Question about the ring $A_P$ of regular functions on a variety $V$
$begingroup$
Note: I am doing an introductory course in Commutative Algebra and not Algebraic Geometry; This is just a quick application I am trying to understand. Also, rings are commutative with $1$.
When discussing localizations $S^{-1}A$ for a ring $A$ and $Ssubset A$, we made the following example.
Def. Let $k$ be a field, and $P$ be a prime ideal in $A=k[X_1,dots, X_n]$. Let $V={(a_1,dots,a_n)in k^n:f(a_1,dots,a_n)=0, forall fin P}$ be the variety of $P$, and let $S=Asetminus P$.
Then $S^{-1}A$ is denoted $A_P$, and we have $A_P={dfrac{f}{g}:fin A,gnotin P}$. In class, this was called the ring of regular functions on $V$ (EDIT: I learned this is apparently not how the ring of regular functions on V is defined, see the comments). These are apparently well defined rational functions $k^nto k$ since $g$ does not vanish on $V$, i.e. it seems we have
$gnotin PRightarrow g(a_1,dots a_n)not =0$ for all $(a_1,dots,a_n)in V$.
But, I think it's also evident that we have
$gin PRightarrow g(a_1,dots,a_n)=0$ for all $(a_1,dots,a_n)in V$
which is logically equivalent to $(gnotin P) vee (g(a_1,dots,a_n)=0$ for all $(a_1,dots,a_n)in V)$.
It just feels like I'm missing something obvious, since I can't seem to deduce a contradiction from $gnotin P$, and $g(a_1,dots,a_n)=0$ for some point of $V$.
Question: Why couldn't we have $gnotin P$, but still have $g(a_1,dots,a_n)=0$, for some $(a_1,dots,a_n)in V$?
abstract-algebra ring-theory commutative-algebra localization
asked Dec 12 '18 at 16:28
Christopher.LChristopher.L
7471317
$endgroup$
add a comment |
$begingroup$
Note: I am doing an introductory course in Commutative Algebra and not Algebraic Geometry; This is just a quick application I am trying to understand. Also, rings are commutative with $1$.
When discussing localizations $S^{-1}A$ for a ring $A$ and $Ssubset A$, we made the following example.
Def. Let $k$ be a field, and $P$ be a prime ideal in $A=k[X_1,dots, X_n]$. Let $V={(a_1,dots,a_n)in k^n:f(a_1,dots,a_n)=0, forall fin P}$ be the variety of $P$, and let $S=Asetminus P$.
Then $S^{-1}A$ is denoted $A_P$, and we have $A_P={dfrac{f}{g}:fin A,gnotin P}$. In class, this was called the ring of regular functions on $V$ (EDIT: I learned this is apparently not how the ring of regular functions on V is defined, see the comments). These are apparently well defined rational functions $k^nto k$ since $g$ does not vanish on $V$, i.e. it seems we have
$gnotin PRightarrow g(a_1,dots a_n)not =0$ for all $(a_1,dots,a_n)in V$.
But, I think it's also evident that we have
$gin PRightarrow g(a_1,dots,a_n)=0$ for all $(a_1,dots,a_n)in V$
which is logically equivalent to $(gnotin P) vee (g(a_1,dots,a_n)=0$ for all $(a_1,dots,a_n)in V)$.
It just feels like I'm missing something obvious, since I can't seem to deduce a contradiction from $gnotin P$, and $g(a_1,dots,a_n)=0$ for some point of $V$.
Question: Why couldn't we have $gnotin P$, but still have $g(a_1,dots,a_n)=0$, for some $(a_1,dots,a_n)in V$?
abstract-algebra ring-theory commutative-algebra localization
asked Dec 12 '18 at 16:28
Christopher.LChristopher.L
7471317
$endgroup$
$begingroup$
This is not how I (or a number of others) would define the ring of regular functions on $V = V(mathfrak{P}).$ I would call $k[V] :=A/mathfrak{P}$ the ring of regular functions, and if $p = (a_1,dots, a_n)in V$ is a point on your variety, I would call $k[V]_{mathfrak{m}_p}$ the ring of regular functions at $p$ ($mathfrak{m}_p$ is the maximal ideal of functions vanishing at the point $p$). These are the functions which do not vanish in some [unspecified] neighborhood around $p.$
$endgroup$
– Stahl
Dec 12 '18 at 17:25
$begingroup$
Notice that $k[V]$ also consists of well defined functions on $V,$ as if $pin V$ and $f$ and $g$ represent the same element of $k[V],$ then $f - ginmathfrak{P},$ so that $f(p) - g(p) = 0.$
$endgroup$
– Stahl
Dec 12 '18 at 17:28
$begingroup$
As others have said $A_P$ is not really the ring of regular functions on $V$. Rather, it is the ring of rational functions on affine space $mathbb{A}^n$ that are regular on the subvariety $V$.
