Denseness of algebraic points within a variety
$begingroup$
I want to use the following statement concerning varieties, but I do not know why it is true.
Claim. Let $V subset mathbb{C}^n$ be a variety defined over $mathbb{Q}$. Then the set $V cap bar{mathbb{Q}}$ is dense in $V$ with respect to the Hausdorff topology (not the Zariski topology).
Here, $bar{mathbb{Q}}$ denotes the algebraic closure of the rationals $mathbb{Q}$.
It was pointed out to me that one can show this using the so-called Tarski-Seidenberg Principle, in particular using Proposition 5.3.5 of Real Algebraic Geometry by Bochnak, Coste, Roy.
Let $R$ be a real closed field, $Asubset R^m$ and $Bsubset R^n$ semialgebraic sets, and $f:Ato B$ a semialgebraic map with graph $Gsubset Atimes B$. Let $K$ be a real closed extension of $R$, and denote the extension of a semialgebraic set $S$ defined over $R$ to $K$ as $S_K$.
Proposition 5.3.5
i) The semialgebraic set $A$ is open (resp. closed) in $R^m$ iff $A_K$ is open (resp. closed) in $K^m$. More generally, $clos(A_K)=(clos(A))_K$.
ii) The semialgebraic mapping $f$ is continuous iff $f_k$ is continuous.
I do not see how my statement follows. Does anyone have experience with applying this principle to this kind of situation? Or can you think of another approach leading to a proof of the claim?
algebraic-geometry real-algebraic-geometry
edited Nov 28 '18 at 2:30
KReiser
9,57721435
asked Nov 27 '18 at 10:38
Levi RyffelLevi Ryffel
313
$endgroup$
|
show 4 more comments
$begingroup$
I want to use the following statement concerning varieties, but I do not know why it is true.
Claim. Let $V subset mathbb{C}^n$ be a variety defined over $mathbb{Q}$. Then the set $V cap bar{mathbb{Q}}$ is dense in $V$ with respect to the Hausdorff topology (not the Zariski topology).
Here, $bar{mathbb{Q}}$ denotes the algebraic closure of the rationals $mathbb{Q}$.
It was pointed out to me that one can show this using the so-called Tarski-Seidenberg Principle, in particular using Proposition 5.3.5 of Real Algebraic Geometry by Bochnak, Coste, Roy.
Let $R$ be a real closed field, $Asubset R^m$ and $Bsubset R^n$ semialgebraic sets, and $f:Ato B$ a semialgebraic map with graph $Gsubset Atimes B$. Let $K$ be a real closed extension of $R$, and denote the extension of a semialgebraic set $S$ defined over $R$ to $K$ as $S_K$.
Proposition 5.3.5
i) The semialgebraic set $A$ is open (resp. closed) in $R^m$ iff $A_K$ is open (resp. closed) in $K^m$. More generally, $clos(A_K)=(clos(A))_K$.
ii) The semialgebraic mapping $f$ is continuous iff $f_k$ is continuous.
I do not see how my statement follows. Does anyone have experience with applying this principle to this kind of situation? Or can you think of another approach leading to a proof of the claim?
algebraic-geometry real-algebraic-geometry
edited Nov 28 '18 at 2:30
KReiser
9,57721435
asked Nov 27 '18 at 10:38
Levi RyffelLevi Ryffel
313
$endgroup$
1
$begingroup$
I've edited the question to include the proposition you reference. In the future, please include information like this in the question statement - it will help other users write good answers to your question (and not everyone has this book immediately on hand!).
$endgroup$
– KReiser
Nov 28 '18 at 0:04
$begingroup$
You can pick a transcendental basis to put the function field in the form $K(V) cong K(u_1,ldots,u_m)[z_1,ldots,z_l]/J$ ?
$endgroup$
– reuns
Nov 28 '18 at 3:29
$begingroup$
@reuns Is $overline{mathbb{Q}}$ even real closed field?. Or one should apply the theorem to different fields?
$endgroup$
– random123
Nov 28 '18 at 5:50
1
$begingroup$
$overline{Bbb Q}$ is not real closed (it has $i$, for instance). Techniques from semialgebraic geometry will need a slight adjustment here. The right way around this is to decompose in to real and imaginary parts, then show that the $Bbb R_{alg}$ points are dense in the $Bbb R$ points of $V$ regarded as a semialgebraic set in $Bbb R^{2n}$. This is a place where you may use an extension of real closed fields: $Bbb R_{alg} subset Bbb R$. Unfortunately I don't have time to write a full solution right now.
$endgroup$
– KReiser
Nov 28 '18 at 6:10
1
$begingroup$
@random123 what? I don't see how that's relevant here. What I'm saying is that one writes the coordinates $z_j$ of $Bbb C^n$ as $z_j=x_j+iy_j$ and considers $V$ as a semialgebraic set in $Bbb R^{2n}$ with coordinates $x_j$ and $y_j$. Now the $overline{Bbb Q}$ points of the original $V$ are points in the semialgebraic set in $Bbb R^{2n}$ which have all their coordinates in $Bbb R_{alg}$ while the $Bbb C$ points of the original $V$ are all points in the semialgebraic set.
