Mixing
Definition of $mathbb{P}_A^n$
$begingroup$
I feel confused with these words: "$B=A[x_0,dots,x_n]$. Then $B_+=(x_0,dots,x_n)$ but this will certainly not give a point on $textrm{Proj }B=mathbb P^n_A$. (The point $(0:cdots:0)$ is not defined)."
What does the point $(0:cdots:0)$ mean in $mathbb P^n_A$? By definition every point in $textrm{Proj }B=mathbb P^n_A$ is a prime ideal which doesn't contain $B_+$ right? How can we get them in the coordinate form?
algebraic-geometry
asked Jan 30 at 5:45
KatherineKatherine
438310
$endgroup$
add a comment |
$begingroup$
I feel confused with these words: "$B=A[x_0,dots,x_n]$. Then $B_+=(x_0,dots,x_n)$ but this will certainly not give a point on $textrm{Proj }B=mathbb P^n_A$. (The point $(0:cdots:0)$ is not defined)."
What does the point $(0:cdots:0)$ mean in $mathbb P^n_A$? By definition every point in $textrm{Proj }B=mathbb P^n_A$ is a prime ideal which doesn't contain $B_+$ right? How can we get them in the coordinate form?
algebraic-geometry
asked Jan 30 at 5:45
KatherineKatherine
438310
$endgroup$
add a comment |
$begingroup$
I feel confused with these words: "$B=A[x_0,dots,x_n]$. Then $B_+=(x_0,dots,x_n)$ but this will certainly not give a point on $textrm{Proj }B=mathbb P^n_A$. (The point $(0:cdots:0)$ is not defined)."
What does the point $(0:cdots:0)$ mean in $mathbb P^n_A$? By definition every point in $textrm{Proj }B=mathbb P^n_A$ is a prime ideal which doesn't contain $B_+$ right? How can we get them in the coordinate form?
algebraic-geometry
asked Jan 30 at 5:45
KatherineKatherine
438310
$endgroup$
I feel confused with these words: "$B=A[x_0,dots,x_n]$. Then $B_+=(x_0,dots,x_n)$ but this will certainly not give a point on $textrm{Proj }B=mathbb P^n_A$. (The point $(0:cdots:0)$ is not defined)."
What does the point $(0:cdots:0)$ mean in $mathbb P^n_A$? By definition every point in $textrm{Proj }B=mathbb P^n_A$ is a prime ideal which doesn't contain $B_+$ right? How can we get them in the coordinate form?
algebraic-geometry
algebraic-geometry
asked Jan 30 at 5:45
KatherineKatherine
438310
asked Jan 30 at 5:45
KatherineKatherine
438310
asked Jan 30 at 5:45
KatherineKatherine
438310
asked Jan 30 at 5:45
KatherineKatherine
438310
asked Jan 30 at 5:45
KatherineKatherine
438310
438310
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The class of projective spaces you want to keep in mind here is where $A$ is an algebraically closed field, $k$. For $mathbb{A}^n_k$, the Nullstellensatz gives a correspondence between (closed) points $(a_1,dots,a_n)$ and maximal ideals $(x_1 - a_1,dots,x_n - a_n)$.
For $mathbb{P}^n_k$, a point $a = (a_0: cdots : a_n)$ corresponds to the line
$$ ell_a = { (ta_0, dots, ta_n) : t in k } subseteq k^{n + 1}$$
Each line contains $(0,dots,0)$ and $ell_a = ell_{a'}$ if and only if $(a_0,dots,a_n) = t(a_0',dots,a_n')$ for some $t in k^*$.
So the closed points of $mathbb{P}^n_k$ correspond to elements of $k^{n + 1} setminus {(0,dots,0)}$ and modulo non-zero scalar multiplication. This is what the notation $(a_0:dots:a_n)$ is: it's the equivalence class of $(a_0,dots,a_n)$ modulo non-zero scalar multiplication. This is how $mathbb{P}^n_k$ is defined as a variety.