$endgroup$
– André 3000
Dec 12 '18 at 20:04
$begingroup$
Ok, this was just what it was explicitly called by the professor, but I also guess he may have been a bit informal and quick there, since it was just an example. I edited in a comment about this into my answer.
$endgroup$
– Christopher.L
Dec 13 '18 at 9:20
add a comment |
$begingroup$
Note: I am doing an introductory course in Commutative Algebra and not Algebraic Geometry; This is just a quick application I am trying to understand. Also, rings are commutative with $1$.
When discussing localizations $S^{-1}A$ for a ring $A$ and $Ssubset A$, we made the following example.
Def. Let $k$ be a field, and $P$ be a prime ideal in $A=k[X_1,dots, X_n]$. Let $V={(a_1,dots,a_n)in k^n:f(a_1,dots,a_n)=0, forall fin P}$ be the variety of $P$, and let $S=Asetminus P$.
Then $S^{-1}A$ is denoted $A_P$, and we have $A_P={dfrac{f}{g}:fin A,gnotin P}$. In class, this was called the ring of regular functions on $V$ (EDIT: I learned this is apparently not how the ring of regular functions on V is defined, see the comments). These are apparently well defined rational functions $k^nto k$ since $g$ does not vanish on $V$, i.e. it seems we have
$gnotin PRightarrow g(a_1,dots a_n)not =0$ for all $(a_1,dots,a_n)in V$.
But, I think it's also evident that we have
$gin PRightarrow g(a_1,dots,a_n)=0$ for all $(a_1,dots,a_n)in V$
which is logically equivalent to $(gnotin P) vee (g(a_1,dots,a_n)=0$ for all $(a_1,dots,a_n)in V)$.
It just feels like I'm missing something obvious, since I can't seem to deduce a contradiction from $gnotin P$, and $g(a_1,dots,a_n)=0$ for some point of $V$.
Question: Why couldn't we have $gnotin P$, but still have $g(a_1,dots,a_n)=0$, for some $(a_1,dots,a_n)in V$?
abstract-algebra ring-theory commutative-algebra localization
asked Dec 12 '18 at 16:28
Christopher.LChristopher.L
7471317
$endgroup$
Note: I am doing an introductory course in Commutative Algebra and not Algebraic Geometry; This is just a quick application I am trying to understand. Also, rings are commutative with $1$.
When discussing localizations $S^{-1}A$ for a ring $A$ and $Ssubset A$, we made the following example.
Def. Let $k$ be a field, and $P$ be a prime ideal in $A=k[X_1,dots, X_n]$. Let $V={(a_1,dots,a_n)in k^n:f(a_1,dots,a_n)=0, forall fin P}$ be the variety of $P$, and let $S=Asetminus P$.
Then $S^{-1}A$ is denoted $A_P$, and we have $A_P={dfrac{f}{g}:fin A,gnotin P}$. In class, this was called the ring of regular functions on $V$ (EDIT: I learned this is apparently not how the ring of regular functions on V is defined, see the comments). These are apparently well defined rational functions $k^nto k$ since $g$ does not vanish on $V$, i.e. it seems we have
$gnotin PRightarrow g(a_1,dots a_n)not =0$ for all $(a_1,dots,a_n)in V$.
But, I think it's also evident that we have
$gin PRightarrow g(a_1,dots,a_n)=0$ for all $(a_1,dots,a_n)in V$
which is logically equivalent to $(gnotin P) vee (g(a_1,dots,a_n)=0$ for all $(a_1,dots,a_n)in V)$.
It just feels like I'm missing something obvious, since I can't seem to deduce a contradiction from $gnotin P$, and $g(a_1,dots,a_n)=0$ for some point of $V$.
