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Is a projective algebraic manifold irreducible algebraic set in $P^n$?
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A compact complex manifold $X$ which admits an embedding into $P^n(mathbb{C})$ (for some $n$) is called a projective algebraic manifold. And by a theorem of Chow, every complex submanifold $V$ of $P^n(mathbb{C})$ is actually an algebraic submanifold. However, how to determine if the embedded complex submanifold is an irreducible algebraic set?
algebraic-geometry complex-geometry
asked Jan 17 at 16:28
DannyDanny
1,138412
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add a comment |
$begingroup$
A compact complex manifold $X$ which admits an embedding into $P^n(mathbb{C})$ (for some $n$) is called a projective algebraic manifold. And by a theorem of Chow, every complex submanifold $V$ of $P^n(mathbb{C})$ is actually an algebraic submanifold. However, how to determine if the embedded complex submanifold is an irreducible algebraic set?
algebraic-geometry complex-geometry
asked Jan 17 at 16:28
DannyDanny
1,138412
$endgroup$
$begingroup$
If $X$ is connected its ring of meromorphic functions $M(X)$ is a field. Choose the embedding such that $X' = X$ minus a few points is affine and open in $X$. $mathbb{C}[X']$ its ring of polynomials (inherited from $A^n(mathbb{C})$) is a subring of $M(X)$ thus an integral domain and $X'$ is an affine variety.
$endgroup$
– reuns
Jan 17 at 18:26
add a comment |
$begingroup$
A compact complex manifold $X$ which admits an embedding into $P^n(mathbb{C})$ (for some $n$) is called a projective algebraic manifold. And by a theorem of Chow, every complex submanifold $V$ of $P^n(mathbb{C})$ is actually an algebraic submanifold. However, how to determine if the embedded complex submanifold is an irreducible algebraic set?
algebraic-geometry complex-geometry
asked Jan 17 at 16:28
DannyDanny
1,138412
$endgroup$
A compact complex manifold $X$ which admits an embedding into $P^n(mathbb{C})$ (for some $n$) is called a projective algebraic manifold. And by a theorem of Chow, every complex submanifold $V$ of $P^n(mathbb{C})$ is actually an algebraic submanifold. However, how to determine if the embedded complex submanifold is an irreducible algebraic set?
algebraic-geometry complex-geometry
algebraic-geometry complex-geometry
asked Jan 17 at 16:28
DannyDanny
1,138412
asked Jan 17 at 16:28
DannyDanny
1,138412
asked Jan 17 at 16:28
DannyDanny
1,138412
asked Jan 17 at 16:28
DannyDanny
1,138412
asked Jan 17 at 16:28
DannyDanny
1,138412
1,138412
$begingroup$
If $X$ is connected its ring of meromorphic functions $M(X)$ is a field. Choose the embedding such that $X' = X$ minus a few points is affine and open in $X$. $mathbb{C}[X']$ its ring of polynomials (inherited from $A^n(mathbb{C})$) is a subring of $M(X)$ thus an integral domain and $X'$ is an affine variety.
$endgroup$
– reuns
Jan 17 at 18:26
add a comment |
$begingroup$
If $X$ is connected its ring of meromorphic functions $M(X)$ is a field. Choose the embedding such that $X' = X$ minus a few points is affine and open in $X$. $mathbb{C}[X']$ its ring of polynomials (inherited from $A^n(mathbb{C})$) is a subring of $M(X)$ thus an integral domain and $X'$ is an affine variety.
$endgroup$
– reuns
Jan 17 at 18:26
$begingroup$
If $X$ is connected its ring of meromorphic functions $M(X)$ is a field. Choose the embedding such that $X' = X$ minus a few points is affine and open in $X$. $mathbb{C}[X']$ its ring of polynomials (inherited from $A^n(mathbb{C})$) is a subring of $M(X)$ thus an integral domain and $X'$ is an affine variety.
$endgroup$
– reuns
Jan 17 at 18:26
$begingroup$
If $X$ is connected its ring of meromorphic functions $M(X)$ is a field. Choose the embedding such that $X' = X$ minus a few points is affine and open in $X$. $mathbb{C}[X']$ its ring of polynomials (inherited from $A^n(mathbb{C})$) is a subring of $M(X)$ thus an integral domain and $X'$ is an affine variety.
$endgroup$
– reuns
Jan 17 at 18:26
add a comment |
1 Answer
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$begingroup$
Assuming you mean embedding to mean injective on points, this question has nothing to do with the embedding. For a smooth locally noetherian scheme over a field, irreducible components are connected components - these conditions imply that all the local rings of the variety are regular local rings, which in turn implies that each point is on exactly one irreducible component (by the correspondence between minimal primes of the local ring and irreducible components containing the point). So $X$ is irreducible iff it's connected.
answered Jan 17 at 20:01
KReiserKReiser
9,72721435
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add a comment |
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1 Answer
1
active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Assuming you mean embedding to mean injective on points, this question has nothing to do with the embedding. For a smooth locally noetherian scheme over a field, irreducible components are connected components - these conditions imply that all the local rings of the variety are regular local rings, which in turn implies that each point is on exactly one irreducible component (by the correspondence between minimal primes of the local ring and irreducible components containing the point). So $X$ is irreducible iff it's connected.
answered Jan 17 at 20:01
KReiserKReiser
9,72721435
$endgroup$
add a comment |
$begingroup$
Assuming you mean embedding to mean injective on points, this question has nothing to do with the embedding. For a smooth locally noetherian scheme over a field, irreducible components are connected components - these conditions imply that all the local rings of the variety are regular local rings, which in turn implies that each point is on exactly one irreducible component (by the correspondence between minimal primes of the local ring and irreducible components containing the point). So $X$ is irreducible iff it's connected.
answered Jan 17 at 20:01
KReiserKReiser
9,72721435
$endgroup$
add a comment |
$begingroup$
Assuming you mean embedding to mean injective on points, this question has nothing to do with the embedding. For a smooth locally noetherian scheme over a field, irreducible components are connected components - these conditions imply that all the local rings of the variety are regular local rings, which in turn implies that each point is on exactly one irreducible component (by the correspondence between minimal primes of the local ring and irreducible components containing the point). So $X$ is irreducible iff it's connected.
answered Jan 17 at 20:01
KReiserKReiser
9,72721435
$endgroup$
Assuming you mean embedding to mean injective on points, this question has nothing to do with the embedding. For a smooth locally noetherian scheme over a field, irreducible components are connected components - these conditions imply that all the local rings of the variety are regular local rings, which in turn implies that each point is on exactly one irreducible component (by the correspondence between minimal primes of the local ring and irreducible components containing the point). So $X$ is irreducible iff it's connected.
answered Jan 17 at 20:01
KReiserKReiser
9,72721435
answered Jan 17 at 20:01
KReiserKReiser
9,72721435
answered Jan 17 at 20:01
KReiserKReiser
9,72721435
answered Jan 17 at 20:01
KReiserKReiser
9,72721435
9,72721435
add a comment |
add a comment |
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$begingroup$
If $X$ is connected its ring of meromorphic functions $M(X)$ is a field. Choose the embedding such that $X' = X$ minus a few points is affine and open in $X$. $mathbb{C}[X']$ its ring of polynomials (inherited from $A^n(mathbb{C})$) is a subring of $M(X)$ thus an integral domain and $X'$ is an affine variety.
$endgroup$
– reuns
Jan 17 at 18:26