$endgroup$
– KReiser
Nov 28 '18 at 9:18
|
show 4 more comments
$begingroup$
I want to use the following statement concerning varieties, but I do not know why it is true.
Claim. Let $V subset mathbb{C}^n$ be a variety defined over $mathbb{Q}$. Then the set $V cap bar{mathbb{Q}}$ is dense in $V$ with respect to the Hausdorff topology (not the Zariski topology).
Here, $bar{mathbb{Q}}$ denotes the algebraic closure of the rationals $mathbb{Q}$.
It was pointed out to me that one can show this using the so-called Tarski-Seidenberg Principle, in particular using Proposition 5.3.5 of Real Algebraic Geometry by Bochnak, Coste, Roy.
Let $R$ be a real closed field, $Asubset R^m$ and $Bsubset R^n$ semialgebraic sets, and $f:Ato B$ a semialgebraic map with graph $Gsubset Atimes B$. Let $K$ be a real closed extension of $R$, and denote the extension of a semialgebraic set $S$ defined over $R$ to $K$ as $S_K$.
Proposition 5.3.5
i) The semialgebraic set $A$ is open (resp. closed) in $R^m$ iff $A_K$ is open (resp. closed) in $K^m$. More generally, $clos(A_K)=(clos(A))_K$.
ii) The semialgebraic mapping $f$ is continuous iff $f_k$ is continuous.
I do not see how my statement follows. Does anyone have experience with applying this principle to this kind of situation? Or can you think of another approach leading to a proof of the claim?
algebraic-geometry real-algebraic-geometry
edited Nov 28 '18 at 2:30
KReiser
9,57721435
asked Nov 27 '18 at 10:38
Levi RyffelLevi Ryffel
313
$endgroup$
I want to use the following statement concerning varieties, but I do not know why it is true.
Claim. Let $V subset mathbb{C}^n$ be a variety defined over $mathbb{Q}$. Then the set $V cap bar{mathbb{Q}}$ is dense in $V$ with respect to the Hausdorff topology (not the Zariski topology).
Here, $bar{mathbb{Q}}$ denotes the algebraic closure of the rationals $mathbb{Q}$.
It was pointed out to me that one can show this using the so-called Tarski-Seidenberg Principle, in particular using Proposition 5.3.5 of Real Algebraic Geometry by Bochnak, Coste, Roy.
Let $R$ be a real closed field, $Asubset R^m$ and $Bsubset R^n$ semialgebraic sets, and $f:Ato B$ a semialgebraic map with graph $Gsubset Atimes B$. Let $K$ be a real closed extension of $R$, and denote the extension of a semialgebraic set $S$ defined over $R$ to $K$ as $S_K$.
Proposition 5.3.5
i) The semialgebraic set $A$ is open (resp. closed) in $R^m$ iff $A_K$ is open (resp. closed) in $K^m$. More generally, $clos(A_K)=(clos(A))_K$.
ii) The semialgebraic mapping $f$ is continuous iff $f_k$ is continuous.
I do not see how my statement follows. Does anyone have experience with applying this principle to this kind of situation? Or can you think of another approach leading to a proof of the claim?
algebraic-geometry real-algebraic-geometry
algebraic-geometry real-algebraic-geometry
edited Nov 28 '18 at 2:30
KReiser
9,57721435
asked Nov 27 '18 at 10:38
Levi RyffelLevi Ryffel
313
edited Nov 28 '18 at 2:30
KReiser
9,57721435
asked Nov 27 '18 at 10:38
Levi RyffelLevi Ryffel
313
edited Nov 28 '18 at 2:30
KReiser
9,57721435
edited Nov 28 '18 at 2:30
KReiser
9,57721435
edited Nov 28 '18 at 2:30
KReiser
9,57721435
9,57721435
asked Nov 27 '18 at 10:38
Levi RyffelLevi Ryffel
313
asked Nov 27 '18 at 10:38
Levi RyffelLevi Ryffel
313
asked Nov 27 '18 at 10:38
Levi RyffelLevi Ryffel
313
313
1
$begingroup$
I've edited the question to include the proposition you reference. In the future, please include information like this in the question statement - it will help other users write good answers to your question (and not everyone has this book immediately on hand!).
$endgroup$
– KReiser
Nov 28 '18 at 0:04
$begingroup$
You can pick a transcendental basis to put the function field in the form $K(V) cong K(u_1,ldots,u_m)[z_1,ldots,z_l]/J$ ?
$endgroup$
– reuns
Nov 28 '18 at 3:29
$begingroup$
@reuns Is $overline{mathbb{Q}}$ even real closed field?. Or one should apply the theorem to different fields?