Question: Why couldn't we have $gnotin P$, but still have $g(a_1,dots,a_n)=0$, for some $(a_1,dots,a_n)in V$?
abstract-algebra ring-theory commutative-algebra localization
abstract-algebra ring-theory commutative-algebra localization
asked Dec 12 '18 at 16:28
Christopher.LChristopher.L
7471317
asked Dec 12 '18 at 16:28
Christopher.LChristopher.L
7471317
edited Dec 13 '18 at 9:17
Christopher.L
asked Dec 12 '18 at 16:28
Christopher.LChristopher.L
7471317
asked Dec 12 '18 at 16:28
Christopher.LChristopher.L
7471317
asked Dec 12 '18 at 16:28
Christopher.LChristopher.L
7471317
7471317
$begingroup$
This is not how I (or a number of others) would define the ring of regular functions on $V = V(mathfrak{P}).$ I would call $k[V] :=A/mathfrak{P}$ the ring of regular functions, and if $p = (a_1,dots, a_n)in V$ is a point on your variety, I would call $k[V]_{mathfrak{m}_p}$ the ring of regular functions at $p$ ($mathfrak{m}_p$ is the maximal ideal of functions vanishing at the point $p$). These are the functions which do not vanish in some [unspecified] neighborhood around $p.$
$endgroup$
– Stahl
Dec 12 '18 at 17:25
$begingroup$
Notice that $k[V]$ also consists of well defined functions on $V,$ as if $pin V$ and $f$ and $g$ represent the same element of $k[V],$ then $f - ginmathfrak{P},$ so that $f(p) - g(p) = 0.$
$endgroup$
– Stahl
Dec 12 '18 at 17:28
$begingroup$
As others have said $A_P$ is not really the ring of regular functions on $V$. Rather, it is the ring of rational functions on affine space $mathbb{A}^n$ that are regular on the subvariety $V$.
$endgroup$
– André 3000
Dec 12 '18 at 20:04
$begingroup$
Ok, this was just what it was explicitly called by the professor, but I also guess he may have been a bit informal and quick there, since it was just an example. I edited in a comment about this into my answer.
$endgroup$
– Christopher.L
Dec 13 '18 at 9:20
add a comment |
$begingroup$
This is not how I (or a number of others) would define the ring of regular functions on $V = V(mathfrak{P}).$ I would call $k[V] :=A/mathfrak{P}$ the ring of regular functions, and if $p = (a_1,dots, a_n)in V$ is a point on your variety, I would call $k[V]_{mathfrak{m}_p}$ the ring of regular functions at $p$ ($mathfrak{m}_p$ is the maximal ideal of functions vanishing at the point $p$). These are the functions which do not vanish in some [unspecified] neighborhood around $p.$
$endgroup$
– Stahl
Dec 12 '18 at 17:25
$begingroup$
Notice that $k[V]$ also consists of well defined functions on $V,$ as if $pin V$ and $f$ and $g$ represent the same element of $k[V],$ then $f - ginmathfrak{P},$ so that $f(p) - g(p) = 0.$
$endgroup$
– Stahl
Dec 12 '18 at 17:28
$begingroup$
As others have said $A_P$ is not really the ring of regular functions on $V$. Rather, it is the ring of rational functions on affine space $mathbb{A}^n$ that are regular on the subvariety $V$.
$endgroup$
– André 3000
Dec 12 '18 at 20:04
$begingroup$
Ok, this was just what it was explicitly called by the professor, but I also guess he may have been a bit informal and quick there, since it was just an example. I edited in a comment about this into my answer.