What is the ideal of $ell_a$? It is generated by relations of the form $a_jx_i = a_ix_j$. If all the $a_i$'s are zero, then we have no equations and $ell_a$ isn't a line. So at least one $a_i$ is non-zero and WLOG, suppose that $a_0 ne 0$. Then the ideal is generated by $x_j = frac{a_j}{a_0}x_0$ for $j in {1,dots,n}$. So we have $n$ linear relations, meaning we get a linear space of codimension $n$, which is indeed a line.
All the closed points of $mathbb{P}^n_k$ are ideals of this form. They are homogeneous ideals.
Conversely, given any homogeneous ideal $I$ in the variables $x_0,dots,x_n$, the variety it defines in $mathbb{A}^{n+1}_k$ is always a union of lines through the origin. Because if $a = (a_0,dots,a_n) in V(I)$ then $ta in V(I)$ for any $t in k$ and hence $ell_a subseteq V(I)$.
There's one exception to this, and that's the ideal $I = (x_0,dots,x_n)$ because then $V(I)$ doesn't contain any point other than $0 = (0,dots,0)$ and $ell_0$ isn't a line. The ideals we want to ignore are those ideals whose vanishing set is ${0}$. By the Nullstellensatz, $V(I) = {0}$ if and only if $sqrt{I} = (x_0,dots,x_n)$. Hence the condition of not containing $(x_0,dots,x_n)$.
As an exercise: using the Nullstellensatz, classify all the closed points of $mathbb{P}^n_k$. That is, homogeneous, radical ideals of $k[x_0,dots,x_n]$ which are maximal with respect to not containing $(x_0,dots,x_n)$.
answered Jan 30 at 6:19
Trevor GunnTrevor Gunn
15k32047
$endgroup$
$begingroup$
Then why do we define Proj $B$ to be the prime ideals rpt not containing $(x_0,...,x_n)$ instead of maximal ideals rpt not containing $(x_0,...,x_n)$?
$endgroup$
– Katherine
Jan 30 at 21:08
1
$begingroup$
@Katherine For varieties (over an algebraically closed field) you can do everything with just the maximal ideals. For more general rings, you need to use every maximal ideal. For example over $mathbb{Z}$, the points we had of the form $(1:a_1:cdots:a_n)$ give us the ideal $(y_1 - a_1, dots, y_n - a_n)$ of $mathbb{Z}[y_1,dots,y_n]$, where $y_i = x_i/x_0$. (Here I'm looking at an affine piece of the full projective space.
$endgroup$
– Trevor Gunn
Jan 30 at 22:32
1
$begingroup$
If $mathbb{Z}$ were a field, then this would be maximal, but because $mathbb{Z}$ is not a field, it is only prime and a larger ideal is (for example) $(2,y_1 - a_1,dots, y_n - a_n)$. So one reason is we have more prime ideals in the ring $A$. The other reason, which is hard to understand at first, is that it is often helpful to think of subvarieties as points themselves. Like the line $y = x $ is the prime ideal $(y - x)$ of $k[x,y]$. One reason is that it allows us to make sense of the doubled line as the ideal $(y - x)^2$.
$endgroup$
– Trevor Gunn
Jan 30 at 22:33
add a comment |
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1 Answer
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$begingroup$
The class of projective spaces you want to keep in mind here is where $A$ is an algebraically closed field, $k$. For $mathbb{A}^n_k$, the Nullstellensatz gives a correspondence between (closed) points $(a_1,dots,a_n)$ and maximal ideals $(x_1 - a_1,dots,x_n - a_n)$.
For $mathbb{P}^n_k$, a point $a = (a_0: cdots : a_n)$ corresponds to the line
$$ ell_a = { (ta_0, dots, ta_n) : t in k } subseteq k^{n + 1}$$
Each line contains $(0,dots,0)$ and $ell_a = ell_{a'}$ if and only if $(a_0,dots,a_n) = t(a_0',dots,a_n')$ for some $t in k^*$.
