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Question about a Variety as a Representable Functor
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Let $k$ be a unital commutative ring. Let $(X_alpha)_alpha in J$ be a family of indeterminates, let $I$ be an ideal in the ring $K[ (X_alpha)_alpha in J ],$ and set $A := K[ (X_alpha)_alpha in J ]/ I.$ For each $alpha in J,$ set $x_alpha = X_alpha + I.$
I know that we can define a functor from (commutative, unital) $k$-algebras to Sets by associating a $k$-algebra $R$ with the set of all points $V(R) subseteq R^J$ that vanish when evaluated on each element of $I.$ Moreover, this functor is representable because we can identify $(r_alpha) in V(R)$ with the $k$-algebra map $x_alpha mapsto r_alpha.$ Thus, $mathrmHom_k(A, - ) cong V(-).$
What I was confused about is the following. Suppose that $R$ is a $k$-algebra, and suppose that $(s_alpha) in V(R).$ Suppose also that there exists a unital subring $S$ of $R$ such that $(s_alpha) in S^J$ but such that $S$ is NOT a $k$-algebra. Then how do we speak of $S$-points algebraically? (For example, we could have $k$ is the rationals, $R$ is the reals, and $S$ is the integers.) So, I was wondering, is the correct way to speak of the $S$-points of $V$ algebraically by identifying the $S$-points of $V$ with $mathrmHom_k(A, k otimes_mathbbZ S),$ where we identify $(s_alpha) in V(R) cap S^J$ with the map $x_alpha mapsto 1_k otimes_mathbbZ s_alpha$?
algebraic-geometry commutative-algebra
asked Jul 30 at 20:37
Mishel Skenderi
1398
add a comment |Â
up vote
0
down vote
favorite
Let $k$ be a unital commutative ring. Let $(X_alpha)_alpha in J$ be a family of indeterminates, let $I$ be an ideal in the ring $K[ (X_alpha)_alpha in J ],$ and set $A := K[ (X_alpha)_alpha in J ]/ I.$ For each $alpha in J,$ set $x_alpha = X_alpha + I.$
I know that we can define a functor from (commutative, unital) $k$-algebras to Sets by associating a $k$-algebra $R$ with the set of all points $V(R) subseteq R^J$ that vanish when evaluated on each element of $I.$ Moreover, this functor is representable because we can identify $(r_alpha) in V(R)$ with the $k$-algebra map $x_alpha mapsto r_alpha.$ Thus, $mathrmHom_k(A, - ) cong V(-).$
What I was confused about is the following. Suppose that $R$ is a $k$-algebra, and suppose that $(s_alpha) in V(R).$ Suppose also that there exists a unital subring $S$ of $R$ such that $(s_alpha) in S^J$ but such that $S$ is NOT a $k$-algebra. Then how do we speak of $S$-points algebraically? (For example, we could have $k$ is the rationals, $R$ is the reals, and $S$ is the integers.) So, I was wondering, is the correct way to speak of the $S$-points of $V$ algebraically by identifying the $S$-points of $V$ with $mathrmHom_k(A, k otimes_mathbbZ S),$ where we identify $(s_alpha) in V(R) cap S^J$ with the map $x_alpha mapsto 1_k otimes_mathbbZ s_alpha$?
algebraic-geometry commutative-algebra
asked Jul 30 at 20:37
Mishel Skenderi
1398
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $k$ be a unital commutative ring. Let $(X_alpha)_alpha in J$ be a family of indeterminates, let $I$ be an ideal in the ring $K[ (X_alpha)_alpha in J ],$ and set $A := K[ (X_alpha)_alpha in J ]/ I.$ For each $alpha in J,$ set $x_alpha = X_alpha + I.$
I know that we can define a functor from (commutative, unital) $k$-algebras to Sets by associating a $k$-algebra $R$ with the set of all points $V(R) subseteq R^J$ that vanish when evaluated on each element of $I.$ Moreover, this functor is representable because we can identify $(r_alpha) in V(R)$ with the $k$-algebra map $x_alpha mapsto r_alpha.$ Thus, $mathrmHom_k(A, - ) cong V(-).$
What I was confused about is the following. Suppose that $R$ is a $k$-algebra, and suppose that $(s_alpha) in V(R).$ Suppose also that there exists a unital subring $S$ of $R$ such that $(s_alpha) in S^J$ but such that $S$ is NOT a $k$-algebra. Then how do we speak of $S$-points algebraically? (For example, we could have $k$ is the rationals, $R$ is the reals, and $S$ is the integers.) So, I was wondering, is the correct way to speak of the $S$-points of $V$ algebraically by identifying the $S$-points of $V$ with $mathrmHom_k(A, k otimes_mathbbZ S),$ where we identify $(s_alpha) in V(R) cap S^J$ with the map $x_alpha mapsto 1_k otimes_mathbbZ s_alpha$?
