Why the ring of regular functions on a variety at a point is a local ringQuestions about the local ring of a point on a variety.The maximal ideal of a point of a variety is a principal ideal.Localization of Coordinate Ring isomorphic to ring of local regular functionsRegular Local RingLocalization of a regular local ring is regularRing of regular functions on a pointQuotient field of ring of regular functions at some point in affine variety is the field of rational functions on the varietycompletion morphism for $mathscrH_x$, the local ring of germs of holomorphic functions in a neighborhood of xIntuition of local ring $O_p,X,pin X$ prime ideals correspondence with subvarieties of variety $X$Isomorphism between ring of regular functions and coordinate ring of an affine variety.
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Why the ring of regular functions on a variety at a point is a local ring
Questions about the local ring of a point on a variety.The maximal ideal of a point of a variety is a principal ideal.Localization of Coordinate Ring isomorphic to ring of local regular functionsRegular Local RingLocalization of a regular local ring is regularRing of regular functions on a pointQuotient field of ring of regular functions at some point in affine variety is the field of rational functions on the varietycompletion morphism for $mathscrH_x$, the local ring of germs of holomorphic functions in a neighborhood of xIntuition of local ring $O_p,X,pin X$ prime ideals correspondence with subvarieties of variety $X$Isomorphism between ring of regular functions and coordinate ring of an affine variety.
$begingroup$
In the Hartshorne's book on Algebraic Geometry one reads the following, '' Let Y be a variety. Denote by $O(Y) $ the ring of all regular functions on $Y. $ For $pin Y, $ we define the local ring of $P$ on $Y, O_P, $ to be the ring of germs of regular functions on $Y$ near $P$''. He then gives an argument, why $O_P $ is a local ring: '' its maximal ideal $m$ is the set of germs of regular functions which vanish at $P. $ For if $f(P) neq 0,$ then $1/f$ is regular in some neighborhood of $P$ ''. I unfortunately don't understand the argument, why $O_P$ contains a maximal ideal. Can somebody give some more explanations on that? Many thanks for your comment.
abstract-algebra algebraic-geometry
asked Mar 13 at 9:00
user249018user249018
435137
$endgroup$
add a comment |
$begingroup$
In the Hartshorne's book on Algebraic Geometry one reads the following, '' Let Y be a variety. Denote by $O(Y) $ the ring of all regular functions on $Y. $ For $pin Y, $ we define the local ring of $P$ on $Y, O_P, $ to be the ring of germs of regular functions on $Y$ near $P$''. He then gives an argument, why $O_P $ is a local ring: '' its maximal ideal $m$ is the set of germs of regular functions which vanish at $P. $ For if $f(P) neq 0,$ then $1/f$ is regular in some neighborhood of $P$ ''. I unfortunately don't understand the argument, why $O_P$ contains a maximal ideal. Can somebody give some more explanations on that? Many thanks for your comment.
abstract-algebra algebraic-geometry
asked Mar 13 at 9:00
user249018user249018
435137
$endgroup$
1
$begingroup$
The point is that $m$ is an ideal and every other element not in $m$ is invertible. So $m$ is maximal. Cheers.
$endgroup$
– Wuestenfux
Mar 13 at 9:08
add a comment |
$begingroup$
In the Hartshorne's book on Algebraic Geometry one reads the following, '' Let Y be a variety. Denote by $O(Y) $ the ring of all regular functions on $Y. $ For $pin Y, $ we define the local ring of $P$ on $Y, O_P, $ to be the ring of germs of regular functions on $Y$ near $P$''. He then gives an argument, why $O_P $ is a local ring: '' its maximal ideal $m$ is the set of germs of regular functions which vanish at $P. $ For if $f(P) neq 0,$ then $1/f$ is regular in some neighborhood of $P$ ''. I unfortunately don't understand the argument, why $O_P$ contains a maximal ideal. Can somebody give some more explanations on that? Many thanks for your comment.
abstract-algebra algebraic-geometry
asked Mar 13 at 9:00
user249018user249018
435137
$endgroup$
In the Hartshorne's book on Algebraic Geometry one reads the following, '' Let Y be a variety. Denote by $O(Y) $ the ring of all regular functions on $Y. $ For $pin Y, $ we define the local ring of $P$ on $Y, O_P, $ to be the ring of germs of regular functions on $Y$ near $P$''. He then gives an argument, why $O_P $ is a local ring: '' its maximal ideal $m$ is the set of germs of regular functions which vanish at $P. $ For if $f(P) neq 0,$ then $1/f$ is regular in some neighborhood of $P$ ''. I unfortunately don't understand the argument, why $O_P$ contains a maximal ideal. Can somebody give some more explanations on that? Many thanks for your comment.
