Affine charts are dense in projective space
$begingroup$
Given a field $k$, we define the scheme-theoretic $n$-th affine space over $k$ by $mathbb{A}^n_k=text{Spec}(k[X_1,dots,X_n])$ and the $n$-th projective space over $k$ by $mathbb{P}^n_k=text{Proj}(k[X_0,dots, X_n])$. We know $mathbb{P}^n_k$ is covered by $n+1$ affine charts given by $D_+(X_i)=mathbb{P}^n_ksmallsetminus V_+(X_i)$ for $i=0,dots, n$, each isomorphic to $mathbb{A}^n_k$.
I was asking myself if - as it happens in the naive case - each of those charts is dense in $mathbb{P}^n_k$.
Here is my attempt. Take e.g. $D_+(X_0)$. Then $D_+(X_0)$ being dense in $mathbb{P}^n_k$ is equivalent to $V_+(X_0)$ having empty interior. Suppose there exists $fin (X_0,dots, X_n)$ s.t. $D_+(f)subset V_+(X_0)$. Then every prime $mathfrak{p}in mathbb{P}^n_k$ s.t. $fnotin mathfrak{p}$ is s.t. $X_0in mathfrak{p}$, which means $D_+(fX_0)=emptyset$, i.e. $V_+(fX_0)=V_+(0)=mathbb{P}^n_k$, and thus $f=0$ since $k[X_0,dots, X_n]$ is an integral domain.
Is this proof correct? Does anyone know a shorter way to prove it?
algebraic-geometry schemes projective-space projective-schemes
asked Jan 11 at 14:10
OromisOromis
404413
$endgroup$
add a comment |
$begingroup$
Given a field $k$, we define the scheme-theoretic $n$-th affine space over $k$ by $mathbb{A}^n_k=text{Spec}(k[X_1,dots,X_n])$ and the $n$-th projective space over $k$ by $mathbb{P}^n_k=text{Proj}(k[X_0,dots, X_n])$. We know $mathbb{P}^n_k$ is covered by $n+1$ affine charts given by $D_+(X_i)=mathbb{P}^n_ksmallsetminus V_+(X_i)$ for $i=0,dots, n$, each isomorphic to $mathbb{A}^n_k$.
I was asking myself if - as it happens in the naive case - each of those charts is dense in $mathbb{P}^n_k$.
Here is my attempt. Take e.g. $D_+(X_0)$. Then $D_+(X_0)$ being dense in $mathbb{P}^n_k$ is equivalent to $V_+(X_0)$ having empty interior. Suppose there exists $fin (X_0,dots, X_n)$ s.t. $D_+(f)subset V_+(X_0)$. Then every prime $mathfrak{p}in mathbb{P}^n_k$ s.t. $fnotin mathfrak{p}$ is s.t. $X_0in mathfrak{p}$, which means $D_+(fX_0)=emptyset$, i.e. $V_+(fX_0)=V_+(0)=mathbb{P}^n_k$, and thus $f=0$ since $k[X_0,dots, X_n]$ is an integral domain.
Is this proof correct? Does anyone know a shorter way to prove it?
algebraic-geometry schemes projective-space projective-schemes
asked Jan 11 at 14:10
OromisOromis
404413
$endgroup$
$begingroup$
You can also prove it by observing the fact that $D_+(X_0)$ is dense in every chart, i.e. $D_+(X_0)cap D_+(X_i)$ is dense in $D_+(X_i)$.
$endgroup$
– Levent
Jan 11 at 14:57
add a comment |
$begingroup$
Given a field $k$, we define the scheme-theoretic $n$-th affine space over $k$ by $mathbb{A}^n_k=text{Spec}(k[X_1,dots,X_n])$ and the $n$-th projective space over $k$ by $mathbb{P}^n_k=text{Proj}(k[X_0,dots, X_n])$. We know $mathbb{P}^n_k$ is covered by $n+1$ affine charts given by $D_+(X_i)=mathbb{P}^n_ksmallsetminus V_+(X_i)$ for $i=0,dots, n$, each isomorphic to $mathbb{A}^n_k$.
I was asking myself if - as it happens in the naive case - each of those charts is dense in $mathbb{P}^n_k$.