$endgroup$
– Christopher.L
Dec 13 '18 at 9:20
$begingroup$
This is not how I (or a number of others) would define the ring of regular functions on $V = V(mathfrak{P}).$ I would call $k[V] :=A/mathfrak{P}$ the ring of regular functions, and if $p = (a_1,dots, a_n)in V$ is a point on your variety, I would call $k[V]_{mathfrak{m}_p}$ the ring of regular functions at $p$ ($mathfrak{m}_p$ is the maximal ideal of functions vanishing at the point $p$). These are the functions which do not vanish in some [unspecified] neighborhood around $p.$
$endgroup$
– Stahl
Dec 12 '18 at 17:25
$begingroup$
This is not how I (or a number of others) would define the ring of regular functions on $V = V(mathfrak{P}).$ I would call $k[V] :=A/mathfrak{P}$ the ring of regular functions, and if $p = (a_1,dots, a_n)in V$ is a point on your variety, I would call $k[V]_{mathfrak{m}_p}$ the ring of regular functions at $p$ ($mathfrak{m}_p$ is the maximal ideal of functions vanishing at the point $p$). These are the functions which do not vanish in some [unspecified] neighborhood around $p.$
$endgroup$
– Stahl
Dec 12 '18 at 17:25
$begingroup$
Notice that $k[V]$ also consists of well defined functions on $V,$ as if $pin V$ and $f$ and $g$ represent the same element of $k[V],$ then $f - ginmathfrak{P},$ so that $f(p) - g(p) = 0.$
$endgroup$
– Stahl
Dec 12 '18 at 17:28
$begingroup$
Notice that $k[V]$ also consists of well defined functions on $V,$ as if $pin V$ and $f$ and $g$ represent the same element of $k[V],$ then $f - ginmathfrak{P},$ so that $f(p) - g(p) = 0.$
$endgroup$
– Stahl
Dec 12 '18 at 17:28
$begingroup$
As others have said $A_P$ is not really the ring of regular functions on $V$. Rather, it is the ring of rational functions on affine space $mathbb{A}^n$ that are regular on the subvariety $V$.
$endgroup$
– André 3000
Dec 12 '18 at 20:04
$begingroup$
As others have said $A_P$ is not really the ring of regular functions on $V$. Rather, it is the ring of rational functions on affine space $mathbb{A}^n$ that are regular on the subvariety $V$.
$endgroup$
– André 3000
Dec 12 '18 at 20:04
$begingroup$
Ok, this was just what it was explicitly called by the professor, but I also guess he may have been a bit informal and quick there, since it was just an example. I edited in a comment about this into my answer.
$endgroup$
– Christopher.L
Dec 13 '18 at 9:20
$begingroup$
Ok, this was just what it was explicitly called by the professor, but I also guess he may have been a bit informal and quick there, since it was just an example. I edited in a comment about this into my answer.
$endgroup$
– Christopher.L
Dec 13 '18 at 9:20
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We absolutely can have $gnotin P$ and $g(a_1,dots,a_n)=0$ for some $(a_1,dots,a_n)in V$.
Take $n=2$ and the prime ideal $P=(x)subset k[x,y]$, so that $V={(a,b)in k^2mid a=0}$. The function $g=y$ will vanish at the point $(0,0)in V$, but it will not vanish at any other point $(0,b)$ with $bneq0$.
answered Dec 12 '18 at 16:44
Alex MathersAlex Mathers
10.9k21344
$endgroup$
$begingroup$
I think I got a bit confused by a statement in the lectures sayin "If $gnotin P$, then $g$ is non-vanishing on V". So, that is incorrect I suppose? Can we technically even have $g(a)=0$, for all $ain V$ then?
$endgroup$
– Christopher.L
Dec 12 '18 at 16:56
$begingroup$
@Christopher.L You would typically say $g$ is "vanishing on $V$" whenever $g(a)=0$ for all points $ain V$, which is by definition if and only if $gin P$. Your professor probably then meant by "non-vanishing" that $g$ does not vanish on at least one point of $V$.
$endgroup$
– Alex Mathers
Dec 12 '18 at 17:00
$begingroup$
Ok, but perhaps Im misunderstanding something: As I've defined V above, I only say that $gin P$ implies $g(a)=0$ for all $ain V$, right? I dont say the converse. So, as I said, it feels we could have g vanishing on V.
$endgroup$
– Christopher.L
Dec 12 '18 at 17:09
$begingroup$
@Christopher.L yes you are right, the converse is not obvious a priori. However, this is actually true, and will likely be proved soon in your class as a theorem (sometimes called the Nullstellensatz).
$endgroup$
– Alex Mathers
Dec 12 '18 at 17:30
$begingroup$
Ah, ok, I've only heard the name of the theorem, have not gotten there yet; but I think I understand it better. And perhaps the professor took a little bit too much for granted since it was however just a quick application.
