Show that ${Z(p):pin mathscr P}$ is a basis for the closed sets of some topology (Called the Zariski...












0












$begingroup$


Let $nin mathbb N$, and let $mathscr P$ denote the collection of all polynomials in $n$ variables. For $pin mathscr P$, let $Z(p)={(x_1,x_2,...,x_n)|p(x_1,x_2,...,x_n)=0}$.
Show that ${Z(p):pin mathscr P}$ is a basis for the closed sets of some topology (Called the Zariski topology) on $mathbb R^n$.



My attempt:-



I used the result for Basis for the closed sets of topology on X.



(I) We need to prove that $bigcap_{pin mathscr P}Z(p)=emptyset.$ $Z(p(x_1,x_2,...,x_n)=1)={(x_1,x_2,...,x_n)|p(x_1,x_2,...,x_n)=0}=emptyset$. Which implies $bigcap_{pin mathscr P}Z(p) subset Z(p(x_1,x_2,...,x_n)=1)=emptyset implies bigcap_{pin mathscr P}Z(p)=emptyset.$



(II) if $Z(p_0),Z(p_1)inmathscr X$ and $(x_1,x_2,...,x_n) notin Z(p_0)cup Z(p_1)$, then there is a $Z(p_2)inmathscr X$ such that $(x_1,x_2,...,x_n) notin Z(p_2)supseteq Z(p_0)cup Z(p_1)$.



$(x_1,x_2,...,x_n) notin Z(p_0)cup Z(p_1)implies p_0(x_1,x_2,...,x_n)neq0$ and $p_1(x_1,x_2,...,x_n)neq 0.$ We can choose $p_2=p_0 p_1$ such that $(x_1,x_2,...,x_n)notin Z(p_2)$. We know that $Z(p_0)cup Z(p_1) subset Z(p_2)$.



Have I done the proof correctly?






general-topology zariski-topology







share|cite|improve this question











$endgroup$












  • $begingroup$
    Define: Z(p = 1).
    $endgroup$
    – William Elliot
    Jan 13 at 12:28










  • $begingroup$
    $Z(p)={(x_1,x_2,...,x_n)|p(x_1,x_2,...,x_n)=0}$
    $endgroup$
    – Math geek
    Jan 13 at 12:33










  • $begingroup$
    When we substitute $p(x_1,x_2,...,x_n)=1$ we get $Z(p=1)$. which is an empty set.
    $endgroup$
    – Math geek
    Jan 13 at 12:34










  • $begingroup$
    If I define like this, is my proof correct?
    $endgroup$
    – Math geek
    Jan 13 at 12:48










  • $begingroup$
    No, Z is defined for polynomials, it is not defined for statements like p = 1. Thus Z(p = 1) is meaningless.
    $endgroup$
    – William Elliot
    Jan 13 at 23:21
















0












$begingroup$


Let $nin mathbb N$, and let $mathscr P$ denote the collection of all polynomials in $n$ variables. For $pin mathscr P$, let $Z(p)={(x_1,x_2,...,x_n)|p(x_1,x_2,...,x_n)=0}$.
Show that ${Z(p):pin mathscr P}$ is a basis for the closed sets of some topology (Called the Zariski topology) on $mathbb R^n$.



My attempt:-



I used the result for Basis for the closed sets of topology on X.



(I) We need to prove that $bigcap_{pin mathscr P}Z(p)=emptyset.$ $Z(p(x_1,x_2,...,x_n)=1)={(x_1,x_2,...,x_n)|p(x_1,x_2,...,x_n)=0}=emptyset$. Which implies $bigcap_{pin mathscr P}Z(p) subset Z(p(x_1,x_2,...,x_n)=1)=emptyset implies bigcap_{pin mathscr P}Z(p)=emptyset.$



(II) if $Z(p_0),Z(p_1)inmathscr X$ and $(x_1,x_2,...,x_n) notin Z(p_0)cup Z(p_1)$, then there is a $Z(p_2)inmathscr X$ such that $(x_1,x_2,...,x_n) notin Z(p_2)supseteq Z(p_0)cup Z(p_1)$.