So the closed points of $mathbb{P}^n_k$ correspond to elements of $k^{n + 1} setminus {(0,dots,0)}$ and modulo non-zero scalar multiplication. This is what the notation $(a_0:dots:a_n)$ is: it's the equivalence class of $(a_0,dots,a_n)$ modulo non-zero scalar multiplication. This is how $mathbb{P}^n_k$ is defined as a variety.
What is the ideal of $ell_a$? It is generated by relations of the form $a_jx_i = a_ix_j$. If all the $a_i$'s are zero, then we have no equations and $ell_a$ isn't a line. So at least one $a_i$ is non-zero and WLOG, suppose that $a_0 ne 0$. Then the ideal is generated by $x_j = frac{a_j}{a_0}x_0$ for $j in {1,dots,n}$. So we have $n$ linear relations, meaning we get a linear space of codimension $n$, which is indeed a line.
All the closed points of $mathbb{P}^n_k$ are ideals of this form. They are homogeneous ideals.
Conversely, given any homogeneous ideal $I$ in the variables $x_0,dots,x_n$, the variety it defines in $mathbb{A}^{n+1}_k$ is always a union of lines through the origin. Because if $a = (a_0,dots,a_n) in V(I)$ then $ta in V(I)$ for any $t in k$ and hence $ell_a subseteq V(I)$.
There's one exception to this, and that's the ideal $I = (x_0,dots,x_n)$ because then $V(I)$ doesn't contain any point other than $0 = (0,dots,0)$ and $ell_0$ isn't a line. The ideals we want to ignore are those ideals whose vanishing set is ${0}$. By the Nullstellensatz, $V(I) = {0}$ if and only if $sqrt{I} = (x_0,dots,x_n)$. Hence the condition of not containing $(x_0,dots,x_n)$.
As an exercise: using the Nullstellensatz, classify all the closed points of $mathbb{P}^n_k$. That is, homogeneous, radical ideals of $k[x_0,dots,x_n]$ which are maximal with respect to not containing $(x_0,dots,x_n)$.
answered Jan 30 at 6:19
Trevor GunnTrevor Gunn
15k32047
$endgroup$
$begingroup$
Then why do we define Proj $B$ to be the prime ideals rpt not containing $(x_0,...,x_n)$ instead of maximal ideals rpt not containing $(x_0,...,x_n)$?
$endgroup$
– Katherine
Jan 30 at 21:08
1
$begingroup$
@Katherine For varieties (over an algebraically closed field) you can do everything with just the maximal ideals. For more general rings, you need to use every maximal ideal. For example over $mathbb{Z}$, the points we had of the form $(1:a_1:cdots:a_n)$ give us the ideal $(y_1 - a_1, dots, y_n - a_n)$ of $mathbb{Z}[y_1,dots,y_n]$, where $y_i = x_i/x_0$. (Here I'm looking at an affine piece of the full projective space.
$endgroup$
– Trevor Gunn
Jan 30 at 22:32
1
$begingroup$
If $mathbb{Z}$ were a field, then this would be maximal, but because $mathbb{Z}$ is not a field, it is only prime and a larger ideal is (for example) $(2,y_1 - a_1,dots, y_n - a_n)$. So one reason is we have more prime ideals in the ring $A$. The other reason, which is hard to understand at first, is that it is often helpful to think of subvarieties as points themselves. Like the line $y = x $ is the prime ideal $(y - x)$ of $k[x,y]$. One reason is that it allows us to make sense of the doubled line as the ideal $(y - x)^2$.
$endgroup$
– Trevor Gunn
Jan 30 at 22:33
add a comment |
$begingroup$
The class of projective spaces you want to keep in mind here is where $A$ is an algebraically closed field, $k$. For $mathbb{A}^n_k$, the Nullstellensatz gives a correspondence between (closed) points $(a_1,dots,a_n)$ and maximal ideals $(x_1 - a_1,dots,x_n - a_n)$.