algebraic-geometry commutative-algebra
asked Jul 30 at 20:37
Mishel Skenderi
1398
Let $k$ be a unital commutative ring. Let $(X_alpha)_alpha in J$ be a family of indeterminates, let $I$ be an ideal in the ring $K[ (X_alpha)_alpha in J ],$ and set $A := K[ (X_alpha)_alpha in J ]/ I.$ For each $alpha in J,$ set $x_alpha = X_alpha + I.$
I know that we can define a functor from (commutative, unital) $k$-algebras to Sets by associating a $k$-algebra $R$ with the set of all points $V(R) subseteq R^J$ that vanish when evaluated on each element of $I.$ Moreover, this functor is representable because we can identify $(r_alpha) in V(R)$ with the $k$-algebra map $x_alpha mapsto r_alpha.$ Thus, $mathrmHom_k(A, - ) cong V(-).$
What I was confused about is the following. Suppose that $R$ is a $k$-algebra, and suppose that $(s_alpha) in V(R).$ Suppose also that there exists a unital subring $S$ of $R$ such that $(s_alpha) in S^J$ but such that $S$ is NOT a $k$-algebra. Then how do we speak of $S$-points algebraically? (For example, we could have $k$ is the rationals, $R$ is the reals, and $S$ is the integers.) So, I was wondering, is the correct way to speak of the $S$-points of $V$ algebraically by identifying the $S$-points of $V$ with $mathrmHom_k(A, k otimes_mathbbZ S),$ where we identify $(s_alpha) in V(R) cap S^J$ with the map $x_alpha mapsto 1_k otimes_mathbbZ s_alpha$?
algebraic-geometry commutative-algebra
asked Jul 30 at 20:37
Mishel Skenderi
1398
asked Jul 30 at 20:37
Mishel Skenderi
1398
asked Jul 30 at 20:37
Mishel Skenderi
1398
asked Jul 30 at 20:37
Mishel Skenderi
1398
1398
add a comment |Â
add a comment |Â
1 Answer
1
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up vote
1
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The correct way to speak about $S$-points is to not speak about them at all. Notice in particular that the condition that some $(s_alpha)in S^J$ is in $V(R)$ depends on how $S$ is embedded in a $k$-algebra $R$, since without such an embedding we cannot evaluate elements of $I$ at the $s_alpha$. So you could embed $S$ into $k$-algebras in two different ways and get different sets of $S$-points. Moreover, this notion of $S$-points also depends on the presentation chosen for the algebra $A$, since you are requiring only the generators $x_alpha$ to map to $S$ (this is not an issue if $S$ is a $k$-algebra since then if the generators map to a $k$-algebra so does all of $A$).
answered Jul 30 at 20:54
Eric Wofsey
162k12188297
This is a bit disappointing, but I guess we can still consider special cases. In the example that I mentioned above, would my description of $S$-points accurately capture the notion of integer points in the usual naïve sense?
– Mishel Skenderi
Jul 30 at 21:01
You mean $mathrmHom_k(A, k otimes_mathbbZ S)$? No, that doesn't work. For example, when $A$ is just $mathbbQ[X]$ (with $I=0$), then the integer points should be just $mathbbZ$ (the homomorphisms $mathbbQ[X]tomathbbR$ which send $X$ to an integer), but your proposal would give all of $mathbbQ$. This is related to the second issue I mentioned--the "$S$-points" depend on the chosen presentation of $A$, and so you won't be able to get a correct definition without referring to the presentation.