abstract-algebra algebraic-geometry
abstract-algebra algebraic-geometry
asked Mar 13 at 9:00
user249018user249018
435137
asked Mar 13 at 9:00
user249018user249018
435137
asked Mar 13 at 9:00
user249018user249018
435137
asked Mar 13 at 9:00
user249018user249018
435137
asked Mar 13 at 9:00
user249018user249018
435137
435137
1
$begingroup$
The point is that $m$ is an ideal and every other element not in $m$ is invertible. So $m$ is maximal. Cheers.
$endgroup$
– Wuestenfux
Mar 13 at 9:08
add a comment |
1
$begingroup$
The point is that $m$ is an ideal and every other element not in $m$ is invertible. So $m$ is maximal. Cheers.
$endgroup$
– Wuestenfux
Mar 13 at 9:08
1
1
$begingroup$
The point is that $m$ is an ideal and every other element not in $m$ is invertible. So $m$ is maximal. Cheers.
$endgroup$
– Wuestenfux
Mar 13 at 9:08
$begingroup$
The point is that $m$ is an ideal and every other element not in $m$ is invertible. So $m$ is maximal. Cheers.
$endgroup$
– Wuestenfux
Mar 13 at 9:08
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hartshorne is making use of the following commutative algebra characterisation of maximal ideals.
Lemma 1. Let $A$ be a commutative ring with unity. Then the set of elements which are not invertible is a maximal ideal in $A$.
In your case, you have that the group of invertible germs is
$$mathscrU(O_P) :=[f]in O_P : f(P)neq 0 $$
and this follows from Hartshorne's observation: if $f(P)neq 0$ and $f$ is regular around $P$, then $1/f$ is so and $[1/f]in O_P$ as germ; thus such germs have inverses.
Hence $mathfrak m = A setminus mathscrU(A) = [f]in O_P : f(P) = 0$ is a maximal ideal.
Now such a ring must be local: if there were another maximal ideal $mathfrak m^prime$ and if $mathfrak m^prime neq mathfrak m$ then there would exist $[f]in mathfrak m^prime notin mathfrak m$, so $[f]$ is invertible in $O_P$ and so $mathfrak m^prime = O_P$.
answered Mar 13 at 11:45
CaligulaCaligula
1,241416
$endgroup$
1
$begingroup$
Did you mean to assume that $A$ is local in your lemma? I don't think it's true otherwise. Or maybe you wanted to say that if the set of non-units is an ideal, then it is the unique maximal ideal.
$endgroup$
– André 3000
Mar 13 at 18:38
add a comment |
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$begingroup$
Hartshorne is making use of the following commutative algebra characterisation of maximal ideals.
Lemma 1. Let $A$ be a commutative ring with unity. Then the set of elements which are not invertible is a maximal ideal in $A$.
In your case, you have that the group of invertible germs is
$$mathscrU(O_P) :=[f]in O_P : f(P)neq 0 $$
and this follows from Hartshorne's observation: if $f(P)neq 0$ and $f$ is regular around $P$, then $1/f$ is so and $[1/f]in O_P$ as germ; thus such germs have inverses.
Hence $mathfrak m = A setminus mathscrU(A) = [f]in O_P : f(P) = 0$ is a maximal ideal.
Now such a ring must be local: if there were another maximal ideal $mathfrak m^prime$ and if $mathfrak m^prime neq mathfrak m$ then there would exist $[f]in mathfrak m^prime notin mathfrak m$, so $[f]$ is invertible in $O_P$ and so $mathfrak m^prime = O_P$.
answered Mar 13 at 11:45
CaligulaCaligula
1,241416
$endgroup$
1
$begingroup$
Did you mean to assume that $A$ is local in your lemma? I don't think it's true otherwise. Or maybe you wanted to say that if the set of non-units is an ideal, then it is the unique maximal ideal.
$endgroup$
– André 3000
Mar 13 at 18:38
add a comment |
$begingroup$
Hartshorne is making use of the following commutative algebra characterisation of maximal ideals.
Lemma 1. Let $A$ be a commutative ring with unity. Then the set of elements which are not invertible is a maximal ideal in $A$.
In your case, you have that the group of invertible germs is
$$mathscrU(O_P) :=[f]in O_P : f(P)neq 0 $$
and this follows from Hartshorne's observation: if $f(P)neq 0$ and $f$ is regular around $P$, then $1/f$ is so and $[1/f]in O_P$ as germ; thus such germs have inverses.
