nonsingular variety is a smooth manifold, do we need to require the nonsingular variety to be irreducible? I am thinking that each irreducible component is a smooth manifold and different components might have different dimensions, which might lead to a problem of the dimension of the manifold.
algebraic-geometry smooth-manifolds
asked Dec 27 '18 at 12:33
66666666
1,353621
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Varieties are usually required to be irreducible.
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– Moishe Kohan
Dec 27 '18 at 12:35
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@MoisheCohen this heavily depends on who's paper you're reading.
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– KReiser
Dec 27 '18 at 19:25
add a comment |
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For the claim that a nonsingular variety is a smooth manifold, do we need to require the nonsingular variety to be irreducible? I am thinking that each irreducible component is a smooth manifold and different components might have different dimensions, which might lead to a problem of the dimension of the manifold.
algebraic-geometry smooth-manifolds
asked Dec 27 '18 at 12:33
66666666
1,353621
$endgroup$
For the claim that a nonsingular variety is a smooth manifold, do we need to require the nonsingular variety to be irreducible? I am thinking that each irreducible component is a smooth manifold and different components might have different dimensions, which might lead to a problem of the dimension of the manifold.
algebraic-geometry smooth-manifolds
algebraic-geometry smooth-manifolds
asked Dec 27 '18 at 12:33
66666666
1,353621
asked Dec 27 '18 at 12:33
66666666
1,353621
asked Dec 27 '18 at 12:33
66666666
1,353621
asked Dec 27 '18 at 12:33
66666666
1,353621
asked Dec 27 '18 at 12:33
66666666
1,353621
1,353621
$begingroup$
Varieties are usually required to be irreducible.
$endgroup$
– Moishe Kohan
Dec 27 '18 at 12:35
$begingroup$
@MoisheCohen this heavily depends on who's paper you're reading.
$endgroup$
– KReiser
Dec 27 '18 at 19:25
add a comment |
$begingroup$
Varieties are usually required to be irreducible.
$endgroup$
– Moishe Kohan
Dec 27 '18 at 12:35
$begingroup$
@MoisheCohen this heavily depends on who's paper you're reading.
$endgroup$
– KReiser
Dec 27 '18 at 19:25
$begingroup$
Varieties are usually required to be irreducible.
$endgroup$
– Moishe Kohan
Dec 27 '18 at 12:35
$begingroup$
Varieties are usually required to be irreducible.
$endgroup$
– Moishe Kohan
Dec 27 '18 at 12:35
$begingroup$
@MoisheCohen this heavily depends on who's paper you're reading.
$endgroup$
– KReiser
Dec 27 '18 at 19:25
$begingroup$
@MoisheCohen this heavily depends on who's paper you're reading.
$endgroup$
– KReiser
Dec 27 '18 at 19:25
add a comment |
1 Answer
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If your definition of a manifold does not require that all connected components have the same dimension, you need not worry about irreducibilty. In a smooth variety, irreducible components are connected components, so one may just deal with each component separately.
answered Dec 27 '18 at 19:29
KReiserKReiser
9,87121435
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If your definition of a manifold does not require that all connected components have the same dimension, you need not worry about irreducibilty. In a smooth variety, irreducible components are connected components, so one may just deal with each component separately.
answered Dec 27 '18 at 19:29
KReiserKReiser
9,87121435
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add a comment |
$begingroup$
If your definition of a manifold does not require that all connected components have the same dimension, you need not worry about irreducibilty. In a smooth variety, irreducible components are connected components, so one may just deal with each component separately.
answered Dec 27 '18 at 19:29
KReiserKReiser
9,87121435
$endgroup$
add a comment |
$begingroup$
If your definition of a manifold does not require that all connected components have the same dimension, you need not worry about irreducibilty. In a smooth variety, irreducible components are connected components, so one may just deal with each component separately.
answered Dec 27 '18 at 19:29
KReiserKReiser
9,87121435
$endgroup$
If your definition of a manifold does not require that all connected components have the same dimension, you need not worry about irreducibilty. In a smooth variety, irreducible components are connected components, so one may just deal with each component separately.
answered Dec 27 '18 at 19:29
KReiserKReiser
9,87121435
answered Dec 27 '18 at 19:29
KReiserKReiser
9,87121435
answered Dec 27 '18 at 19:29
KReiserKReiser
9,87121435
answered Dec 27 '18 at 19:29
KReiserKReiser
9,87121435
9,87121435
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$begingroup$
Varieties are usually required to be irreducible.
$endgroup$
– Moishe Kohan
Dec 27 '18 at 12:35
$begingroup$
@MoisheCohen this heavily depends on who's paper you're reading.
$endgroup$
– KReiser
Dec 27 '18 at 19:25