$endgroup$
– random123
Nov 28 '18 at 5:50
1
$begingroup$
$overline{Bbb Q}$ is not real closed (it has $i$, for instance). Techniques from semialgebraic geometry will need a slight adjustment here. The right way around this is to decompose in to real and imaginary parts, then show that the $Bbb R_{alg}$ points are dense in the $Bbb R$ points of $V$ regarded as a semialgebraic set in $Bbb R^{2n}$. This is a place where you may use an extension of real closed fields: $Bbb R_{alg} subset Bbb R$. Unfortunately I don't have time to write a full solution right now.
$endgroup$
– KReiser
Nov 28 '18 at 6:10
1
$begingroup$
@random123 what? I don't see how that's relevant here. What I'm saying is that one writes the coordinates $z_j$ of $Bbb C^n$ as $z_j=x_j+iy_j$ and considers $V$ as a semialgebraic set in $Bbb R^{2n}$ with coordinates $x_j$ and $y_j$. Now the $overline{Bbb Q}$ points of the original $V$ are points in the semialgebraic set in $Bbb R^{2n}$ which have all their coordinates in $Bbb R_{alg}$ while the $Bbb C$ points of the original $V$ are all points in the semialgebraic set.
$endgroup$
– KReiser
Nov 28 '18 at 9:18
|
show 4 more comments
1
$begingroup$
I've edited the question to include the proposition you reference. In the future, please include information like this in the question statement - it will help other users write good answers to your question (and not everyone has this book immediately on hand!).
$endgroup$
– KReiser
Nov 28 '18 at 0:04
$begingroup$
You can pick a transcendental basis to put the function field in the form $K(V) cong K(u_1,ldots,u_m)[z_1,ldots,z_l]/J$ ?
$endgroup$
– reuns
Nov 28 '18 at 3:29
$begingroup$
@reuns Is $overline{mathbb{Q}}$ even real closed field?. Or one should apply the theorem to different fields?
$endgroup$
– random123
Nov 28 '18 at 5:50
1
$begingroup$
$overline{Bbb Q}$ is not real closed (it has $i$, for instance). Techniques from semialgebraic geometry will need a slight adjustment here. The right way around this is to decompose in to real and imaginary parts, then show that the $Bbb R_{alg}$ points are dense in the $Bbb R$ points of $V$ regarded as a semialgebraic set in $Bbb R^{2n}$. This is a place where you may use an extension of real closed fields: $Bbb R_{alg} subset Bbb R$. Unfortunately I don't have time to write a full solution right now.
$endgroup$
– KReiser
Nov 28 '18 at 6:10
1
$begingroup$
@random123 what? I don't see how that's relevant here. What I'm saying is that one writes the coordinates $z_j$ of $Bbb C^n$ as $z_j=x_j+iy_j$ and considers $V$ as a semialgebraic set in $Bbb R^{2n}$ with coordinates $x_j$ and $y_j$. Now the $overline{Bbb Q}$ points of the original $V$ are points in the semialgebraic set in $Bbb R^{2n}$ which have all their coordinates in $Bbb R_{alg}$ while the $Bbb C$ points of the original $V$ are all points in the semialgebraic set.
$endgroup$
– KReiser
Nov 28 '18 at 9:18
1
1
$begingroup$
I've edited the question to include the proposition you reference. In the future, please include information like this in the question statement - it will help other users write good answers to your question (and not everyone has this book immediately on hand!).
$endgroup$
– KReiser
Nov 28 '18 at 0:04
$begingroup$
I've edited the question to include the proposition you reference. In the future, please include information like this in the question statement - it will help other users write good answers to your question (and not everyone has this book immediately on hand!).
$endgroup$
– KReiser
Nov 28 '18 at 0:04
$begingroup$
You can pick a transcendental basis to put the function field in the form $K(V) cong K(u_1,ldots,u_m)[z_1,ldots,z_l]/J$ ?
$endgroup$
– reuns
Nov 28 '18 at 3:29
$begingroup$
You can pick a transcendental basis to put the function field in the form $K(V) cong K(u_1,ldots,u_m)[z_1,ldots,z_l]/J$ ?
$endgroup$
– reuns
Nov 28 '18 at 3:29
$begingroup$
@reuns Is $overline{mathbb{Q}}$ even real closed field?. Or one should apply the theorem to different fields?
$endgroup$
– random123
Nov 28 '18 at 5:50
$begingroup$
@reuns Is $overline{mathbb{Q}}$ even real closed field?. Or one should apply the theorem to different fields?
$endgroup$
– random123
Nov 28 '18 at 5:50
1
1
$begingroup$
$overline{Bbb Q}$ is not real closed (it has $i$, for instance). Techniques from semialgebraic geometry will need a slight adjustment here. The right way around this is to decompose in to real and imaginary parts, then show that the $Bbb R_{alg}$ points are dense in the $Bbb R$ points of $V$ regarded as a semialgebraic set in $Bbb R^{2n}$. This is a place where you may use an extension of real closed fields: $Bbb R_{alg} subset Bbb R$. Unfortunately I don't have time to write a full solution right now.
