Irreducible of finite Krull dimension implies quasi-compact? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)A closed point in the closure of any point in the closure of any point of an irreducible schemeWhen is an irreducible scheme quasi-compact?not locally of finite type implies not universally closed?If $X,Y$ are regular and of finite type over $S$, can $Xtimes _S Y$ be embedded into a regular $S$-scheme? Krull dimension in equivariant cohomologyKrull dimension and Morley rankQuasi-finite morphisms of stacksQuasi-compactness of irreducible separated scheme locally of finite typeWhen does the image of a morphism of schemes support scheme structure?Sufficient condition for quasi-compactness of scheme morphismSchemes monomorphing into affine scheme of dimension 1

Irreducible of finite Krull dimension implies quasi-compact?



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)A closed point in the closure of any point in the closure of any point of an irreducible schemeWhen is an irreducible scheme quasi-compact?not locally of finite type implies not universally closed?If $X,Y$ are regular and of finite type over $S$, can $Xtimes _S Y$ be embedded into a regular $S$-scheme? Krull dimension in equivariant cohomologyKrull dimension and Morley rankQuasi-finite morphisms of stacksQuasi-compactness of irreducible separated scheme locally of finite typeWhen does the image of a morphism of schemes support scheme structure?Sufficient condition for quasi-compactness of scheme morphismSchemes monomorphing into affine scheme of dimension 1










5












$begingroup$


Let $X$ be the underlying space of a scheme.



  • If $X$ is irreducible of finite Krull dimension, is it necessarily
    quasi-compact?

  • Is it necessarily Noetherian?

  • What if we assume not
    only that Krull dimension is finite but also that it is 1?









share|cite|improve this question









New contributor




schematic_ftm is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
















    5












    $begingroup$


    Let $X$ be the underlying space of a scheme.



    • If $X$ is irreducible of finite Krull dimension, is it necessarily
      quasi-compact?

    • Is it necessarily Noetherian?

    • What if we assume not
      only that Krull dimension is finite but also that it is 1?









    share|cite|improve this question









    New contributor




    schematic_ftm is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$














      5












      5








      5





      $begingroup$


      Let $X$ be the underlying space of a scheme.



      • If $X$ is irreducible of finite Krull dimension, is it necessarily
        quasi-compact?

      • Is it necessarily Noetherian?

      • What if we assume not
        only that Krull dimension is finite but also that it is 1?









      share|cite|improve this question









      New contributor




      schematic_ftm is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      Let $X$ be the underlying space of a scheme.



      • If $X$ is irreducible of finite Krull dimension, is it necessarily
        quasi-compact?

      • Is it necessarily Noetherian?

      • What if we assume not
        only that Krull dimension is finite but also that it is 1?






      at.algebraic-topology gn.general-topology schemes






      share|cite|improve this question









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      schematic_ftm is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question









      New contributor




      schematic_ftm is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      share|cite|improve this question




      share|cite|improve this question








      edited 3 hours ago









      András Bátkai

      3,84142342




      3,84142342






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      asked 3 hours ago









      schematic_ftmschematic_ftm

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          1 Answer
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          active

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          10












          $begingroup$

          The answer is no for all these questions. Take the line with infinite origins: the scheme obtained by gluing an infinite amount of copies of $mathbbA^1$ along the open subsets $mathbbG_m$. This has Krull dimension 1 (there are only closed points and the unique generic point) and it is irreducible (the only proper nonempty closed subsets are the closed points) but it is not quasi-compact (it contains an infinite discrete set), and so in particular not Noetherian.






          share|cite|improve this answer











          $endgroup$








          • 1




            $begingroup$
            Denis Nardin, I am not sure about the second paragraph. I thought nilpotent thickening never affects the underlying topological space. So while your nilpotent thickening gives a non-locally Noetherian scheme, does it give a non-locally Noetherian space (about which is the question)?
            $endgroup$
            – Stepan Banach
            2 hours ago






          • 1




            $begingroup$
            @StepanBanach Ah sorry, I was thinking of a non-Noetherian scheme. Let me correct it.
            $endgroup$
            – Denis Nardin
            2 hours ago











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          1 Answer
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          active

          oldest

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          10












          $begingroup$

          The answer is no for all these questions. Take the line with infinite origins: the scheme obtained by gluing an infinite amount of copies of $mathbbA^1$ along the open subsets $mathbbG_m$. This has Krull dimension 1 (there are only closed points and the unique generic point) and it is irreducible (the only proper nonempty closed subsets are the closed points) but it is not quasi-compact (it contains an infinite discrete set), and so in particular not Noetherian.