$endgroup$
– Christopher.L
Dec 12 '18 at 17:31
add a comment |
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$begingroup$
We absolutely can have $gnotin P$ and $g(a_1,dots,a_n)=0$ for some $(a_1,dots,a_n)in V$.
Take $n=2$ and the prime ideal $P=(x)subset k[x,y]$, so that $V={(a,b)in k^2mid a=0}$. The function $g=y$ will vanish at the point $(0,0)in V$, but it will not vanish at any other point $(0,b)$ with $bneq0$.
answered Dec 12 '18 at 16:44
Alex MathersAlex Mathers
10.9k21344
$endgroup$
$begingroup$
I think I got a bit confused by a statement in the lectures sayin "If $gnotin P$, then $g$ is non-vanishing on V". So, that is incorrect I suppose? Can we technically even have $g(a)=0$, for all $ain V$ then?
$endgroup$
– Christopher.L
Dec 12 '18 at 16:56
$begingroup$
@Christopher.L You would typically say $g$ is "vanishing on $V$" whenever $g(a)=0$ for all points $ain V$, which is by definition if and only if $gin P$. Your professor probably then meant by "non-vanishing" that $g$ does not vanish on at least one point of $V$.
$endgroup$
– Alex Mathers
Dec 12 '18 at 17:00
$begingroup$
Ok, but perhaps Im misunderstanding something: As I've defined V above, I only say that $gin P$ implies $g(a)=0$ for all $ain V$, right? I dont say the converse. So, as I said, it feels we could have g vanishing on V.
$endgroup$
– Christopher.L
Dec 12 '18 at 17:09
$begingroup$
@Christopher.L yes you are right, the converse is not obvious a priori. However, this is actually true, and will likely be proved soon in your class as a theorem (sometimes called the Nullstellensatz).
$endgroup$
– Alex Mathers
Dec 12 '18 at 17:30
$begingroup$
Ah, ok, I've only heard the name of the theorem, have not gotten there yet; but I think I understand it better. And perhaps the professor took a little bit too much for granted since it was however just a quick application.
$endgroup$
– Christopher.L
Dec 12 '18 at 17:31
add a comment |
$begingroup$
We absolutely can have $gnotin P$ and $g(a_1,dots,a_n)=0$ for some $(a_1,dots,a_n)in V$.
Take $n=2$ and the prime ideal $P=(x)subset k[x,y]$, so that $V={(a,b)in k^2mid a=0}$. The function $g=y$ will vanish at the point $(0,0)in V$, but it will not vanish at any other point $(0,b)$ with $bneq0$.
answered Dec 12 '18 at 16:44
Alex MathersAlex Mathers
10.9k21344
$endgroup$
$begingroup$
I think I got a bit confused by a statement in the lectures sayin "If $gnotin P$, then $g$ is non-vanishing on V". So, that is incorrect I suppose? Can we technically even have $g(a)=0$, for all $ain V$ then?
$endgroup$
– Christopher.L
Dec 12 '18 at 16:56
$begingroup$
@Christopher.L You would typically say $g$ is "vanishing on $V$" whenever $g(a)=0$ for all points $ain V$, which is by definition if and only if $gin P$. Your professor probably then meant by "non-vanishing" that $g$ does not vanish on at least one point of $V$.
$endgroup$
– Alex Mathers
Dec 12 '18 at 17:00
$begingroup$
Ok, but perhaps Im misunderstanding something: As I've defined V above, I only say that $gin P$ implies $g(a)=0$ for all $ain V$, right? I dont say the converse. So, as I said, it feels we could have g vanishing on V.
$endgroup$
– Christopher.L
Dec 12 '18 at 17:09
$begingroup$
@Christopher.L yes you are right, the converse is not obvious a priori. However, this is actually true, and will likely be proved soon in your class as a theorem (sometimes called the Nullstellensatz).
$endgroup$
– Alex Mathers
Dec 12 '18 at 17:30
$begingroup$
Ah, ok, I've only heard the name of the theorem, have not gotten there yet; but I think I understand it better. And perhaps the professor took a little bit too much for granted since it was however just a quick application.
$endgroup$
– Christopher.L
Dec 12 '18 at 17:31
add a comment |
$begingroup$
We absolutely can have $gnotin P$ and $g(a_1,dots,a_n)=0$ for some $(a_1,dots,a_n)in V$.