$endgroup$
– KReiser
Nov 28 '18 at 6:10
$begingroup$
$overline{Bbb Q}$ is not real closed (it has $i$, for instance). Techniques from semialgebraic geometry will need a slight adjustment here. The right way around this is to decompose in to real and imaginary parts, then show that the $Bbb R_{alg}$ points are dense in the $Bbb R$ points of $V$ regarded as a semialgebraic set in $Bbb R^{2n}$. This is a place where you may use an extension of real closed fields: $Bbb R_{alg} subset Bbb R$. Unfortunately I don't have time to write a full solution right now.
$endgroup$
– KReiser
Nov 28 '18 at 6:10
1
1
$begingroup$
@random123 what? I don't see how that's relevant here. What I'm saying is that one writes the coordinates $z_j$ of $Bbb C^n$ as $z_j=x_j+iy_j$ and considers $V$ as a semialgebraic set in $Bbb R^{2n}$ with coordinates $x_j$ and $y_j$. Now the $overline{Bbb Q}$ points of the original $V$ are points in the semialgebraic set in $Bbb R^{2n}$ which have all their coordinates in $Bbb R_{alg}$ while the $Bbb C$ points of the original $V$ are all points in the semialgebraic set.
$endgroup$
– KReiser
Nov 28 '18 at 9:18
$begingroup$
@random123 what? I don't see how that's relevant here. What I'm saying is that one writes the coordinates $z_j$ of $Bbb C^n$ as $z_j=x_j+iy_j$ and considers $V$ as a semialgebraic set in $Bbb R^{2n}$ with coordinates $x_j$ and $y_j$. Now the $overline{Bbb Q}$ points of the original $V$ are points in the semialgebraic set in $Bbb R^{2n}$ which have all their coordinates in $Bbb R_{alg}$ while the $Bbb C$ points of the original $V$ are all points in the semialgebraic set.
$endgroup$
– KReiser
Nov 28 '18 at 9:18
|
show 4 more comments
1 Answer
1
active
oldest
votes
$begingroup$
A proof originally developed in the comments:
Let $Bbb R_{alg}$ denote the set of real algebraic numbers. Let $z_j$ be the coordinates on $Bbb C^n$. Write $z_j=x_j+iy_j$ which identifies $Bbb C^n$ with $Bbb R^{2n}$, and identifies the $overline{Bbb Q}$ points of $V$ with the $Bbb R_{alg}$ points of the semialgebraic set given by all points of $V$ under the identification of $Bbb C^n$ with $Bbb R^{2n}$.
From here, if one can show that the $Bbb R_{alg}$ points of a semialgebraic set $Ssubset Bbb R_{alg}^n$ are dense in the $Bbb R$ points, then everything is fine. By a cell decomposition, we may obtain a finite list of semialgebraic homeomorphisms of $(0,1)^d$ with semialgebraic subsets of $S$ with each map defined over $Bbb R_{alg}$. Then the problem reduces to showing that the $Bbb R_{alg}$ points of $(0,1)^d$ are dense in the $Bbb R$ points of $(0,1)^d$, which is clear as $Bbb Qsubset Bbb R_{alg}$ and the $Bbb Q$ points are clearly dense in $(0,1)^d$.
This feels a little inelegant, but it works.
Alternatively, picking up from the point where we need to prove that the $Bbb R_{alg}$ points are dense in the $Bbb R$ points: Suppose there was a point in $S(Bbb R)$ which was not in the closure of $S(Bbb R_{alg})$. Then this point cannot be in any of the semialgebraic components of $S(Bbb R_{alg})$, so by careful selecting of components of $S$, one would obtain a set which is semialgebraically connected over $Bbb R_{alg}$ but semialgebraically disconnected over $Bbb R$, which violates Proposition 5.3.6, a direct corollary of 5.3.5. (I should also mention the authors of the book prove this claim essentially by cell decomposition, so it's the same argument as the first part of this post, just with a different ending.)
answered Nov 28 '18 at 9:43
KReiserKReiser
9,57721435
$endgroup$
1
$begingroup$
"and and the ℚ-points are clearly dense in (0,1)^d." Why?
$endgroup$
– Asal Beag Dubh
Nov 28 '18 at 15:40
$begingroup$
@AsalBeagDubh because $Bbb Q$ is dense in $Bbb R$? This is a standard fact from real analysis.
$endgroup$
– KReiser
Nov 28 '18 at 18:58
$begingroup$
An affine elliptic curve can have no rational points (but a continuum of real points)
$endgroup$
– reuns
Nov 29 '18 at 4:51
$begingroup$
@reuns and...? The rational points in $(0,1)^d$ are not necessarily sent to rational points by the homeomorphisms from the cell decomposition: as stated the maps are defined over $Bbb R_{alg}$ (and even if one defines them over $Bbb Q$, they aren't even necessarily polynomial mappings, so the result you quote is not necessarily a contradiction). But it is true that the image of a dense set under a homeomorphism is again a dense set.