$(x_1,x_2,...,x_n) notin Z(p_0)cup Z(p_1)implies p_0(x_1,x_2,...,x_n)neq0$ and $p_1(x_1,x_2,...,x_n)neq 0.$ We can choose $p_2=p_0 p_1$ such that $(x_1,x_2,...,x_n)notin Z(p_2)$. We know that $Z(p_0)cup Z(p_1) subset Z(p_2)$.



Have I done the proof correctly?






general-topology zariski-topology







share|cite|improve this question











$endgroup$












  • $begingroup$
    Define: Z(p = 1).
    $endgroup$
    – William Elliot
    Jan 13 at 12:28










  • $begingroup$
    $Z(p)={(x_1,x_2,...,x_n)|p(x_1,x_2,...,x_n)=0}$
    $endgroup$
    – Math geek
    Jan 13 at 12:33










  • $begingroup$
    When we substitute $p(x_1,x_2,...,x_n)=1$ we get $Z(p=1)$. which is an empty set.
    $endgroup$
    – Math geek
    Jan 13 at 12:34










  • $begingroup$
    If I define like this, is my proof correct?
    $endgroup$
    – Math geek
    Jan 13 at 12:48










  • $begingroup$
    No, Z is defined for polynomials, it is not defined for statements like p = 1. Thus Z(p = 1) is meaningless.
    $endgroup$
    – William Elliot
    Jan 13 at 23:21














0












0








0





$begingroup$


Let $nin mathbb N$, and let $mathscr P$ denote the collection of all polynomials in $n$ variables. For $pin mathscr P$, let $Z(p)={(x_1,x_2,...,x_n)|p(x_1,x_2,...,x_n)=0}$.
Show that ${Z(p):pin mathscr P}$ is a basis for the closed sets of some topology (Called the Zariski topology) on $mathbb R^n$.



My attempt:-



I used the result for Basis for the closed sets of topology on X.



(I) We need to prove that $bigcap_{pin mathscr P}Z(p)=emptyset.$ $Z(p(x_1,x_2,...,x_n)=1)={(x_1,x_2,...,x_n)|p(x_1,x_2,...,x_n)=0}=emptyset$. Which implies $bigcap_{pin mathscr P}Z(p) subset Z(p(x_1,x_2,...,x_n)=1)=emptyset implies bigcap_{pin mathscr P}Z(p)=emptyset.$



(II) if $Z(p_0),Z(p_1)inmathscr X$ and $(x_1,x_2,...,x_n) notin Z(p_0)cup Z(p_1)$, then there is a $Z(p_2)inmathscr X$ such that $(x_1,x_2,...,x_n) notin Z(p_2)supseteq Z(p_0)cup Z(p_1)$.



$(x_1,x_2,...,x_n) notin Z(p_0)cup Z(p_1)implies p_0(x_1,x_2,...,x_n)neq0$ and $p_1(x_1,x_2,...,x_n)neq 0.$ We can choose $p_2=p_0 p_1$ such that $(x_1,x_2,...,x_n)notin Z(p_2)$. We know that $Z(p_0)cup Z(p_1) subset Z(p_2)$.



Have I done the proof correctly?






general-topology zariski-topology







share|cite|improve this question











$endgroup$




Let $nin mathbb N$, and let $mathscr P$ denote the collection of all polynomials in $n$ variables. For $pin mathscr P$, let $Z(p)={(x_1,x_2,...,x_n)|p(x_1,x_2,...,x_n)=0}$.
Show that ${Z(p):pin mathscr P}$ is a basis for the closed sets of some topology (Called the Zariski topology) on $mathbb R^n$.



My attempt:-



I used the result for Basis for the closed sets of topology on X.