Here is my attempt. Take e.g. $D_+(X_0)$. Then $D_+(X_0)$ being dense in $mathbb{P}^n_k$ is equivalent to $V_+(X_0)$ having empty interior. Suppose there exists $fin (X_0,dots, X_n)$ s.t. $D_+(f)subset V_+(X_0)$. Then every prime $mathfrak{p}in mathbb{P}^n_k$ s.t. $fnotin mathfrak{p}$ is s.t. $X_0in mathfrak{p}$, which means $D_+(fX_0)=emptyset$, i.e. $V_+(fX_0)=V_+(0)=mathbb{P}^n_k$, and thus $f=0$ since $k[X_0,dots, X_n]$ is an integral domain.
Is this proof correct? Does anyone know a shorter way to prove it?
algebraic-geometry schemes projective-space projective-schemes
asked Jan 11 at 14:10
OromisOromis
404413
$endgroup$
Given a field $k$, we define the scheme-theoretic $n$-th affine space over $k$ by $mathbb{A}^n_k=text{Spec}(k[X_1,dots,X_n])$ and the $n$-th projective space over $k$ by $mathbb{P}^n_k=text{Proj}(k[X_0,dots, X_n])$. We know $mathbb{P}^n_k$ is covered by $n+1$ affine charts given by $D_+(X_i)=mathbb{P}^n_ksmallsetminus V_+(X_i)$ for $i=0,dots, n$, each isomorphic to $mathbb{A}^n_k$.
I was asking myself if - as it happens in the naive case - each of those charts is dense in $mathbb{P}^n_k$.
Here is my attempt. Take e.g. $D_+(X_0)$. Then $D_+(X_0)$ being dense in $mathbb{P}^n_k$ is equivalent to $V_+(X_0)$ having empty interior. Suppose there exists $fin (X_0,dots, X_n)$ s.t. $D_+(f)subset V_+(X_0)$. Then every prime $mathfrak{p}in mathbb{P}^n_k$ s.t. $fnotin mathfrak{p}$ is s.t. $X_0in mathfrak{p}$, which means $D_+(fX_0)=emptyset$, i.e. $V_+(fX_0)=V_+(0)=mathbb{P}^n_k$, and thus $f=0$ since $k[X_0,dots, X_n]$ is an integral domain.
Is this proof correct? Does anyone know a shorter way to prove it?
algebraic-geometry schemes projective-space projective-schemes
algebraic-geometry schemes projective-space projective-schemes
asked Jan 11 at 14:10
OromisOromis
404413
asked Jan 11 at 14:10
OromisOromis
404413
asked Jan 11 at 14:10
OromisOromis
404413
asked Jan 11 at 14:10
OromisOromis
404413
asked Jan 11 at 14:10
OromisOromis
404413
404413
$begingroup$
You can also prove it by observing the fact that $D_+(X_0)$ is dense in every chart, i.e. $D_+(X_0)cap D_+(X_i)$ is dense in $D_+(X_i)$.
$endgroup$
– Levent
Jan 11 at 14:57
add a comment |
$begingroup$
You can also prove it by observing the fact that $D_+(X_0)$ is dense in every chart, i.e. $D_+(X_0)cap D_+(X_i)$ is dense in $D_+(X_i)$.
$endgroup$
– Levent
Jan 11 at 14:57
$begingroup$
You can also prove it by observing the fact that $D_+(X_0)$ is dense in every chart, i.e. $D_+(X_0)cap D_+(X_i)$ is dense in $D_+(X_i)$.
$endgroup$
– Levent
Jan 11 at 14:57
$begingroup$
You can also prove it by observing the fact that $D_+(X_0)$ is dense in every chart, i.e. $D_+(X_0)cap D_+(X_i)$ is dense in $D_+(X_i)$.
$endgroup$
– Levent
Jan 11 at 14:57
add a comment |
1 Answer
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$begingroup$
Your argument is correct (you might just want to emphasize that the $f$ you introduce is homogeneous). The point is that the equality $V_+(fX_0)=mathbf{P}_k^n$ is equivalent to $(fX_0)cap(X_0,ldots,X_n)subseteqsqrt{0}$; the lefthand side of this inclusion is just the ideal $(fX_0)$ since $fX_0in(X_0,ldots,X_n)$, while the righthand side is $0$ (the zero ideal) since $k[X_0,ldots,X_n]$ is reduced.