$endgroup$
– KReiser
Nov 29 '18 at 5:17
$begingroup$
What is it ? To me the density of algebraic points is obtained from a $overline{mathbb{Q}}$ birational map obtained from $overline{mathbb{Q}}(V) cong overline{mathbb{Q}}(u_1,ldots,u_m)[z_1,ldots,z_l]/J= overline{mathbb{Q}}(S)$ and the points in $S(overline{mathbb{Q}})$ are easy to find and clearly dense in $S(mathbb{C})$.
$endgroup$
– reuns
Nov 29 '18 at 5:22
|
show 6 more comments
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1 Answer
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$begingroup$
A proof originally developed in the comments:
Let $Bbb R_{alg}$ denote the set of real algebraic numbers. Let $z_j$ be the coordinates on $Bbb C^n$. Write $z_j=x_j+iy_j$ which identifies $Bbb C^n$ with $Bbb R^{2n}$, and identifies the $overline{Bbb Q}$ points of $V$ with the $Bbb R_{alg}$ points of the semialgebraic set given by all points of $V$ under the identification of $Bbb C^n$ with $Bbb R^{2n}$.
From here, if one can show that the $Bbb R_{alg}$ points of a semialgebraic set $Ssubset Bbb R_{alg}^n$ are dense in the $Bbb R$ points, then everything is fine. By a cell decomposition, we may obtain a finite list of semialgebraic homeomorphisms of $(0,1)^d$ with semialgebraic subsets of $S$ with each map defined over $Bbb R_{alg}$. Then the problem reduces to showing that the $Bbb R_{alg}$ points of $(0,1)^d$ are dense in the $Bbb R$ points of $(0,1)^d$, which is clear as $Bbb Qsubset Bbb R_{alg}$ and the $Bbb Q$ points are clearly dense in $(0,1)^d$.
This feels a little inelegant, but it works.
Alternatively, picking up from the point where we need to prove that the $Bbb R_{alg}$ points are dense in the $Bbb R$ points: Suppose there was a point in $S(Bbb R)$ which was not in the closure of $S(Bbb R_{alg})$. Then this point cannot be in any of the semialgebraic components of $S(Bbb R_{alg})$, so by careful selecting of components of $S$, one would obtain a set which is semialgebraically connected over $Bbb R_{alg}$ but semialgebraically disconnected over $Bbb R$, which violates Proposition 5.3.6, a direct corollary of 5.3.5. (I should also mention the authors of the book prove this claim essentially by cell decomposition, so it's the same argument as the first part of this post, just with a different ending.)
answered Nov 28 '18 at 9:43
KReiserKReiser
9,57721435
$endgroup$
1
$begingroup$
"and and the ℚ-points are clearly dense in (0,1)^d." Why?
$endgroup$
– Asal Beag Dubh
Nov 28 '18 at 15:40
$begingroup$
@AsalBeagDubh because $Bbb Q$ is dense in $Bbb R$? This is a standard fact from real analysis.
$endgroup$
– KReiser
Nov 28 '18 at 18:58
$begingroup$
An affine elliptic curve can have no rational points (but a continuum of real points)
$endgroup$
– reuns
Nov 29 '18 at 4:51
$begingroup$
@reuns and...? The rational points in $(0,1)^d$ are not necessarily sent to rational points by the homeomorphisms from the cell decomposition: as stated the maps are defined over $Bbb R_{alg}$ (and even if one defines them over $Bbb Q$, they aren't even necessarily polynomial mappings, so the result you quote is not necessarily a contradiction). But it is true that the image of a dense set under a homeomorphism is again a dense set.
$endgroup$
– KReiser
Nov 29 '18 at 5:17
$begingroup$
What is it ? To me the density of algebraic points is obtained from a $overline{mathbb{Q}}$ birational map obtained from $overline{mathbb{Q}}(V) cong overline{mathbb{Q}}(u_1,ldots,u_m)[z_1,ldots,z_l]/J= overline{mathbb{Q}}(S)$ and the points in $S(overline{mathbb{Q}})$ are easy to find and clearly dense in $S(mathbb{C})$.
$endgroup$
– reuns
Nov 29 '18 at 5:22
|
show 6 more comments
$begingroup$
A proof originally developed in the comments:
Let $Bbb R_{alg}$ denote the set of real algebraic numbers. Let $z_j$ be the coordinates on $Bbb C^n$. Write $z_j=x_j+iy_j$ which identifies $Bbb C^n$ with $Bbb R^{2n}$, and identifies the $overline{Bbb Q}$ points of $V$ with the $Bbb R_{alg}$ points of the semialgebraic set given by all points of $V$ under the identification of $Bbb C^n$ with $Bbb R^{2n}$.
