Showing that the Zariski topology with polynomials in $Q[x_1,…,x_n]$ over $Bbb R^n$ is $T_0$
up vote
1
down vote
favorite
I am considering the Zariski topology over $Q[x_1,...,x_n$] as a topology over $Bbb R^n$, so that a set is Zariski-closed if there exists a set of polynomials $I$ such that $V= {r in Bbb R^n | f(r)=0 $ for all $f in I }$
First off a topological space is $T_0$ if for distinct $x,y$ we have an open set that contains one and not the other, which I believe is equivalent to having a closed set that does not contain one and not the other.
Next I consider three cases
1) If $x,y$ have rational components, then I just take a polynomial $f(x)=t-x_i$ for some $i$ where $x_i,y_i$ are different.
2) If one of $x,y$ has all rational components and the other isn't , then again I take $f(x)=t-x_i$ for some $i$ where $x_i,y_i$ are different.
3) If both $x,y$ have irrational components, I am stuck. I would need to find a polynomial that is zero for $x$ but not $y$ (or vice versa) which is tripping me up. Any hints appreciated.
Edited to reflect that this is over $Bbb R_n$
general-topology polynomials
asked Nov 17 at 22:20
IntegrateThis
1,6971717
add a comment |
up vote
1
down vote
favorite
I am considering the Zariski topology over $Q[x_1,...,x_n$] as a topology over $Bbb R^n$, so that a set is Zariski-closed if there exists a set of polynomials $I$ such that $V= {r in Bbb R^n | f(r)=0 $ for all $f in I }$
First off a topological space is $T_0$ if for distinct $x,y$ we have an open set that contains one and not the other, which I believe is equivalent to having a closed set that does not contain one and not the other.
Next I consider three cases
1) If $x,y$ have rational components, then I just take a polynomial $f(x)=t-x_i$ for some $i$ where $x_i,y_i$ are different.
2) If one of $x,y$ has all rational components and the other isn't , then again I take $f(x)=t-x_i$ for some $i$ where $x_i,y_i$ are different.
3) If both $x,y$ have irrational components, I am stuck. I would need to find a polynomial that is zero for $x$ but not $y$ (or vice versa) which is tripping me up. Any hints appreciated.
Edited to reflect that this is over $Bbb R_n$
general-topology polynomials
asked Nov 17 at 22:20
IntegrateThis
1,6971717
@SaucyO'Path yes I know, I just phrased it badly, will edit.
– IntegrateThis
Nov 17 at 22:35
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am considering the Zariski topology over $Q[x_1,...,x_n$] as a topology over $Bbb R^n$, so that a set is Zariski-closed if there exists a set of polynomials $I$ such that $V= {r in Bbb R^n | f(r)=0 $ for all $f in I }$
First off a topological space is $T_0$ if for distinct $x,y$ we have an open set that contains one and not the other, which I believe is equivalent to having a closed set that does not contain one and not the other.
Next I consider three cases
1) If $x,y$ have rational components, then I just take a polynomial $f(x)=t-x_i$ for some $i$ where $x_i,y_i$ are different.
2) If one of $x,y$ has all rational components and the other isn't , then again I take $f(x)=t-x_i$ for some $i$ where $x_i,y_i$ are different.
3) If both $x,y$ have irrational components, I am stuck. I would need to find a polynomial that is zero for $x$ but not $y$ (or vice versa) which is tripping me up. Any hints appreciated.
Edited to reflect that this is over $Bbb R_n$
general-topology polynomials
asked Nov 17 at 22:20
IntegrateThis
1,6971717
I am considering the Zariski topology over $Q[x_1,...,x_n$] as a topology over $Bbb R^n$, so that a set is Zariski-closed if there exists a set of polynomials $I$ such that $V= {r in Bbb R^n | f(r)=0 $ for all $f in I }$
First off a topological space is $T_0$ if for distinct $x,y$ we have an open set that contains one and not the other, which I believe is equivalent to having a closed set that does not contain one and not the other.
Next I consider three cases
1) If $x,y$ have rational components, then I just take a polynomial $f(x)=t-x_i$ for some $i$ where $x_i,y_i$ are different.
2) If one of $x,y$ has all rational components and the other isn't , then again I take $f(x)=t-x_i$ for some $i$ where $x_i,y_i$ are different.
3) If both $x,y$ have irrational components, I am stuck. I would need to find a polynomial that is zero for $x$ but not $y$ (or vice versa) which is tripping me up. Any hints appreciated.
Edited to reflect that this is over $Bbb R_n$
general-topology polynomials
general-topology polynomials
asked Nov 17 at 22:20
IntegrateThis
1,6971717
asked Nov 17 at 22:20
IntegrateThis
1,6971717
edited Nov 17 at 22:43
asked Nov 17 at 22:20
IntegrateThis
1,6971717
asked Nov 17 at 22:20
IntegrateThis
1,6971717
asked Nov 17 at 22:20
IntegrateThis
1,6971717
1,6971717
@SaucyO'Path yes I know, I just phrased it badly, will edit.
– IntegrateThis
Nov 17 at 22:35
add a comment |
@SaucyO'Path yes I know, I just phrased it badly, will edit.