– Eric Wofsey
Jul 30 at 21:08
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The correct way to speak about $S$-points is to not speak about them at all. Notice in particular that the condition that some $(s_alpha)in S^J$ is in $V(R)$ depends on how $S$ is embedded in a $k$-algebra $R$, since without such an embedding we cannot evaluate elements of $I$ at the $s_alpha$. So you could embed $S$ into $k$-algebras in two different ways and get different sets of $S$-points. Moreover, this notion of $S$-points also depends on the presentation chosen for the algebra $A$, since you are requiring only the generators $x_alpha$ to map to $S$ (this is not an issue if $S$ is a $k$-algebra since then if the generators map to a $k$-algebra so does all of $A$).
answered Jul 30 at 20:54
Eric Wofsey
162k12188297
This is a bit disappointing, but I guess we can still consider special cases. In the example that I mentioned above, would my description of $S$-points accurately capture the notion of integer points in the usual naïve sense?
– Mishel Skenderi
Jul 30 at 21:01
You mean $mathrmHom_k(A, k otimes_mathbbZ S)$? No, that doesn't work. For example, when $A$ is just $mathbbQ[X]$ (with $I=0$), then the integer points should be just $mathbbZ$ (the homomorphisms $mathbbQ[X]tomathbbR$ which send $X$ to an integer), but your proposal would give all of $mathbbQ$. This is related to the second issue I mentioned--the "$S$-points" depend on the chosen presentation of $A$, and so you won't be able to get a correct definition without referring to the presentation.
– Eric Wofsey
Jul 30 at 21:08
add a comment |Â
up vote
1
down vote
accepted
The correct way to speak about $S$-points is to not speak about them at all. Notice in particular that the condition that some $(s_alpha)in S^J$ is in $V(R)$ depends on how $S$ is embedded in a $k$-algebra $R$, since without such an embedding we cannot evaluate elements of $I$ at the $s_alpha$. So you could embed $S$ into $k$-algebras in two different ways and get different sets of $S$-points. Moreover, this notion of $S$-points also depends on the presentation chosen for the algebra $A$, since you are requiring only the generators $x_alpha$ to map to $S$ (this is not an issue if $S$ is a $k$-algebra since then if the generators map to a $k$-algebra so does all of $A$).
answered Jul 30 at 20:54
Eric Wofsey
162k12188297
This is a bit disappointing, but I guess we can still consider special cases. In the example that I mentioned above, would my description of $S$-points accurately capture the notion of integer points in the usual naïve sense?
– Mishel Skenderi
Jul 30 at 21:01
You mean $mathrmHom_k(A, k otimes_mathbbZ S)$? No, that doesn't work. For example, when $A$ is just $mathbbQ[X]$ (with $I=0$), then the integer points should be just $mathbbZ$ (the homomorphisms $mathbbQ[X]tomathbbR$ which send $X$ to an integer), but your proposal would give all of $mathbbQ$. This is related to the second issue I mentioned--the "$S$-points" depend on the chosen presentation of $A$, and so you won't be able to get a correct definition without referring to the presentation.
– Eric Wofsey
Jul 30 at 21:08
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The correct way to speak about $S$-points is to not speak about them at all. Notice in particular that the condition that some $(s_alpha)in S^J$ is in $V(R)$ depends on how $S$ is embedded in a $k$-algebra $R$, since without such an embedding we cannot evaluate elements of $I$ at the $s_alpha$. So you could embed $S$ into $k$-algebras in two different ways and get different sets of $S$-points. Moreover, this notion of $S$-points also depends on the presentation chosen for the algebra $A$, since you are requiring only the generators $x_alpha$ to map to $S$ (this is not an issue if $S$ is a $k$-algebra since then if the generators map to a $k$-algebra so does all of $A$).