For $mathbb{P}^n_k$, a point $a = (a_0: cdots : a_n)$ corresponds to the line
$$ ell_a = { (ta_0, dots, ta_n) : t in k } subseteq k^{n + 1}$$
Each line contains $(0,dots,0)$ and $ell_a = ell_{a'}$ if and only if $(a_0,dots,a_n) = t(a_0',dots,a_n')$ for some $t in k^*$.
So the closed points of $mathbb{P}^n_k$ correspond to elements of $k^{n + 1} setminus {(0,dots,0)}$ and modulo non-zero scalar multiplication. This is what the notation $(a_0:dots:a_n)$ is: it's the equivalence class of $(a_0,dots,a_n)$ modulo non-zero scalar multiplication. This is how $mathbb{P}^n_k$ is defined as a variety.
What is the ideal of $ell_a$? It is generated by relations of the form $a_jx_i = a_ix_j$. If all the $a_i$'s are zero, then we have no equations and $ell_a$ isn't a line. So at least one $a_i$ is non-zero and WLOG, suppose that $a_0 ne 0$. Then the ideal is generated by $x_j = frac{a_j}{a_0}x_0$ for $j in {1,dots,n}$. So we have $n$ linear relations, meaning we get a linear space of codimension $n$, which is indeed a line.
All the closed points of $mathbb{P}^n_k$ are ideals of this form. They are homogeneous ideals.
Conversely, given any homogeneous ideal $I$ in the variables $x_0,dots,x_n$, the variety it defines in $mathbb{A}^{n+1}_k$ is always a union of lines through the origin. Because if $a = (a_0,dots,a_n) in V(I)$ then $ta in V(I)$ for any $t in k$ and hence $ell_a subseteq V(I)$.
There's one exception to this, and that's the ideal $I = (x_0,dots,x_n)$ because then $V(I)$ doesn't contain any point other than $0 = (0,dots,0)$ and $ell_0$ isn't a line. The ideals we want to ignore are those ideals whose vanishing set is ${0}$. By the Nullstellensatz, $V(I) = {0}$ if and only if $sqrt{I} = (x_0,dots,x_n)$. Hence the condition of not containing $(x_0,dots,x_n)$.
As an exercise: using the Nullstellensatz, classify all the closed points of $mathbb{P}^n_k$. That is, homogeneous, radical ideals of $k[x_0,dots,x_n]$ which are maximal with respect to not containing $(x_0,dots,x_n)$.
answered Jan 30 at 6:19
Trevor GunnTrevor Gunn
15k32047
$endgroup$
$begingroup$
Then why do we define Proj $B$ to be the prime ideals rpt not containing $(x_0,...,x_n)$ instead of maximal ideals rpt not containing $(x_0,...,x_n)$?
$endgroup$
– Katherine
Jan 30 at 21:08
1
$begingroup$
@Katherine For varieties (over an algebraically closed field) you can do everything with just the maximal ideals. For more general rings, you need to use every maximal ideal. For example over $mathbb{Z}$, the points we had of the form $(1:a_1:cdots:a_n)$ give us the ideal $(y_1 - a_1, dots, y_n - a_n)$ of $mathbb{Z}[y_1,dots,y_n]$, where $y_i = x_i/x_0$. (Here I'm looking at an affine piece of the full projective space.
$endgroup$
– Trevor Gunn
Jan 30 at 22:32
1
$begingroup$
If $mathbb{Z}$ were a field, then this would be maximal, but because $mathbb{Z}$ is not a field, it is only prime and a larger ideal is (for example) $(2,y_1 - a_1,dots, y_n - a_n)$. So one reason is we have more prime ideals in the ring $A$. The other reason, which is hard to understand at first, is that it is often helpful to think of subvarieties as points themselves. Like the line $y = x $ is the prime ideal $(y - x)$ of $k[x,y]$. One reason is that it allows us to make sense of the doubled line as the ideal $(y - x)^2$.
