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What does dense mean in algebraic geometry?

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$begingroup$

• If $f, g$ are regular functions on a variety X, then the set of points where $f-g=0$ is closed and dense, hence equal to $X. $ Why is it dense and why does the closure of such a set satisfy the equation?




• An open subset of a variety is dense (which means that the closure of an open subset is the whole variety). Why is it dense? Many thanks for your comment.

algebraic-geometry

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edited Mar 13 at 23:29

Dzoooks

905416

asked Mar 13 at 21:52

user249018user249018

435137

$endgroup$

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  $begingroup$
  To the latter question: this is because a variety is irreducible as a topological space. If $U$ is open, then $overline U$ and $Xsetminus U$ are closed and cover $X$, so either $U$ is empty or its closure is $X$. For the former, this is clearly not the case for arbitrary $f,g$. Are you perhaps working through a proof of the statement "if $f,g$ are equal on a dense set, they are equal everywhere"? If so, this set is dense by assumption.
  $endgroup$
  – Wojowu
  Mar 13 at 21:55
  

  
  
  
  


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  2
  

  
  
  
  
  
  

  
  $begingroup$
  Your title and question are mismatched - are you actually interested in the meaning of dense in addition to the questions in your post? Secondly, you're missing important assumptions about $X$ and the locus where $f=g$ in your first question. If $g=f+1$, certainly they can't be equal anywhere.
  $endgroup$
  – KReiser
  Mar 13 at 22:00
  

  
  
  
  

add a comment | 

1

$begingroup$

• If $f, g$ are regular functions on a variety X, then the set of points where $f-g=0$ is closed and dense, hence equal to $X. $ Why is it dense and why does the closure of such a set satisfy the equation?




• An open subset of a variety is dense (which means that the closure of an open subset is the whole variety). Why is it dense? Many thanks for your comment.

algebraic-geometry

share|cite|improve this question

edited Mar 13 at 23:29

Dzoooks

905416

asked Mar 13 at 21:52

user249018user249018

435137

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  To the latter question: this is because a variety is irreducible as a topological space. If $U$ is open, then $overline U$ and $Xsetminus U$ are closed and cover $X$, so either $U$ is empty or its closure is $X$. For the former, this is clearly not the case for arbitrary $f,g$. Are you perhaps working through a proof of the statement "if $f,g$ are equal on a dense set, they are equal everywhere"? If so, this set is dense by assumption.
  $endgroup$
  – Wojowu
  Mar 13 at 21:55
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Your title and question are mismatched - are you actually interested in the meaning of dense in addition to the questions in your post? Secondly, you're missing important assumptions about $X$ and the locus where $f=g$ in your first question. If $g=f+1$, certainly they can't be equal anywhere.
  $endgroup$
  – KReiser
  Mar 13 at 22:00
  

  
  
  
  

add a comment | 

$begingroup$

• If $f, g$ are regular functions on a variety X, then the set of points where $f-g=0$ is closed and dense, hence equal to $X. $ Why is it dense and why does the closure of such a set satisfy the equation?




• An open subset of a variety is dense (which means that the closure of an open subset is the whole variety). Why is it dense? Many thanks for your comment.

algebraic-geometry

share|cite|improve this question

edited Mar 13 at 23:29

Dzoooks

905416

asked Mar 13 at 21:52

user249018user249018

435137

$endgroup$

share|cite|improve this question

edited Mar 13 at 23:29

Dzoooks

905416

asked Mar 13 at 21:52

user249018user249018

435137

share|cite|improve this question

edited Mar 13 at 23:29

Dzoooks

905416

asked Mar 13 at 21:52

user249018user249018

435137

edited Mar 13 at 23:29

Dzoooks

905416

edited Mar 13 at 23:29

Dzoooks

905416

asked Mar 13 at 21:52

user249018user249018

435137

asked Mar 13 at 21:52

user249018user249018

435137