(I) We need to prove that $bigcap_{pin mathscr P}Z(p)=emptyset.$ $Z(p(x_1,x_2,...,x_n)=1)={(x_1,x_2,...,x_n)|p(x_1,x_2,...,x_n)=0}=emptyset$. Which implies $bigcap_{pin mathscr P}Z(p) subset Z(p(x_1,x_2,...,x_n)=1)=emptyset implies bigcap_{pin mathscr P}Z(p)=emptyset.$



(II) if $Z(p_0),Z(p_1)inmathscr X$ and $(x_1,x_2,...,x_n) notin Z(p_0)cup Z(p_1)$, then there is a $Z(p_2)inmathscr X$ such that $(x_1,x_2,...,x_n) notin Z(p_2)supseteq Z(p_0)cup Z(p_1)$.



$(x_1,x_2,...,x_n) notin Z(p_0)cup Z(p_1)implies p_0(x_1,x_2,...,x_n)neq0$ and $p_1(x_1,x_2,...,x_n)neq 0.$ We can choose $p_2=p_0 p_1$ such that $(x_1,x_2,...,x_n)notin Z(p_2)$. We know that $Z(p_0)cup Z(p_1) subset Z(p_2)$.



Have I done the proof correctly?






general-topology zariski-topology




general-topology zariski-topology






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 13 at 14:48









Martin Sleziak

45k10123277




45k10123277










asked Jan 13 at 11:07









Math geekMath geek

69111




69111












  • $begingroup$
    Define: Z(p = 1).
    $endgroup$
    – William Elliot
    Jan 13 at 12:28










  • $begingroup$
    $Z(p)={(x_1,x_2,...,x_n)|p(x_1,x_2,...,x_n)=0}$
    $endgroup$
    – Math geek
    Jan 13 at 12:33










  • $begingroup$
    When we substitute $p(x_1,x_2,...,x_n)=1$ we get $Z(p=1)$. which is an empty set.
    $endgroup$
    – Math geek
    Jan 13 at 12:34










  • $begingroup$
    If I define like this, is my proof correct?
    $endgroup$
    – Math geek
    Jan 13 at 12:48










  • $begingroup$
    No, Z is defined for polynomials, it is not defined for statements like p = 1. Thus Z(p = 1) is meaningless.
    $endgroup$
    – William Elliot
    Jan 13 at 23:21


















  • $begingroup$
    Define: Z(p = 1).
    $endgroup$
    – William Elliot
    Jan 13 at 12:28










  • $begingroup$
    $Z(p)={(x_1,x_2,...,x_n)|p(x_1,x_2,...,x_n)=0}$
    $endgroup$
    – Math geek
    Jan 13 at 12:33










  • $begingroup$
    When we substitute $p(x_1,x_2,...,x_n)=1$ we get $Z(p=1)$. which is an empty set.
    $endgroup$
    – Math geek
    Jan 13 at 12:34










  • $begingroup$
    If I define like this, is my proof correct?
    $endgroup$
    – Math geek
    Jan 13 at 12:48










  • $begingroup$
    No, Z is defined for polynomials, it is not defined for statements like p = 1. Thus Z(p = 1) is meaningless.
    $endgroup$
    – William Elliot
    Jan 13 at 23:21
















$begingroup$
Define: Z(p = 1).
$endgroup$
– William Elliot
Jan 13 at 12:28




$begingroup$
Define: Z(p = 1).
$endgroup$
– William Elliot
Jan 13 at 12:28












$begingroup$
$Z(p)={(x_1,x_2,...,x_n)|p(x_1,x_2,...,x_n)=0}$
$endgroup$
– Math geek
Jan 13 at 12:33




$begingroup$
$Z(p)={(x_1,x_2,...,x_n)|p(x_1,x_2,...,x_n)=0}$
$endgroup$
– Math geek
Jan 13 at 12:33












$begingroup$
When we substitute $p(x_1,x_2,...,x_n)=1$ we get $Z(p=1)$. which is an empty set.
$endgroup$
– Math geek
Jan 13 at 12:34




$begingroup$
When we substitute $p(x_1,x_2,...,x_n)=1$ we get $Z(p=1)$. which is an empty set.
$endgroup$
– Math geek
Jan 13 at 12:34