$endgroup$
– Trevor Gunn
Jan 30 at 22:33
add a comment |
$begingroup$
The class of projective spaces you want to keep in mind here is where $A$ is an algebraically closed field, $k$. For $mathbb{A}^n_k$, the Nullstellensatz gives a correspondence between (closed) points $(a_1,dots,a_n)$ and maximal ideals $(x_1 - a_1,dots,x_n - a_n)$.
For $mathbb{P}^n_k$, a point $a = (a_0: cdots : a_n)$ corresponds to the line
$$ ell_a = { (ta_0, dots, ta_n) : t in k } subseteq k^{n + 1}$$
Each line contains $(0,dots,0)$ and $ell_a = ell_{a'}$ if and only if $(a_0,dots,a_n) = t(a_0',dots,a_n')$ for some $t in k^*$.
So the closed points of $mathbb{P}^n_k$ correspond to elements of $k^{n + 1} setminus {(0,dots,0)}$ and modulo non-zero scalar multiplication. This is what the notation $(a_0:dots:a_n)$ is: it's the equivalence class of $(a_0,dots,a_n)$ modulo non-zero scalar multiplication. This is how $mathbb{P}^n_k$ is defined as a variety.
What is the ideal of $ell_a$? It is generated by relations of the form $a_jx_i = a_ix_j$. If all the $a_i$'s are zero, then we have no equations and $ell_a$ isn't a line. So at least one $a_i$ is non-zero and WLOG, suppose that $a_0 ne 0$. Then the ideal is generated by $x_j = frac{a_j}{a_0}x_0$ for $j in {1,dots,n}$. So we have $n$ linear relations, meaning we get a linear space of codimension $n$, which is indeed a line.
All the closed points of $mathbb{P}^n_k$ are ideals of this form. They are homogeneous ideals.
Conversely, given any homogeneous ideal $I$ in the variables $x_0,dots,x_n$, the variety it defines in $mathbb{A}^{n+1}_k$ is always a union of lines through the origin. Because if $a = (a_0,dots,a_n) in V(I)$ then $ta in V(I)$ for any $t in k$ and hence $ell_a subseteq V(I)$.
There's one exception to this, and that's the ideal $I = (x_0,dots,x_n)$ because then $V(I)$ doesn't contain any point other than $0 = (0,dots,0)$ and $ell_0$ isn't a line. The ideals we want to ignore are those ideals whose vanishing set is ${0}$. By the Nullstellensatz, $V(I) = {0}$ if and only if $sqrt{I} = (x_0,dots,x_n)$. Hence the condition of not containing $(x_0,dots,x_n)$.
As an exercise: using the Nullstellensatz, classify all the closed points of $mathbb{P}^n_k$. That is, homogeneous, radical ideals of $k[x_0,dots,x_n]$ which are maximal with respect to not containing $(x_0,dots,x_n)$.
answered Jan 30 at 6:19
Trevor GunnTrevor Gunn
15k32047
$endgroup$
The class of projective spaces you want to keep in mind here is where $A$ is an algebraically closed field, $k$. For $mathbb{A}^n_k$, the Nullstellensatz gives a correspondence between (closed) points $(a_1,dots,a_n)$ and maximal ideals $(x_1 - a_1,dots,x_n - a_n)$.
For $mathbb{P}^n_k$, a point $a = (a_0: cdots : a_n)$ corresponds to the line
$$ ell_a = { (ta_0, dots, ta_n) : t in k } subseteq k^{n + 1}$$
Each line contains $(0,dots,0)$ and $ell_a = ell_{a'}$ if and only if $(a_0,dots,a_n) = t(a_0',dots,a_n')$ for some $t in k^*$.
So the closed points of $mathbb{P}^n_k$ correspond to elements of $k^{n + 1} setminus {(0,dots,0)}$ and modulo non-zero scalar multiplication. This is what the notation $(a_0:dots:a_n)$ is: it's the equivalence class of $(a_0,dots,a_n)$ modulo non-zero scalar multiplication. This is how $mathbb{P}^n_k$ is defined as a variety.