Hence $mathfrak m = A setminus mathscrU(A) = [f]in O_P : f(P) = 0$ is a maximal ideal.
Now such a ring must be local: if there were another maximal ideal $mathfrak m^prime$ and if $mathfrak m^prime neq mathfrak m$ then there would exist $[f]in mathfrak m^prime notin mathfrak m$, so $[f]$ is invertible in $O_P$ and so $mathfrak m^prime = O_P$.
answered Mar 13 at 11:45
CaligulaCaligula
1,241416
$endgroup$
1
$begingroup$
Did you mean to assume that $A$ is local in your lemma? I don't think it's true otherwise. Or maybe you wanted to say that if the set of non-units is an ideal, then it is the unique maximal ideal.
$endgroup$
– André 3000
Mar 13 at 18:38
add a comment |
$begingroup$
Hartshorne is making use of the following commutative algebra characterisation of maximal ideals.
Lemma 1. Let $A$ be a commutative ring with unity. Then the set of elements which are not invertible is a maximal ideal in $A$.
In your case, you have that the group of invertible germs is
$$mathscrU(O_P) :=[f]in O_P : f(P)neq 0 $$
and this follows from Hartshorne's observation: if $f(P)neq 0$ and $f$ is regular around $P$, then $1/f$ is so and $[1/f]in O_P$ as germ; thus such germs have inverses.
Hence $mathfrak m = A setminus mathscrU(A) = [f]in O_P : f(P) = 0$ is a maximal ideal.
Now such a ring must be local: if there were another maximal ideal $mathfrak m^prime$ and if $mathfrak m^prime neq mathfrak m$ then there would exist $[f]in mathfrak m^prime notin mathfrak m$, so $[f]$ is invertible in $O_P$ and so $mathfrak m^prime = O_P$.
answered Mar 13 at 11:45
CaligulaCaligula
1,241416
$endgroup$
Hartshorne is making use of the following commutative algebra characterisation of maximal ideals.
Lemma 1. Let $A$ be a commutative ring with unity. Then the set of elements which are not invertible is a maximal ideal in $A$.
In your case, you have that the group of invertible germs is
$$mathscrU(O_P) :=[f]in O_P : f(P)neq 0 $$
and this follows from Hartshorne's observation: if $f(P)neq 0$ and $f$ is regular around $P$, then $1/f$ is so and $[1/f]in O_P$ as germ; thus such germs have inverses.
Hence $mathfrak m = A setminus mathscrU(A) = [f]in O_P : f(P) = 0$ is a maximal ideal.
Now such a ring must be local: if there were another maximal ideal $mathfrak m^prime$ and if $mathfrak m^prime neq mathfrak m$ then there would exist $[f]in mathfrak m^prime notin mathfrak m$, so $[f]$ is invertible in $O_P$ and so $mathfrak m^prime = O_P$.
answered Mar 13 at 11:45
CaligulaCaligula
1,241416
answered Mar 13 at 11:45
CaligulaCaligula
1,241416
answered Mar 13 at 11:45
CaligulaCaligula
1,241416
answered Mar 13 at 11:45
CaligulaCaligula
1,241416
1,241416
1
$begingroup$
Did you mean to assume that $A$ is local in your lemma? I don't think it's true otherwise. Or maybe you wanted to say that if the set of non-units is an ideal, then it is the unique maximal ideal.
$endgroup$
– André 3000
Mar 13 at 18:38
add a comment |
1
$begingroup$
Did you mean to assume that $A$ is local in your lemma? I don't think it's true otherwise. Or maybe you wanted to say that if the set of non-units is an ideal, then it is the unique maximal ideal.
$endgroup$
– André 3000
Mar 13 at 18:38
1
1
$begingroup$
Did you mean to assume that $A$ is local in your lemma? I don't think it's true otherwise. Or maybe you wanted to say that if the set of non-units is an ideal, then it is the unique maximal ideal.
$endgroup$
– André 3000
Mar 13 at 18:38
$begingroup$
Did you mean to assume that $A$ is local in your lemma? I don't think it's true otherwise. Or maybe you wanted to say that if the set of non-units is an ideal, then it is the unique maximal ideal.
$endgroup$
– André 3000
Mar 13 at 18:38
add a comment |
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The point is that $m$ is an ideal and every other element not in $m$ is invertible. So $m$ is maximal. Cheers.
$endgroup$
– Wuestenfux
Mar 13 at 9:08