$begingroup$
If I define like this, is my proof correct?
$endgroup$
– Math geek
Jan 13 at 12:48




$begingroup$
If I define like this, is my proof correct?
$endgroup$
– Math geek
Jan 13 at 12:48












$begingroup$
No, Z is defined for polynomials, it is not defined for statements like p = 1. Thus Z(p = 1) is meaningless.
$endgroup$
– William Elliot
Jan 13 at 23:21




$begingroup$
No, Z is defined for polynomials, it is not defined for statements like p = 1. Thus Z(p = 1) is meaningless.
$endgroup$
– William Elliot
Jan 13 at 23:21










2 Answers
2






active

oldest

votes


















0












$begingroup$

The set of ${Z(I): I subseteq mathcal{P}}$, where $Z(I)=bigcap_{p in I} Z(p)$ is in fact the whole topology, not just a closed base for it, as I showed here.
The basis fact immediately follows, by the definition of a closed base (you can write all closed sets as intersection of base elements).






share|cite|improve this answer











$endgroup$













  • $begingroup$
    No, We want to prove ${Z(p_1),Z(p_2),...,Z(p_n),....}$ is a basic set.
    $endgroup$
    – Math geek
    Jan 14 at 0:46










  • $begingroup$
    @Mathgeek if $C$ is closed, then $C = Z(I)= bigcap {Z(p): p in I}$. BTW, don’t think the base is countable, it is not, though you seem to suggest so.
    $endgroup$
    – Henno Brandsma
    Jan 14 at 5:56



















0












$begingroup$

$Z(p) cup Z(q) = Z(pq)$.

Let 1 be the polynomial of $n$ variables with constant value 1.
$Z(1)$ is empty.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    what about the second property?
    $endgroup$
    – Math geek
    Jan 13 at 16:02










  • $begingroup$
    The first line proves it.
    $endgroup$
    – William Elliot
    Jan 13 at 23:16










  • $begingroup$
    @Mathgeek the product of two polynomials is a polynomial. So define $p_2=p_1 p_2$ in your proof attempt.
    $endgroup$
    – Henno Brandsma
    Jan 14 at 6:01










  • $begingroup$
    That is how I defined @HennoBrandsma. I defined $p_2=p_1p_0$.
    $endgroup$
    – Math geek
    Jan 14 at 6:20










  • $begingroup$
    In the second part of the proof.
    $endgroup$
    – Math geek
    Jan 14 at 6:20












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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

The set of ${Z(I): I subseteq mathcal{P}}$, where $Z(I)=bigcap_{p in I} Z(p)$ is in fact the whole topology, not just a closed base for it, as I showed here.
The basis fact immediately follows, by the definition of a closed base (you can write all closed sets as intersection of base elements).






share|cite|improve this answer











$endgroup$













  • $begingroup$
    No, We want to prove ${Z(p_1),Z(p_2),...,Z(p_n),....}$ is a basic set.
    $endgroup$
    – Math geek
    Jan 14 at 0:46










  • $begingroup$
    @Mathgeek if $C$ is closed, then $C = Z(I)= bigcap {Z(p): p in I}$. BTW, don’t think the base is countable, it is not, though you seem to suggest so.
    $endgroup$
    – Henno Brandsma
    Jan 14 at 5:56
















0












$begingroup$

The set of ${Z(I): I subseteq mathcal{P}}$, where $Z(I)=bigcap_{p in I} Z(p)$ is in fact the whole topology, not just a closed base for it, as I showed here.
The basis fact immediately follows, by the definition of a closed base (you can write all closed sets as intersection of base elements).






share|cite|improve this answer











$endgroup$













  • $begingroup$
    No, We want to prove ${Z(p_1),Z(p_2),...,Z(p_n),....}$ is a basic set.
    $endgroup$
    – Math geek
    Jan 14 at 0:46










  • $begingroup$
    @Mathgeek if $C$ is closed, then $C = Z(I)= bigcap {Z(p): p in I}$. BTW, don’t think the base is countable, it is not, though you seem to suggest so.
    $endgroup$
    – Henno Brandsma
    Jan 14 at 5:56