Take $n=2$ and the prime ideal $P=(x)subset k[x,y]$, so that $V={(a,b)in k^2mid a=0}$. The function $g=y$ will vanish at the point $(0,0)in V$, but it will not vanish at any other point $(0,b)$ with $bneq0$.
answered Dec 12 '18 at 16:44
Alex MathersAlex Mathers
10.9k21344
$endgroup$
We absolutely can have $gnotin P$ and $g(a_1,dots,a_n)=0$ for some $(a_1,dots,a_n)in V$.
Take $n=2$ and the prime ideal $P=(x)subset k[x,y]$, so that $V={(a,b)in k^2mid a=0}$. The function $g=y$ will vanish at the point $(0,0)in V$, but it will not vanish at any other point $(0,b)$ with $bneq0$.
answered Dec 12 '18 at 16:44
Alex MathersAlex Mathers
10.9k21344
edited Dec 12 '18 at 16:50
answered Dec 12 '18 at 16:44
Alex MathersAlex Mathers
10.9k21344
answered Dec 12 '18 at 16:44
Alex MathersAlex Mathers
10.9k21344
answered Dec 12 '18 at 16:44
Alex MathersAlex Mathers
10.9k21344
10.9k21344
$begingroup$
I think I got a bit confused by a statement in the lectures sayin "If $gnotin P$, then $g$ is non-vanishing on V". So, that is incorrect I suppose? Can we technically even have $g(a)=0$, for all $ain V$ then?
$endgroup$
– Christopher.L
Dec 12 '18 at 16:56
$begingroup$
@Christopher.L You would typically say $g$ is "vanishing on $V$" whenever $g(a)=0$ for all points $ain V$, which is by definition if and only if $gin P$. Your professor probably then meant by "non-vanishing" that $g$ does not vanish on at least one point of $V$.
$endgroup$
– Alex Mathers
Dec 12 '18 at 17:00
$begingroup$
Ok, but perhaps Im misunderstanding something: As I've defined V above, I only say that $gin P$ implies $g(a)=0$ for all $ain V$, right? I dont say the converse. So, as I said, it feels we could have g vanishing on V.
$endgroup$
– Christopher.L
Dec 12 '18 at 17:09
$begingroup$
@Christopher.L yes you are right, the converse is not obvious a priori. However, this is actually true, and will likely be proved soon in your class as a theorem (sometimes called the Nullstellensatz).
$endgroup$
– Alex Mathers
Dec 12 '18 at 17:30
$begingroup$
Ah, ok, I've only heard the name of the theorem, have not gotten there yet; but I think I understand it better. And perhaps the professor took a little bit too much for granted since it was however just a quick application.
$endgroup$
– Christopher.L
Dec 12 '18 at 17:31
add a comment |
$begingroup$
I think I got a bit confused by a statement in the lectures sayin "If $gnotin P$, then $g$ is non-vanishing on V". So, that is incorrect I suppose? Can we technically even have $g(a)=0$, for all $ain V$ then?
$endgroup$
– Christopher.L
Dec 12 '18 at 16:56
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@Christopher.L You would typically say $g$ is "vanishing on $V$" whenever $g(a)=0$ for all points $ain V$, which is by definition if and only if $gin P$. Your professor probably then meant by "non-vanishing" that $g$ does not vanish on at least one point of $V$.
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– Alex Mathers
Dec 12 '18 at 17:00
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Ok, but perhaps Im misunderstanding something: As I've defined V above, I only say that $gin P$ implies $g(a)=0$ for all $ain V$, right? I dont say the converse. So, as I said, it feels we could have g vanishing on V.
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– Christopher.L
Dec 12 '18 at 17:09
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@Christopher.L yes you are right, the converse is not obvious a priori. However, this is actually true, and will likely be proved soon in your class as a theorem (sometimes called the Nullstellensatz).
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– Alex Mathers
Dec 12 '18 at 17:30
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Ah, ok, I've only heard the name of the theorem, have not gotten there yet; but I think I understand it better. And perhaps the professor took a little bit too much for granted since it was however just a quick application.