From here, if one can show that the $Bbb R_{alg}$ points of a semialgebraic set $Ssubset Bbb R_{alg}^n$ are dense in the $Bbb R$ points, then everything is fine. By a cell decomposition, we may obtain a finite list of semialgebraic homeomorphisms of $(0,1)^d$ with semialgebraic subsets of $S$ with each map defined over $Bbb R_{alg}$. Then the problem reduces to showing that the $Bbb R_{alg}$ points of $(0,1)^d$ are dense in the $Bbb R$ points of $(0,1)^d$, which is clear as $Bbb Qsubset Bbb R_{alg}$ and the $Bbb Q$ points are clearly dense in $(0,1)^d$.
This feels a little inelegant, but it works.
Alternatively, picking up from the point where we need to prove that the $Bbb R_{alg}$ points are dense in the $Bbb R$ points: Suppose there was a point in $S(Bbb R)$ which was not in the closure of $S(Bbb R_{alg})$. Then this point cannot be in any of the semialgebraic components of $S(Bbb R_{alg})$, so by careful selecting of components of $S$, one would obtain a set which is semialgebraically connected over $Bbb R_{alg}$ but semialgebraically disconnected over $Bbb R$, which violates Proposition 5.3.6, a direct corollary of 5.3.5. (I should also mention the authors of the book prove this claim essentially by cell decomposition, so it's the same argument as the first part of this post, just with a different ending.)
answered Nov 28 '18 at 9:43
KReiserKReiser
9,57721435
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1
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"and and the ℚ-points are clearly dense in (0,1)^d." Why?
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– Asal Beag Dubh
Nov 28 '18 at 15:40
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@AsalBeagDubh because $Bbb Q$ is dense in $Bbb R$? This is a standard fact from real analysis.
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– KReiser
Nov 28 '18 at 18:58
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An affine elliptic curve can have no rational points (but a continuum of real points)
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– reuns
Nov 29 '18 at 4:51
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@reuns and...? The rational points in $(0,1)^d$ are not necessarily sent to rational points by the homeomorphisms from the cell decomposition: as stated the maps are defined over $Bbb R_{alg}$ (and even if one defines them over $Bbb Q$, they aren't even necessarily polynomial mappings, so the result you quote is not necessarily a contradiction). But it is true that the image of a dense set under a homeomorphism is again a dense set.
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– KReiser
Nov 29 '18 at 5:17
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What is it ? To me the density of algebraic points is obtained from a $overline{mathbb{Q}}$ birational map obtained from $overline{mathbb{Q}}(V) cong overline{mathbb{Q}}(u_1,ldots,u_m)[z_1,ldots,z_l]/J= overline{mathbb{Q}}(S)$ and the points in $S(overline{mathbb{Q}})$ are easy to find and clearly dense in $S(mathbb{C})$.
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– reuns
Nov 29 '18 at 5:22
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show 6 more comments
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A proof originally developed in the comments:
Let $Bbb R_{alg}$ denote the set of real algebraic numbers. Let $z_j$ be the coordinates on $Bbb C^n$. Write $z_j=x_j+iy_j$ which identifies $Bbb C^n$ with $Bbb R^{2n}$, and identifies the $overline{Bbb Q}$ points of $V$ with the $Bbb R_{alg}$ points of the semialgebraic set given by all points of $V$ under the identification of $Bbb C^n$ with $Bbb R^{2n}$.
From here, if one can show that the $Bbb R_{alg}$ points of a semialgebraic set $Ssubset Bbb R_{alg}^n$ are dense in the $Bbb R$ points, then everything is fine. By a cell decomposition, we may obtain a finite list of semialgebraic homeomorphisms of $(0,1)^d$ with semialgebraic subsets of $S$ with each map defined over $Bbb R_{alg}$. Then the problem reduces to showing that the $Bbb R_{alg}$ points of $(0,1)^d$ are dense in the $Bbb R$ points of $(0,1)^d$, which is clear as $Bbb Qsubset Bbb R_{alg}$ and the $Bbb Q$ points are clearly dense in $(0,1)^d$.
This feels a little inelegant, but it works.
Alternatively, picking up from the point where we need to prove that the $Bbb R_{alg}$ points are dense in the $Bbb R$ points: Suppose there was a point in $S(Bbb R)$ which was not in the closure of $S(Bbb R_{alg})$. Then this point cannot be in any of the semialgebraic components of $S(Bbb R_{alg})$, so by careful selecting of components of $S$, one would obtain a set which is semialgebraically connected over $Bbb R_{alg}$ but semialgebraically disconnected over $Bbb R$, which violates Proposition 5.3.6, a direct corollary of 5.3.5. (I should also mention the authors of the book prove this claim essentially by cell decomposition, so it's the same argument as the first part of this post, just with a different ending.)
answered Nov 28 '18 at 9:43
KReiserKReiser
9,57721435
$endgroup$
A proof originally developed in the comments:
Let $Bbb R_{alg}$ denote the set of real algebraic numbers. Let $z_j$ be the coordinates on $Bbb C^n$. Write $z_j=x_j+iy_j$ which identifies $Bbb C^n$ with $Bbb R^{2n}$, and identifies the $overline{Bbb Q}$ points of $V$ with the $Bbb R_{alg}$ points of the semialgebraic set given by all points of $V$ under the identification of $Bbb C^n$ with $Bbb R^{2n}$.