This reasoning basically gets to the heart of what is going on, and I don't think it can really be shortened, although it can be modified slightly so as to be maximally general. What you are showing here is (equivalent to the assertion) that $mathbf{P}_k^n$ is irreducible. This irreducibility is inherited from the irreducibility of the standard opens $D_+(X_i)$ and the manner in which they are glued together (more precisely, that $D_+(X_i)cap D_+(X_j)neqemptyset$ for all $i,j$). In general, if you have a nonempty scheme $X$ that can be written as a union $bigcup_{iin I}X_i$ such that
(1) $I$ is nonempty,
(2) each $X_i$ is an irreducible open subset of $X$, and
(3) $X_icap X_jneqemptyset$ for all $i,jin I$,
then $X$ is irreducible. So the reasoning you are using will apply to show the irreducibility of $mathrm{Proj}(S)$ for a wider class of graded rings $S$ than just the class of polynomial rings over fields. For example, it can be applied to quotients of such rings by homogeneous prime ideals.
answered Jan 11 at 23:15
Keenan KidwellKeenan Kidwell
19.9k13676
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add a comment |
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$begingroup$
Your argument is correct (you might just want to emphasize that the $f$ you introduce is homogeneous). The point is that the equality $V_+(fX_0)=mathbf{P}_k^n$ is equivalent to $(fX_0)cap(X_0,ldots,X_n)subseteqsqrt{0}$; the lefthand side of this inclusion is just the ideal $(fX_0)$ since $fX_0in(X_0,ldots,X_n)$, while the righthand side is $0$ (the zero ideal) since $k[X_0,ldots,X_n]$ is reduced.
This reasoning basically gets to the heart of what is going on, and I don't think it can really be shortened, although it can be modified slightly so as to be maximally general. What you are showing here is (equivalent to the assertion) that $mathbf{P}_k^n$ is irreducible. This irreducibility is inherited from the irreducibility of the standard opens $D_+(X_i)$ and the manner in which they are glued together (more precisely, that $D_+(X_i)cap D_+(X_j)neqemptyset$ for all $i,j$). In general, if you have a nonempty scheme $X$ that can be written as a union $bigcup_{iin I}X_i$ such that
(1) $I$ is nonempty,
(2) each $X_i$ is an irreducible open subset of $X$, and
(3) $X_icap X_jneqemptyset$ for all $i,jin I$,
then $X$ is irreducible. So the reasoning you are using will apply to show the irreducibility of $mathrm{Proj}(S)$ for a wider class of graded rings $S$ than just the class of polynomial rings over fields. For example, it can be applied to quotients of such rings by homogeneous prime ideals.
answered Jan 11 at 23:15
Keenan KidwellKeenan Kidwell
19.9k13676
$endgroup$
add a comment |
$begingroup$
Your argument is correct (you might just want to emphasize that the $f$ you introduce is homogeneous). The point is that the equality $V_+(fX_0)=mathbf{P}_k^n$ is equivalent to $(fX_0)cap(X_0,ldots,X_n)subseteqsqrt{0}$; the lefthand side of this inclusion is just the ideal $(fX_0)$ since $fX_0in(X_0,ldots,X_n)$, while the righthand side is $0$ (the zero ideal) since $k[X_0,ldots,X_n]$ is reduced.
This reasoning basically gets to the heart of what is going on, and I don't think it can really be shortened, although it can be modified slightly so as to be maximally general. What you are showing here is (equivalent to the assertion) that $mathbf{P}_k^n$ is irreducible. This irreducibility is inherited from the irreducibility of the standard opens $D_+(X_i)$ and the manner in which they are glued together (more precisely, that $D_+(X_i)cap D_+(X_j)neqemptyset$ for all $i,j$). In general, if you have a nonempty scheme $X$ that can be written as a union $bigcup_{iin I}X_i$ such that
(1) $I$ is nonempty,
(2) each $X_i$ is an irreducible open subset of $X$, and
(3) $X_icap X_jneqemptyset$ for all $i,jin I$,
then $X$ is irreducible. So the reasoning you are using will apply to show the irreducibility of $mathrm{Proj}(S)$ for a wider class of graded rings $S$ than just the class of polynomial rings over fields. For example, it can be applied to quotients of such rings by homogeneous prime ideals.
