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An affine scheme is integral if and only if it is connected and all stalks are integral domains.
Clash Royale CLAN TAG#URR8PPP
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Let X=Spec A. So far I have shown the easy side, i.e, if the affine scheme is integral, then it's irreducible and so connected. Since all stalks are localization of A at prime ideals, we have that all stalks are integral domains since A is an integral domain.
I am having trouble proving the other way. I'm trying to prove that A is an integral domain by looking at A as the ring of global sections of X. So let f,g be two global sections of X such that fg=0. Then f_x.g_x=0(product of stalks) for all x. Then the points where the stalks of f are zero form an open set. I'm not able to show that it's also closed to get a disconnection.
algebraic-geometry commutative-algebra
asked Jul 21 at 3:24
user567863
61
add a comment |Â
up vote
1
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Let X=Spec A. So far I have shown the easy side, i.e, if the affine scheme is integral, then it's irreducible and so connected. Since all stalks are localization of A at prime ideals, we have that all stalks are integral domains since A is an integral domain.
I am having trouble proving the other way. I'm trying to prove that A is an integral domain by looking at A as the ring of global sections of X. So let f,g be two global sections of X such that fg=0. Then f_x.g_x=0(product of stalks) for all x. Then the points where the stalks of f are zero form an open set. I'm not able to show that it's also closed to get a disconnection.
algebraic-geometry commutative-algebra
asked Jul 21 at 3:24
user567863
61
Stacks project seems to say your question is not true but I haven’t read their example stacks.math.columbia.edu/tag/0568
– usr0192
Jul 21 at 5:17
Maybe you mean irreducible (as opposed to connected) and stalks are domains?
– usr0192
Jul 21 at 5:22
@usr0192 Yes, according to the stacks project, the above is not true. It's true if connected is replaced by irreducible. Thank you!
– user567863
Jul 21 at 10:25
Another fix is to add Noetherian (and non-empty) to the hypotheses.
– Samir Canning
Jul 21 at 19:33
@SamirCanning You are right, in that case X need not necessarily be affine.
– user567863
Jul 22 at 0:23
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let X=Spec A. So far I have shown the easy side, i.e, if the affine scheme is integral, then it's irreducible and so connected. Since all stalks are localization of A at prime ideals, we have that all stalks are integral domains since A is an integral domain.
I am having trouble proving the other way. I'm trying to prove that A is an integral domain by looking at A as the ring of global sections of X. So let f,g be two global sections of X such that fg=0. Then f_x.g_x=0(product of stalks) for all x. Then the points where the stalks of f are zero form an open set. I'm not able to show that it's also closed to get a disconnection.
algebraic-geometry commutative-algebra
asked Jul 21 at 3:24
user567863
61
Let X=Spec A. So far I have shown the easy side, i.e, if the affine scheme is integral, then it's irreducible and so connected. Since all stalks are localization of A at prime ideals, we have that all stalks are integral domains since A is an integral domain.
I am having trouble proving the other way. I'm trying to prove that A is an integral domain by looking at A as the ring of global sections of X. So let f,g be two global sections of X such that fg=0. Then f_x.g_x=0(product of stalks) for all x. Then the points where the stalks of f are zero form an open set. I'm not able to show that it's also closed to get a disconnection.
algebraic-geometry commutative-algebra
asked Jul 21 at 3:24
user567863
61
asked Jul 21 at 3:24
user567863
61
asked Jul 21 at 3:24
user567863
61
asked Jul 21 at 3:24
user567863
61
61
Stacks project seems to say your question is not true but I haven’t read their example stacks.math.columbia.edu/tag/0568
– usr0192
Jul 21 at 5:17
Maybe you mean irreducible (as opposed to connected) and stalks are domains?
– usr0192
Jul 21 at 5:22
@usr0192 Yes, according to the stacks project, the above is not true. It's true if connected is replaced by irreducible. Thank you!
– user567863
Jul 21 at 10:25
Another fix is to add Noetherian (and non-empty) to the hypotheses.
– Samir Canning
Jul 21 at 19:33
@SamirCanning You are right, in that case X need not necessarily be affine.
– user567863
Jul 22 at 0:23
add a comment |Â
Stacks project seems to say your question is not true but I haven’t read their example stacks.math.columbia.edu/tag/0568
– usr0192
Jul 21 at 5:17
Maybe you mean irreducible (as opposed to connected) and stalks are domains?
– usr0192
Jul 21 at 5:22
@usr0192 Yes, according to the stacks project, the above is not true. It's true if connected is replaced by irreducible. Thank you!
– user567863
Jul 21 at 10:25
Another fix is to add Noetherian (and non-empty) to the hypotheses.
– Samir Canning
Jul 21 at 19:33
@SamirCanning You are right, in that case X need not necessarily be affine.
– user567863
Jul 22 at 0:23
Stacks project seems to say your question is not true but I haven’t read their example stacks.math.columbia.edu/tag/0568
– usr0192
Jul 21 at 5:17
Stacks project seems to say your question is not true but I haven’t read their example stacks.math.columbia.edu/tag/0568
– usr0192
Jul 21 at 5:17
Maybe you mean irreducible (as opposed to connected) and stalks are domains?
– usr0192
Jul 21 at 5:22
Maybe you mean irreducible (as opposed to connected) and stalks are domains?
– usr0192
Jul 21 at 5:22
@usr0192 Yes, according to the stacks project, the above is not true. It's true if connected is replaced by irreducible. Thank you!
– user567863
Jul 21 at 10:25
@usr0192 Yes, according to the stacks project, the above is not true. It's true if connected is replaced by irreducible. Thank you!
– user567863
Jul 21 at 10:25
Another fix is to add Noetherian (and non-empty) to the hypotheses.
– Samir Canning
Jul 21 at 19:33
Another fix is to add Noetherian (and non-empty) to the hypotheses.
– Samir Canning
Jul 21 at 19:33
@SamirCanning You are right, in that case X need not necessarily be affine.
– user567863
Jul 22 at 0:23
@SamirCanning You are right, in that case X need not necessarily be affine.
– user567863
Jul 22 at 0:23
add a comment |Â
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Stacks project seems to say your question is not true but I haven’t read their example stacks.math.columbia.edu/tag/0568
– usr0192
Jul 21 at 5:17
Maybe you mean irreducible (as opposed to connected) and stalks are domains?
– usr0192
Jul 21 at 5:22
@usr0192 Yes, according to the stacks project, the above is not true. It's true if connected is replaced by irreducible. Thank you!
– user567863
Jul 21 at 10:25
Another fix is to add Noetherian (and non-empty) to the hypotheses.
– Samir Canning
Jul 21 at 19:33
@SamirCanning You are right, in that case X need not necessarily be affine.
– user567863
Jul 22 at 0:23