answered Jul 30 at 20:54
Eric Wofsey
162k12188297
The correct way to speak about $S$-points is to not speak about them at all. Notice in particular that the condition that some $(s_alpha)in S^J$ is in $V(R)$ depends on how $S$ is embedded in a $k$-algebra $R$, since without such an embedding we cannot evaluate elements of $I$ at the $s_alpha$. So you could embed $S$ into $k$-algebras in two different ways and get different sets of $S$-points. Moreover, this notion of $S$-points also depends on the presentation chosen for the algebra $A$, since you are requiring only the generators $x_alpha$ to map to $S$ (this is not an issue if $S$ is a $k$-algebra since then if the generators map to a $k$-algebra so does all of $A$).
answered Jul 30 at 20:54
Eric Wofsey
162k12188297
answered Jul 30 at 20:54
Eric Wofsey
162k12188297
answered Jul 30 at 20:54
Eric Wofsey
162k12188297
answered Jul 30 at 20:54
Eric Wofsey
162k12188297
162k12188297
This is a bit disappointing, but I guess we can still consider special cases. In the example that I mentioned above, would my description of $S$-points accurately capture the notion of integer points in the usual naïve sense?
– Mishel Skenderi
Jul 30 at 21:01
You mean $mathrmHom_k(A, k otimes_mathbbZ S)$? No, that doesn't work. For example, when $A$ is just $mathbbQ[X]$ (with $I=0$), then the integer points should be just $mathbbZ$ (the homomorphisms $mathbbQ[X]tomathbbR$ which send $X$ to an integer), but your proposal would give all of $mathbbQ$. This is related to the second issue I mentioned--the "$S$-points" depend on the chosen presentation of $A$, and so you won't be able to get a correct definition without referring to the presentation.
– Eric Wofsey
Jul 30 at 21:08
add a comment |Â
This is a bit disappointing, but I guess we can still consider special cases. In the example that I mentioned above, would my description of $S$-points accurately capture the notion of integer points in the usual naïve sense?
– Mishel Skenderi
Jul 30 at 21:01
You mean $mathrmHom_k(A, k otimes_mathbbZ S)$? No, that doesn't work. For example, when $A$ is just $mathbbQ[X]$ (with $I=0$), then the integer points should be just $mathbbZ$ (the homomorphisms $mathbbQ[X]tomathbbR$ which send $X$ to an integer), but your proposal would give all of $mathbbQ$. This is related to the second issue I mentioned--the "$S$-points" depend on the chosen presentation of $A$, and so you won't be able to get a correct definition without referring to the presentation.
– Eric Wofsey
Jul 30 at 21:08
This is a bit disappointing, but I guess we can still consider special cases. In the example that I mentioned above, would my description of $S$-points accurately capture the notion of integer points in the usual naïve sense?
– Mishel Skenderi
Jul 30 at 21:01
This is a bit disappointing, but I guess we can still consider special cases. In the example that I mentioned above, would my description of $S$-points accurately capture the notion of integer points in the usual naïve sense?
– Mishel Skenderi
Jul 30 at 21:01
You mean $mathrmHom_k(A, k otimes_mathbbZ S)$? No, that doesn't work. For example, when $A$ is just $mathbbQ[X]$ (with $I=0$), then the integer points should be just $mathbbZ$ (the homomorphisms $mathbbQ[X]tomathbbR$ which send $X$ to an integer), but your proposal would give all of $mathbbQ$. This is related to the second issue I mentioned--the "$S$-points" depend on the chosen presentation of $A$, and so you won't be able to get a correct definition without referring to the presentation.
– Eric Wofsey
Jul 30 at 21:08
You mean $mathrmHom_k(A, k otimes_mathbbZ S)$? No, that doesn't work. For example, when $A$ is just $mathbbQ[X]$ (with $I=0$), then the integer points should be just $mathbbZ$ (the homomorphisms $mathbbQ[X]tomathbbR$ which send $X$ to an integer), but your proposal would give all of $mathbbQ$. This is related to the second issue I mentioned--the "$S$-points" depend on the chosen presentation of $A$, and so you won't be able to get a correct definition without referring to the presentation.
– Eric Wofsey
Jul 30 at 21:08
add a comment |Â
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