– IntegrateThis
Nov 17 at 22:35
@SaucyO'Path yes I know, I just phrased it badly, will edit.
– IntegrateThis
Nov 17 at 22:35
@SaucyO'Path yes I know, I just phrased it badly, will edit.
– IntegrateThis
Nov 17 at 22:35
add a comment |
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
This is not true. In fact, if $(a_1,cdots,a_n)$ is a point such that its coordinates are algebraically independent over $Bbb Q$, then by definition the only closed set that contains it is the whole $Bbb R^n$. So, any two such points are topologically indistinguishable.
answered Nov 17 at 22:45
Saucy O'Path
5,5911526
Right so in $n=1$ no polynomial (let alone with rational coefficients) would be zero for $pi$ or $e$, so there can't be an open set containing one but not the other?
– IntegrateThis
Nov 17 at 22:49
1
@IntegrateThis There is no polynomial with rational coefficients. With real coefficients there are plenty, namely $X-pi$ and such.
– Saucy O'Path
Nov 17 at 22:50
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
This is not true. In fact, if $(a_1,cdots,a_n)$ is a point such that its coordinates are algebraically independent over $Bbb Q$, then by definition the only closed set that contains it is the whole $Bbb R^n$. So, any two such points are topologically indistinguishable.
answered Nov 17 at 22:45
Saucy O'Path
5,5911526
Right so in $n=1$ no polynomial (let alone with rational coefficients) would be zero for $pi$ or $e$, so there can't be an open set containing one but not the other?
– IntegrateThis
Nov 17 at 22:49
1
@IntegrateThis There is no polynomial with rational coefficients. With real coefficients there are plenty, namely $X-pi$ and such.
– Saucy O'Path
Nov 17 at 22:50
add a comment |
up vote
3
down vote
accepted
This is not true. In fact, if $(a_1,cdots,a_n)$ is a point such that its coordinates are algebraically independent over $Bbb Q$, then by definition the only closed set that contains it is the whole $Bbb R^n$. So, any two such points are topologically indistinguishable.
answered Nov 17 at 22:45
Saucy O'Path
5,5911526
Right so in $n=1$ no polynomial (let alone with rational coefficients) would be zero for $pi$ or $e$, so there can't be an open set containing one but not the other?
– IntegrateThis
Nov 17 at 22:49
1
@IntegrateThis There is no polynomial with rational coefficients. With real coefficients there are plenty, namely $X-pi$ and such.
– Saucy O'Path
Nov 17 at 22:50
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
This is not true. In fact, if $(a_1,cdots,a_n)$ is a point such that its coordinates are algebraically independent over $Bbb Q$, then by definition the only closed set that contains it is the whole $Bbb R^n$. So, any two such points are topologically indistinguishable.
answered Nov 17 at 22:45
Saucy O'Path
5,5911526
This is not true. In fact, if $(a_1,cdots,a_n)$ is a point such that its coordinates are algebraically independent over $Bbb Q$, then by definition the only closed set that contains it is the whole $Bbb R^n$. So, any two such points are topologically indistinguishable.
answered Nov 17 at 22:45
Saucy O'Path
5,5911526
answered Nov 17 at 22:45
Saucy O'Path
5,5911526
answered Nov 17 at 22:45
Saucy O'Path
5,5911526
answered Nov 17 at 22:45
Saucy O'Path
5,5911526
5,5911526
Right so in $n=1$ no polynomial (let alone with rational coefficients) would be zero for $pi$ or $e$, so there can't be an open set containing one but not the other?
– IntegrateThis
Nov 17 at 22:49
1
@IntegrateThis There is no polynomial with rational coefficients. With real coefficients there are plenty, namely $X-pi$ and such.
– Saucy O'Path
Nov 17 at 22:50
add a comment |
Right so in $n=1$ no polynomial (let alone with rational coefficients) would be zero for $pi$ or $e$, so there can't be an open set containing one but not the other?
– IntegrateThis
Nov 17 at 22:49
1
@IntegrateThis There is no polynomial with rational coefficients. With real coefficients there are plenty, namely $X-pi$ and such.
– Saucy O'Path
Nov 17 at 22:50
Right so in $n=1$ no polynomial (let alone with rational coefficients) would be zero for $pi$ or $e$, so there can't be an open set containing one but not the other?
– IntegrateThis
Nov 17 at 22:49
Right so in $n=1$ no polynomial (let alone with rational coefficients) would be zero for $pi$ or $e$, so there can't be an open set containing one but not the other?
– IntegrateThis
Nov 17 at 22:49
1
1
@IntegrateThis There is no polynomial with rational coefficients. With real coefficients there are plenty, namely $X-pi$ and such.
– Saucy O'Path
Nov 17 at 22:50
@IntegrateThis There is no polynomial with rational coefficients. With real coefficients there are plenty, namely $X-pi$ and such.
– Saucy O'Path
Nov 17 at 22:50
add a comment |
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3002888%2fshowing-that-the-zariski-topology-with-polynomials-in-qx-1-x-n-over-bb%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
@SaucyO'Path yes I know, I just phrased it badly, will edit.
– IntegrateThis
Nov 17 at 22:35