0












0








0





$begingroup$

The set of ${Z(I): I subseteq mathcal{P}}$, where $Z(I)=bigcap_{p in I} Z(p)$ is in fact the whole topology, not just a closed base for it, as I showed here.
The basis fact immediately follows, by the definition of a closed base (you can write all closed sets as intersection of base elements).






share|cite|improve this answer











$endgroup$



The set of ${Z(I): I subseteq mathcal{P}}$, where $Z(I)=bigcap_{p in I} Z(p)$ is in fact the whole topology, not just a closed base for it, as I showed here.
The basis fact immediately follows, by the definition of a closed base (you can write all closed sets as intersection of base elements).







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 13 at 15:30

























answered Jan 13 at 14:55









Henno BrandsmaHenno Brandsma

117k349127




117k349127












  • $begingroup$
    No, We want to prove ${Z(p_1),Z(p_2),...,Z(p_n),....}$ is a basic set.
    $endgroup$
    – Math geek
    Jan 14 at 0:46










  • $begingroup$
    @Mathgeek if $C$ is closed, then $C = Z(I)= bigcap {Z(p): p in I}$. BTW, don’t think the base is countable, it is not, though you seem to suggest so.
    $endgroup$
    – Henno Brandsma
    Jan 14 at 5:56


















  • $begingroup$
    No, We want to prove ${Z(p_1),Z(p_2),...,Z(p_n),....}$ is a basic set.
    $endgroup$
    – Math geek
    Jan 14 at 0:46










  • $begingroup$
    @Mathgeek if $C$ is closed, then $C = Z(I)= bigcap {Z(p): p in I}$. BTW, don’t think the base is countable, it is not, though you seem to suggest so.
    $endgroup$
    – Henno Brandsma
    Jan 14 at 5:56
















$begingroup$
No, We want to prove ${Z(p_1),Z(p_2),...,Z(p_n),....}$ is a basic set.
$endgroup$
– Math geek
Jan 14 at 0:46




$begingroup$
No, We want to prove ${Z(p_1),Z(p_2),...,Z(p_n),....}$ is a basic set.
$endgroup$
– Math geek
Jan 14 at 0:46












$begingroup$
@Mathgeek if $C$ is closed, then $C = Z(I)= bigcap {Z(p): p in I}$. BTW, don’t think the base is countable, it is not, though you seem to suggest so.
$endgroup$
– Henno Brandsma
Jan 14 at 5:56




$begingroup$
@Mathgeek if $C$ is closed, then $C = Z(I)= bigcap {Z(p): p in I}$. BTW, don’t think the base is countable, it is not, though you seem to suggest so.
$endgroup$
– Henno Brandsma
Jan 14 at 5:56











0












$begingroup$

$Z(p) cup Z(q) = Z(pq)$.

Let 1 be the polynomial of $n$ variables with constant value 1.
$Z(1)$ is empty.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    what about the second property?
    $endgroup$
    – Math geek
    Jan 13 at 16:02










  • $begingroup$
    The first line proves it.
    $endgroup$
    – William Elliot
    Jan 13 at 23:16










  • $begingroup$
    @Mathgeek the product of two polynomials is a polynomial. So define $p_2=p_1 p_2$ in your proof attempt.
    $endgroup$
    – Henno Brandsma
    Jan 14 at 6:01










  • $begingroup$
    That is how I defined @HennoBrandsma. I defined $p_2=p_1p_0$.
    $endgroup$
    – Math geek
    Jan 14 at 6:20










  • $begingroup$
    In the second part of the proof.
    $endgroup$
    – Math geek
    Jan 14 at 6:20
















0












$begingroup$

$Z(p) cup Z(q) = Z(pq)$.