What is the ideal of $ell_a$? It is generated by relations of the form $a_jx_i = a_ix_j$. If all the $a_i$'s are zero, then we have no equations and $ell_a$ isn't a line. So at least one $a_i$ is non-zero and WLOG, suppose that $a_0 ne 0$. Then the ideal is generated by $x_j = frac{a_j}{a_0}x_0$ for $j in {1,dots,n}$. So we have $n$ linear relations, meaning we get a linear space of codimension $n$, which is indeed a line.
All the closed points of $mathbb{P}^n_k$ are ideals of this form. They are homogeneous ideals.
Conversely, given any homogeneous ideal $I$ in the variables $x_0,dots,x_n$, the variety it defines in $mathbb{A}^{n+1}_k$ is always a union of lines through the origin. Because if $a = (a_0,dots,a_n) in V(I)$ then $ta in V(I)$ for any $t in k$ and hence $ell_a subseteq V(I)$.
There's one exception to this, and that's the ideal $I = (x_0,dots,x_n)$ because then $V(I)$ doesn't contain any point other than $0 = (0,dots,0)$ and $ell_0$ isn't a line. The ideals we want to ignore are those ideals whose vanishing set is ${0}$. By the Nullstellensatz, $V(I) = {0}$ if and only if $sqrt{I} = (x_0,dots,x_n)$. Hence the condition of not containing $(x_0,dots,x_n)$.
As an exercise: using the Nullstellensatz, classify all the closed points of $mathbb{P}^n_k$. That is, homogeneous, radical ideals of $k[x_0,dots,x_n]$ which are maximal with respect to not containing $(x_0,dots,x_n)$.
answered Jan 30 at 6:19
Trevor GunnTrevor Gunn
15k32047
answered Jan 30 at 6:19
Trevor GunnTrevor Gunn
15k32047
answered Jan 30 at 6:19
Trevor GunnTrevor Gunn
15k32047
answered Jan 30 at 6:19
Trevor GunnTrevor Gunn
15k32047
15k32047
$begingroup$
Then why do we define Proj $B$ to be the prime ideals rpt not containing $(x_0,...,x_n)$ instead of maximal ideals rpt not containing $(x_0,...,x_n)$?
$endgroup$
– Katherine
Jan 30 at 21:08
1
$begingroup$
@Katherine For varieties (over an algebraically closed field) you can do everything with just the maximal ideals. For more general rings, you need to use every maximal ideal. For example over $mathbb{Z}$, the points we had of the form $(1:a_1:cdots:a_n)$ give us the ideal $(y_1 - a_1, dots, y_n - a_n)$ of $mathbb{Z}[y_1,dots,y_n]$, where $y_i = x_i/x_0$. (Here I'm looking at an affine piece of the full projective space.
$endgroup$
– Trevor Gunn
Jan 30 at 22:32
1
$begingroup$
If $mathbb{Z}$ were a field, then this would be maximal, but because $mathbb{Z}$ is not a field, it is only prime and a larger ideal is (for example) $(2,y_1 - a_1,dots, y_n - a_n)$. So one reason is we have more prime ideals in the ring $A$. The other reason, which is hard to understand at first, is that it is often helpful to think of subvarieties as points themselves. Like the line $y = x $ is the prime ideal $(y - x)$ of $k[x,y]$. One reason is that it allows us to make sense of the doubled line as the ideal $(y - x)^2$.
$endgroup$
– Trevor Gunn
Jan 30 at 22:33
add a comment |
$begingroup$
Then why do we define Proj $B$ to be the prime ideals rpt not containing $(x_0,...,x_n)$ instead of maximal ideals rpt not containing $(x_0,...,x_n)$?