answered Jan 11 at 23:15
Keenan KidwellKeenan Kidwell
19.9k13676
$endgroup$
add a comment |
$begingroup$
Your argument is correct (you might just want to emphasize that the $f$ you introduce is homogeneous). The point is that the equality $V_+(fX_0)=mathbf{P}_k^n$ is equivalent to $(fX_0)cap(X_0,ldots,X_n)subseteqsqrt{0}$; the lefthand side of this inclusion is just the ideal $(fX_0)$ since $fX_0in(X_0,ldots,X_n)$, while the righthand side is $0$ (the zero ideal) since $k[X_0,ldots,X_n]$ is reduced.
This reasoning basically gets to the heart of what is going on, and I don't think it can really be shortened, although it can be modified slightly so as to be maximally general. What you are showing here is (equivalent to the assertion) that $mathbf{P}_k^n$ is irreducible. This irreducibility is inherited from the irreducibility of the standard opens $D_+(X_i)$ and the manner in which they are glued together (more precisely, that $D_+(X_i)cap D_+(X_j)neqemptyset$ for all $i,j$). In general, if you have a nonempty scheme $X$ that can be written as a union $bigcup_{iin I}X_i$ such that
(1) $I$ is nonempty,
(2) each $X_i$ is an irreducible open subset of $X$, and
(3) $X_icap X_jneqemptyset$ for all $i,jin I$,
then $X$ is irreducible. So the reasoning you are using will apply to show the irreducibility of $mathrm{Proj}(S)$ for a wider class of graded rings $S$ than just the class of polynomial rings over fields. For example, it can be applied to quotients of such rings by homogeneous prime ideals.
answered Jan 11 at 23:15
Keenan KidwellKeenan Kidwell
19.9k13676
$endgroup$
Your argument is correct (you might just want to emphasize that the $f$ you introduce is homogeneous). The point is that the equality $V_+(fX_0)=mathbf{P}_k^n$ is equivalent to $(fX_0)cap(X_0,ldots,X_n)subseteqsqrt{0}$; the lefthand side of this inclusion is just the ideal $(fX_0)$ since $fX_0in(X_0,ldots,X_n)$, while the righthand side is $0$ (the zero ideal) since $k[X_0,ldots,X_n]$ is reduced.
This reasoning basically gets to the heart of what is going on, and I don't think it can really be shortened, although it can be modified slightly so as to be maximally general. What you are showing here is (equivalent to the assertion) that $mathbf{P}_k^n$ is irreducible. This irreducibility is inherited from the irreducibility of the standard opens $D_+(X_i)$ and the manner in which they are glued together (more precisely, that $D_+(X_i)cap D_+(X_j)neqemptyset$ for all $i,j$). In general, if you have a nonempty scheme $X$ that can be written as a union $bigcup_{iin I}X_i$ such that
(1) $I$ is nonempty,
(2) each $X_i$ is an irreducible open subset of $X$, and
(3) $X_icap X_jneqemptyset$ for all $i,jin I$,
then $X$ is irreducible. So the reasoning you are using will apply to show the irreducibility of $mathrm{Proj}(S)$ for a wider class of graded rings $S$ than just the class of polynomial rings over fields. For example, it can be applied to quotients of such rings by homogeneous prime ideals.
answered Jan 11 at 23:15
Keenan KidwellKeenan Kidwell
19.9k13676
answered Jan 11 at 23:15
Keenan KidwellKeenan Kidwell
19.9k13676
answered Jan 11 at 23:15
Keenan KidwellKeenan Kidwell
19.9k13676
answered Jan 11 at 23:15
Keenan KidwellKeenan Kidwell
19.9k13676
19.9k13676
add a comment |
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$begingroup$
You can also prove it by observing the fact that $D_+(X_0)$ is dense in every chart, i.e. $D_+(X_0)cap D_+(X_i)$ is dense in $D_+(X_i)$.
$endgroup$
– Levent
Jan 11 at 14:57