From here, if one can show that the $Bbb R_{alg}$ points of a semialgebraic set $Ssubset Bbb R_{alg}^n$ are dense in the $Bbb R$ points, then everything is fine. By a cell decomposition, we may obtain a finite list of semialgebraic homeomorphisms of $(0,1)^d$ with semialgebraic subsets of $S$ with each map defined over $Bbb R_{alg}$. Then the problem reduces to showing that the $Bbb R_{alg}$ points of $(0,1)^d$ are dense in the $Bbb R$ points of $(0,1)^d$, which is clear as $Bbb Qsubset Bbb R_{alg}$ and the $Bbb Q$ points are clearly dense in $(0,1)^d$.
This feels a little inelegant, but it works.
Alternatively, picking up from the point where we need to prove that the $Bbb R_{alg}$ points are dense in the $Bbb R$ points: Suppose there was a point in $S(Bbb R)$ which was not in the closure of $S(Bbb R_{alg})$. Then this point cannot be in any of the semialgebraic components of $S(Bbb R_{alg})$, so by careful selecting of components of $S$, one would obtain a set which is semialgebraically connected over $Bbb R_{alg}$ but semialgebraically disconnected over $Bbb R$, which violates Proposition 5.3.6, a direct corollary of 5.3.5. (I should also mention the authors of the book prove this claim essentially by cell decomposition, so it's the same argument as the first part of this post, just with a different ending.)
answered Nov 28 '18 at 9:43
KReiserKReiser
9,57721435
answered Nov 28 '18 at 9:43
KReiserKReiser
9,57721435
answered Nov 28 '18 at 9:43
KReiserKReiser
9,57721435
answered Nov 28 '18 at 9:43
KReiserKReiser
9,57721435
9,57721435
1
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"and and the ℚ-points are clearly dense in (0,1)^d." Why?
$endgroup$
– Asal Beag Dubh
Nov 28 '18 at 15:40
$begingroup$
@AsalBeagDubh because $Bbb Q$ is dense in $Bbb R$? This is a standard fact from real analysis.
$endgroup$
– KReiser
Nov 28 '18 at 18:58
$begingroup$
An affine elliptic curve can have no rational points (but a continuum of real points)
$endgroup$
– reuns
Nov 29 '18 at 4:51
$begingroup$
@reuns and...? The rational points in $(0,1)^d$ are not necessarily sent to rational points by the homeomorphisms from the cell decomposition: as stated the maps are defined over $Bbb R_{alg}$ (and even if one defines them over $Bbb Q$, they aren't even necessarily polynomial mappings, so the result you quote is not necessarily a contradiction). But it is true that the image of a dense set under a homeomorphism is again a dense set.
$endgroup$
– KReiser
Nov 29 '18 at 5:17
$begingroup$
What is it ? To me the density of algebraic points is obtained from a $overline{mathbb{Q}}$ birational map obtained from $overline{mathbb{Q}}(V) cong overline{mathbb{Q}}(u_1,ldots,u_m)[z_1,ldots,z_l]/J= overline{mathbb{Q}}(S)$ and the points in $S(overline{mathbb{Q}})$ are easy to find and clearly dense in $S(mathbb{C})$.
$endgroup$
– reuns
Nov 29 '18 at 5:22
|
show 6 more comments
1
$begingroup$
"and and the ℚ-points are clearly dense in (0,1)^d." Why?
$endgroup$
– Asal Beag Dubh
Nov 28 '18 at 15:40
$begingroup$
@AsalBeagDubh because $Bbb Q$ is dense in $Bbb R$? This is a standard fact from real analysis.
$endgroup$
– KReiser
Nov 28 '18 at 18:58
$begingroup$
An affine elliptic curve can have no rational points (but a continuum of real points)
$endgroup$
– reuns
Nov 29 '18 at 4:51
$begingroup$
@reuns and...? The rational points in $(0,1)^d$ are not necessarily sent to rational points by the homeomorphisms from the cell decomposition: as stated the maps are defined over $Bbb R_{alg}$ (and even if one defines them over $Bbb Q$, they aren't even necessarily polynomial mappings, so the result you quote is not necessarily a contradiction). But it is true that the image of a dense set under a homeomorphism is again a dense set.
$endgroup$
– KReiser
Nov 29 '18 at 5:17
$begingroup$
What is it ? To me the density of algebraic points is obtained from a $overline{mathbb{Q}}$ birational map obtained from $overline{mathbb{Q}}(V) cong overline{mathbb{Q}}(u_1,ldots,u_m)[z_1,ldots,z_l]/J= overline{mathbb{Q}}(S)$ and the points in $S(overline{mathbb{Q}})$ are easy to find and clearly dense in $S(mathbb{C})$.