Let 1 be the polynomial of $n$ variables with constant value 1.
$Z(1)$ is empty.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    what about the second property?
    $endgroup$
    – Math geek
    Jan 13 at 16:02










  • $begingroup$
    The first line proves it.
    $endgroup$
    – William Elliot
    Jan 13 at 23:16










  • $begingroup$
    @Mathgeek the product of two polynomials is a polynomial. So define $p_2=p_1 p_2$ in your proof attempt.
    $endgroup$
    – Henno Brandsma
    Jan 14 at 6:01










  • $begingroup$
    That is how I defined @HennoBrandsma. I defined $p_2=p_1p_0$.
    $endgroup$
    – Math geek
    Jan 14 at 6:20










  • $begingroup$
    In the second part of the proof.
    $endgroup$
    – Math geek
    Jan 14 at 6:20














0












0








0





$begingroup$

$Z(p) cup Z(q) = Z(pq)$.

Let 1 be the polynomial of $n$ variables with constant value 1.
$Z(1)$ is empty.






share|cite|improve this answer











$endgroup$



$Z(p) cup Z(q) = Z(pq)$.

Let 1 be the polynomial of $n$ variables with constant value 1.
$Z(1)$ is empty.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 14 at 5:58









Henno Brandsma

117k349127




117k349127










answered Jan 13 at 12:41









William ElliotWilliam Elliot

9,1962820




9,1962820












  • $begingroup$
    what about the second property?
    $endgroup$
    – Math geek
    Jan 13 at 16:02










  • $begingroup$
    The first line proves it.
    $endgroup$
    – William Elliot
    Jan 13 at 23:16










  • $begingroup$
    @Mathgeek the product of two polynomials is a polynomial. So define $p_2=p_1 p_2$ in your proof attempt.
    $endgroup$
    – Henno Brandsma
    Jan 14 at 6:01










  • $begingroup$
    That is how I defined @HennoBrandsma. I defined $p_2=p_1p_0$.
    $endgroup$
    – Math geek
    Jan 14 at 6:20










  • $begingroup$
    In the second part of the proof.
    $endgroup$
    – Math geek
    Jan 14 at 6:20


















  • $begingroup$
    what about the second property?
    $endgroup$
    – Math geek
    Jan 13 at 16:02










  • $begingroup$
    The first line proves it.
    $endgroup$
    – William Elliot
    Jan 13 at 23:16










  • $begingroup$
    @Mathgeek the product of two polynomials is a polynomial. So define $p_2=p_1 p_2$ in your proof attempt.
    $endgroup$
    – Henno Brandsma
    Jan 14 at 6:01










  • $begingroup$
    That is how I defined @HennoBrandsma. I defined $p_2=p_1p_0$.
    $endgroup$
    – Math geek
    Jan 14 at 6:20










  • $begingroup$
    In the second part of the proof.
    $endgroup$
    – Math geek
    Jan 14 at 6:20
















$begingroup$
what about the second property?
$endgroup$
– Math geek
Jan 13 at 16:02




$begingroup$
what about the second property?
$endgroup$
– Math geek
Jan 13 at 16:02












$begingroup$
The first line proves it.
$endgroup$
– William Elliot
Jan 13 at 23:16




$begingroup$
The first line proves it.
$endgroup$
– William Elliot
Jan 13 at 23:16












$begingroup$
@Mathgeek the product of two polynomials is a polynomial. So define $p_2=p_1 p_2$ in your proof attempt.
$endgroup$
– Henno Brandsma
Jan 14 at 6:01




$begingroup$
@Mathgeek the product of two polynomials is a polynomial. So define $p_2=p_1 p_2$ in your proof attempt.
$endgroup$
– Henno Brandsma
Jan 14 at 6:01












$begingroup$
That is how I defined @HennoBrandsma. I defined $p_2=p_1p_0$.
$endgroup$
– Math geek
Jan 14 at 6:20




$begingroup$
That is how I defined @HennoBrandsma. I defined $p_2=p_1p_0$.
$endgroup$
– Math geek
Jan 14 at 6:20












$begingroup$
In the second part of the proof.
$endgroup$
– Math geek
Jan 14 at 6:20




$begingroup$
In the second part of the proof.
$endgroup$
– Math geek
Jan 14 at 6:20


















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