$endgroup$
– Katherine
Jan 30 at 21:08
1
$begingroup$
@Katherine For varieties (over an algebraically closed field) you can do everything with just the maximal ideals. For more general rings, you need to use every maximal ideal. For example over $mathbb{Z}$, the points we had of the form $(1:a_1:cdots:a_n)$ give us the ideal $(y_1 - a_1, dots, y_n - a_n)$ of $mathbb{Z}[y_1,dots,y_n]$, where $y_i = x_i/x_0$. (Here I'm looking at an affine piece of the full projective space.
$endgroup$
– Trevor Gunn
Jan 30 at 22:32
1
$begingroup$
If $mathbb{Z}$ were a field, then this would be maximal, but because $mathbb{Z}$ is not a field, it is only prime and a larger ideal is (for example) $(2,y_1 - a_1,dots, y_n - a_n)$. So one reason is we have more prime ideals in the ring $A$. The other reason, which is hard to understand at first, is that it is often helpful to think of subvarieties as points themselves. Like the line $y = x $ is the prime ideal $(y - x)$ of $k[x,y]$. One reason is that it allows us to make sense of the doubled line as the ideal $(y - x)^2$.
$endgroup$
– Trevor Gunn
Jan 30 at 22:33
$begingroup$
Then why do we define Proj $B$ to be the prime ideals rpt not containing $(x_0,...,x_n)$ instead of maximal ideals rpt not containing $(x_0,...,x_n)$?
$endgroup$
– Katherine
Jan 30 at 21:08
$begingroup$
Then why do we define Proj $B$ to be the prime ideals rpt not containing $(x_0,...,x_n)$ instead of maximal ideals rpt not containing $(x_0,...,x_n)$?
$endgroup$
– Katherine
Jan 30 at 21:08
1
1
$begingroup$
@Katherine For varieties (over an algebraically closed field) you can do everything with just the maximal ideals. For more general rings, you need to use every maximal ideal. For example over $mathbb{Z}$, the points we had of the form $(1:a_1:cdots:a_n)$ give us the ideal $(y_1 - a_1, dots, y_n - a_n)$ of $mathbb{Z}[y_1,dots,y_n]$, where $y_i = x_i/x_0$. (Here I'm looking at an affine piece of the full projective space.
$endgroup$
– Trevor Gunn
Jan 30 at 22:32
$begingroup$
@Katherine For varieties (over an algebraically closed field) you can do everything with just the maximal ideals. For more general rings, you need to use every maximal ideal. For example over $mathbb{Z}$, the points we had of the form $(1:a_1:cdots:a_n)$ give us the ideal $(y_1 - a_1, dots, y_n - a_n)$ of $mathbb{Z}[y_1,dots,y_n]$, where $y_i = x_i/x_0$. (Here I'm looking at an affine piece of the full projective space.
$endgroup$
– Trevor Gunn
Jan 30 at 22:32
1
1
$begingroup$
If $mathbb{Z}$ were a field, then this would be maximal, but because $mathbb{Z}$ is not a field, it is only prime and a larger ideal is (for example) $(2,y_1 - a_1,dots, y_n - a_n)$. So one reason is we have more prime ideals in the ring $A$. The other reason, which is hard to understand at first, is that it is often helpful to think of subvarieties as points themselves. Like the line $y = x $ is the prime ideal $(y - x)$ of $k[x,y]$. One reason is that it allows us to make sense of the doubled line as the ideal $(y - x)^2$.
$endgroup$
– Trevor Gunn
Jan 30 at 22:33
$begingroup$
If $mathbb{Z}$ were a field, then this would be maximal, but because $mathbb{Z}$ is not a field, it is only prime and a larger ideal is (for example) $(2,y_1 - a_1,dots, y_n - a_n)$. So one reason is we have more prime ideals in the ring $A$. The other reason, which is hard to understand at first, is that it is often helpful to think of subvarieties as points themselves. Like the line $y = x $ is the prime ideal $(y - x)$ of $k[x,y]$. One reason is that it allows us to make sense of the doubled line as the ideal $(y - x)^2$.
$endgroup$
– Trevor Gunn
Jan 30 at 22:33
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