          share|cite|improve this answer











          $endgroup$








          • 1




            $begingroup$
            Denis Nardin, I am not sure about the second paragraph. I thought nilpotent thickening never affects the underlying topological space. So while your nilpotent thickening gives a non-locally Noetherian scheme, does it give a non-locally Noetherian space (about which is the question)?
            $endgroup$
            – Stepan Banach
            2 hours ago






          • 1




            $begingroup$
            @StepanBanach Ah sorry, I was thinking of a non-Noetherian scheme. Let me correct it.
            $endgroup$
            – Denis Nardin
            2 hours ago















          10












          $begingroup$

          The answer is no for all these questions. Take the line with infinite origins: the scheme obtained by gluing an infinite amount of copies of $mathbbA^1$ along the open subsets $mathbbG_m$. This has Krull dimension 1 (there are only closed points and the unique generic point) and it is irreducible (the only proper nonempty closed subsets are the closed points) but it is not quasi-compact (it contains an infinite discrete set), and so in particular not Noetherian.






          share|cite|improve this answer











          $endgroup$








          • 1




            $begingroup$
            Denis Nardin, I am not sure about the second paragraph. I thought nilpotent thickening never affects the underlying topological space. So while your nilpotent thickening gives a non-locally Noetherian scheme, does it give a non-locally Noetherian space (about which is the question)?
            $endgroup$
            – Stepan Banach
            2 hours ago






          • 1




            $begingroup$
            @StepanBanach Ah sorry, I was thinking of a non-Noetherian scheme. Let me correct it.
            $endgroup$
            – Denis Nardin
            2 hours ago













          10












          10








          10





          $begingroup$

          The answer is no for all these questions. Take the line with infinite origins: the scheme obtained by gluing an infinite amount of copies of $mathbbA^1$ along the open subsets $mathbbG_m$. This has Krull dimension 1 (there are only closed points and the unique generic point) and it is irreducible (the only proper nonempty closed subsets are the closed points) but it is not quasi-compact (it contains an infinite discrete set), and so in particular not Noetherian.






          share|cite|improve this answer











          $endgroup$



          The answer is no for all these questions. Take the line with infinite origins: the scheme obtained by gluing an infinite amount of copies of $mathbbA^1$ along the open subsets $mathbbG_m$. This has Krull dimension 1 (there are only closed points and the unique generic point) and it is irreducible (the only proper nonempty closed subsets are the closed points) but it is not quasi-compact (it contains an infinite discrete set), and so in particular not Noetherian.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 2 hours ago

























          answered 3 hours ago









          Denis NardinDenis Nardin

          9,16223565




          9,16223565







          • 1




            $begingroup$
            Denis Nardin, I am not sure about the second paragraph. I thought nilpotent thickening never affects the underlying topological space. So while your nilpotent thickening gives a non-locally Noetherian scheme, does it give a non-locally Noetherian space (about which is the question)?
            $endgroup$
            – Stepan Banach
            2 hours ago






          • 1




            $begingroup$
            @StepanBanach Ah sorry, I was thinking of a non-Noetherian scheme. Let me correct it.
            $endgroup$
            – Denis Nardin
            2 hours ago












          • 1




            $begingroup$
            Denis Nardin, I am not sure about the second paragraph. I thought nilpotent thickening never affects the underlying topological space. So while your nilpotent thickening gives a non-locally Noetherian scheme, does it give a non-locally Noetherian space (about which is the question)?
            $endgroup$
            – Stepan Banach
            2 hours ago






          • 1




            $begingroup$
            @StepanBanach Ah sorry, I was thinking of a non-Noetherian scheme. Let me correct it.
            $endgroup$
            – Denis Nardin
            2 hours ago







          1




          1




          $begingroup$
          Denis Nardin, I am not sure about the second paragraph. I thought nilpotent thickening never affects the underlying topological space. So while your nilpotent thickening gives a non-locally Noetherian scheme, does it give a non-locally Noetherian space (about which is the question)?
          $endgroup$
          – Stepan Banach
          2 hours ago




          $begingroup$
          Denis Nardin, I am not sure about the second paragraph. I thought nilpotent thickening never affects the underlying topological space. So while your nilpotent thickening gives a non-locally Noetherian scheme, does it give a non-locally Noetherian space (about which is the question)?
          $endgroup$
          – Stepan Banach
          2 hours ago




          1




          1




          $begingroup$
          @StepanBanach Ah sorry, I was thinking of a non-Noetherian scheme. Let me correct it.
          $endgroup$
          – Denis Nardin
          2 hours ago




          $begingroup$
          @StepanBanach Ah sorry, I was thinking of a non-Noetherian scheme. Let me correct it.
          $endgroup$
          – Denis Nardin
          2 hours ago










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