$endgroup$
– reuns
Nov 29 '18 at 5:22
1
1
$begingroup$
"and and the ℚ-points are clearly dense in (0,1)^d." Why?
$endgroup$
– Asal Beag Dubh
Nov 28 '18 at 15:40
$begingroup$
"and and the ℚ-points are clearly dense in (0,1)^d." Why?
$endgroup$
– Asal Beag Dubh
Nov 28 '18 at 15:40
$begingroup$
@AsalBeagDubh because $Bbb Q$ is dense in $Bbb R$? This is a standard fact from real analysis.
$endgroup$
– KReiser
Nov 28 '18 at 18:58
$begingroup$
@AsalBeagDubh because $Bbb Q$ is dense in $Bbb R$? This is a standard fact from real analysis.
$endgroup$
– KReiser
Nov 28 '18 at 18:58
$begingroup$
An affine elliptic curve can have no rational points (but a continuum of real points)
$endgroup$
– reuns
Nov 29 '18 at 4:51
$begingroup$
An affine elliptic curve can have no rational points (but a continuum of real points)
$endgroup$
– reuns
Nov 29 '18 at 4:51
$begingroup$
@reuns and...? The rational points in $(0,1)^d$ are not necessarily sent to rational points by the homeomorphisms from the cell decomposition: as stated the maps are defined over $Bbb R_{alg}$ (and even if one defines them over $Bbb Q$, they aren't even necessarily polynomial mappings, so the result you quote is not necessarily a contradiction). But it is true that the image of a dense set under a homeomorphism is again a dense set.
$endgroup$
– KReiser
Nov 29 '18 at 5:17
$begingroup$
@reuns and...? The rational points in $(0,1)^d$ are not necessarily sent to rational points by the homeomorphisms from the cell decomposition: as stated the maps are defined over $Bbb R_{alg}$ (and even if one defines them over $Bbb Q$, they aren't even necessarily polynomial mappings, so the result you quote is not necessarily a contradiction). But it is true that the image of a dense set under a homeomorphism is again a dense set.
$endgroup$
– KReiser
Nov 29 '18 at 5:17
$begingroup$
What is it ? To me the density of algebraic points is obtained from a $overline{mathbb{Q}}$ birational map obtained from $overline{mathbb{Q}}(V) cong overline{mathbb{Q}}(u_1,ldots,u_m)[z_1,ldots,z_l]/J= overline{mathbb{Q}}(S)$ and the points in $S(overline{mathbb{Q}})$ are easy to find and clearly dense in $S(mathbb{C})$.
$endgroup$
– reuns
Nov 29 '18 at 5:22
$begingroup$
What is it ? To me the density of algebraic points is obtained from a $overline{mathbb{Q}}$ birational map obtained from $overline{mathbb{Q}}(V) cong overline{mathbb{Q}}(u_1,ldots,u_m)[z_1,ldots,z_l]/J= overline{mathbb{Q}}(S)$ and the points in $S(overline{mathbb{Q}})$ are easy to find and clearly dense in $S(mathbb{C})$.
$endgroup$
– reuns
Nov 29 '18 at 5:22
|
show 6 more comments
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I've edited the question to include the proposition you reference. In the future, please include information like this in the question statement - it will help other users write good answers to your question (and not everyone has this book immediately on hand!).
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– KReiser
Nov 28 '18 at 0:04
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You can pick a transcendental basis to put the function field in the form $K(V) cong K(u_1,ldots,u_m)[z_1,ldots,z_l]/J$ ?
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– reuns
Nov 28 '18 at 3:29
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@reuns Is $overline{mathbb{Q}}$ even real closed field?. Or one should apply the theorem to different fields?
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– random123
Nov 28 '18 at 5:50
1
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$overline{Bbb Q}$ is not real closed (it has $i$, for instance). Techniques from semialgebraic geometry will need a slight adjustment here. The right way around this is to decompose in to real and imaginary parts, then show that the $Bbb R_{alg}$ points are dense in the $Bbb R$ points of $V$ regarded as a semialgebraic set in $Bbb R^{2n}$. This is a place where you may use an extension of real closed fields: $Bbb R_{alg} subset Bbb R$. Unfortunately I don't have time to write a full solution right now.
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– KReiser
Nov 28 '18 at 6:10
1
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@random123 what? I don't see how that's relevant here. What I'm saying is that one writes the coordinates $z_j$ of $Bbb C^n$ as $z_j=x_j+iy_j$ and considers $V$ as a semialgebraic set in $Bbb R^{2n}$ with coordinates $x_j$ and $y_j$. Now the $overline{Bbb Q}$ points of the original $V$ are points in the semialgebraic set in $Bbb R^{2n}$ which have all their coordinates in $Bbb R_{alg}$ while the $Bbb C$ points of the original $V$ are all points in the semialgebraic set.
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– KReiser
Nov 28 '18 at 9:18