What does dense mean in algebraic geometry?Relation between varieties in the sense of Serre's FAC and...

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What does dense mean in algebraic geometry?


Relation between varieties in the sense of Serre's FAC and algebraic schemesAffine algebraic sets are quasi-projective varietiesA doubt in the proof of Prop. 1.10 of Hartshorne's Algebraic GeometryDefinition of quasiprojective variety by ShafarevichZariski dense implies classically dense?A rational function on a dense subset of a varietyWhat does Liu mean by “topological open/closed immersion” in his book “Algebraic Geometry and Arithmetic Curves”?Regular functions on a varietyregular functions are determined only up to open setsQuestion on Hartshorne II.6.1













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  • 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.











share|cite|improve this question











$endgroup$








  • 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
















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.











share|cite|improve this question











$endgroup$








  • 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














1












1








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.











share|cite|improve this question











$endgroup$





  • 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















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 13 at 23:29









Dzoooks

905416




905416










asked Mar 13 at 21:52









user249018user249018

435137




435137








  • 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














  • 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








1




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




$begingroup$
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




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




$begingroup$
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










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