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– Christopher.L
Dec 12 '18 at 17:31
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I think I got a bit confused by a statement in the lectures sayin "If $gnotin P$, then $g$ is non-vanishing on V". So, that is incorrect I suppose? Can we technically even have $g(a)=0$, for all $ain V$ then?
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– Christopher.L
Dec 12 '18 at 16:56
$begingroup$
I think I got a bit confused by a statement in the lectures sayin "If $gnotin P$, then $g$ is non-vanishing on V". So, that is incorrect I suppose? Can we technically even have $g(a)=0$, for all $ain V$ then?
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– Christopher.L
Dec 12 '18 at 16:56
$begingroup$
@Christopher.L You would typically say $g$ is "vanishing on $V$" whenever $g(a)=0$ for all points $ain V$, which is by definition if and only if $gin P$. Your professor probably then meant by "non-vanishing" that $g$ does not vanish on at least one point of $V$.
$endgroup$
– Alex Mathers
Dec 12 '18 at 17:00
$begingroup$
@Christopher.L You would typically say $g$ is "vanishing on $V$" whenever $g(a)=0$ for all points $ain V$, which is by definition if and only if $gin P$. Your professor probably then meant by "non-vanishing" that $g$ does not vanish on at least one point of $V$.
$endgroup$
– Alex Mathers
Dec 12 '18 at 17:00
$begingroup$
Ok, but perhaps Im misunderstanding something: As I've defined V above, I only say that $gin P$ implies $g(a)=0$ for all $ain V$, right? I dont say the converse. So, as I said, it feels we could have g vanishing on V.
$endgroup$
– Christopher.L
Dec 12 '18 at 17:09
$begingroup$
Ok, but perhaps Im misunderstanding something: As I've defined V above, I only say that $gin P$ implies $g(a)=0$ for all $ain V$, right? I dont say the converse. So, as I said, it feels we could have g vanishing on V.
$endgroup$
– Christopher.L
Dec 12 '18 at 17:09
$begingroup$
@Christopher.L yes you are right, the converse is not obvious a priori. However, this is actually true, and will likely be proved soon in your class as a theorem (sometimes called the Nullstellensatz).
$endgroup$
– Alex Mathers
Dec 12 '18 at 17:30
$begingroup$
@Christopher.L yes you are right, the converse is not obvious a priori. However, this is actually true, and will likely be proved soon in your class as a theorem (sometimes called the Nullstellensatz).
$endgroup$
– Alex Mathers
Dec 12 '18 at 17:30
$begingroup$
Ah, ok, I've only heard the name of the theorem, have not gotten there yet; but I think I understand it better. And perhaps the professor took a little bit too much for granted since it was however just a quick application.
$endgroup$
– Christopher.L
Dec 12 '18 at 17:31
$begingroup$
Ah, ok, I've only heard the name of the theorem, have not gotten there yet; but I think I understand it better. And perhaps the professor took a little bit too much for granted since it was however just a quick application.
$endgroup$
– Christopher.L
Dec 12 '18 at 17:31
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$begingroup$
This is not how I (or a number of others) would define the ring of regular functions on $V = V(mathfrak{P}).$ I would call $k[V] :=A/mathfrak{P}$ the ring of regular functions, and if $p = (a_1,dots, a_n)in V$ is a point on your variety, I would call $k[V]_{mathfrak{m}_p}$ the ring of regular functions at $p$ ($mathfrak{m}_p$ is the maximal ideal of functions vanishing at the point $p$). These are the functions which do not vanish in some [unspecified] neighborhood around $p.$
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– Stahl
Dec 12 '18 at 17:25
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Notice that $k[V]$ also consists of well defined functions on $V,$ as if $pin V$ and $f$ and $g$ represent the same element of $k[V],$ then $f - ginmathfrak{P},$ so that $f(p) - g(p) = 0.$
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– Stahl
Dec 12 '18 at 17:28
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As others have said $A_P$ is not really the ring of regular functions on $V$. Rather, it is the ring of rational functions on affine space $mathbb{A}^n$ that are regular on the subvariety $V$.
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– André 3000
Dec 12 '18 at 20:04
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Ok, this was just what it was explicitly called by the professor, but I also guess he may have been a bit informal and quick there, since it was just an example. I edited in a comment about this into my answer.
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– Christopher.L
Dec 13 '18 at 9:20