Schemes 1: Introduction

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  • Опубликовано: 1 янв 2025

Комментарии • 28

  • @bobclarke5543
    @bobclarke5543 3 года назад +26

    Absolutely the clearest, most comprehensible introduction to the subject I've ever seen. Clearer, more business-like, and better motivated than Vakil's attempt (which, I must say, is also very good but has an entirely different character). I highly recommend this series to anyone who wants to learn algebraic geometry.

  • @georgediderrich43
    @georgediderrich43 2 года назад +5

    What a pleasure to view these lectures. Professor Borcherds rocks!

  • @ipudisciple
    @ipudisciple 4 года назад +4

    18:25 I think we have two options. We can choose a dimension d and consider the sheaf of vector fields of dimension d. Or we can lump these all together and work with vectors in the direct sum ⨁ ℝ^n, which is just the countably dimensional vector space. Technically, Richard says the second, but I think he might be thinking of the first.

  • @Ibakecookiess
    @Ibakecookiess 4 года назад +9

    Watching this from Colombia. Thank you!

  • @juliavilageliu6797
    @juliavilageliu6797 Год назад +1

    You really helped me through my final degree project :) Thank you so much

  • @vinayakgupta2368
    @vinayakgupta2368 4 года назад +13

    Hey, i am from india. Thank you for you these free lectures.

  • @domingogomez6999
    @domingogomez6999 3 года назад +2

    Sorry, it might be trivial, I don't see why the "constant" presheaf is not a sheaf. I mean that, if the empty set, U_1, U_2 and their union U = U_1\union U_2 are the only open sets, why does F(U) = AxA hold?

    • @ruinenlust_
      @ruinenlust_ 3 года назад +2

      Constant pre-sheaf assigns A to every U. Clearly, if F were to be a sheaf then F(U) must be AxA. But, since F assigns A to every open set, F cannot be a sheaf in general

    • @domingogomez6999
      @domingogomez6999 3 года назад +1

      @@ruinenlust_ the clearly part is not evident for me. Could you develop more, please?

    • @harshavr
      @harshavr 3 года назад +2

      @@domingogomez6999 If U is partitioned into U1 and U2 with no intersection, then the sheaf condition says that section on U1 and section on U2 uniquely determines a section on U ( the compatibility condition is trivial because U1 and U2 dont intersect), So F(U)=F(U1)xF(U2)=AxA.
      The key here is that space is disconnected, For a connected space, the constant sheaf is a sheaf. (U1 and U2 will intersect above and the constant function should restrict to the same constant on the intersection)
      For any space, the locally constant functions form a sheaf.

    • @yiyangjia1967
      @yiyangjia1967 2 года назад +1

      look at the injectivity of the exact sequence

  • @premkumar-s1
    @premkumar-s1 4 года назад +11

    Please make videos explaining on cohomology of sheaves unit 3 of hartshrone. Thank you

  • @nostalgia_1439
    @nostalgia_1439 Год назад

    2:35 Isn't k[x] a k-algebra? I mean I think it's even a finite k-algebra? What is his definition of k-algebra? I know what an algebra \phi: R\to S is (e.g. in the sense of Altman Kleinman's commeutative algebra text, and "Algebra" by Hungerford)

  • @mathgeeks3598
    @mathgeeks3598 4 года назад +1

    nice, video, i am in uSa. thanks for good videos

  • @donatoebbasta2797
    @donatoebbasta2797 2 года назад

    Does anyone have a clear cut pdfLaTeX-compiled version of these lectures?

  • @禿田与平
    @禿田与平 3 года назад +2

    Thank you so much for your online couse !

  • @markcanedo3
    @markcanedo3 2 года назад

    Thank you for this magistral lessons, professor

  • @okoyoso
    @okoyoso Год назад +1

    I believe his name is pronounced "Hart's horn" (hart being an old word for a stag, type of buck)

    • @sdfdsf4162
      @sdfdsf4162 Год назад

      haha me too

    • @okoyoso
      @okoyoso Год назад

      @@sdfdsf4162 You too as in you pronounce it that way too?

    • @sdfdsf4162
      @sdfdsf4162 Год назад

      @@okoyoso yeeep. dude im an eastern europe dog, of course i suck speaking eng =)

    • @okoyoso
      @okoyoso Год назад

      @@sdfdsf4162 I mean a lot of people call him "hart shorne"

  • @ac-dp3jk
    @ac-dp3jk 3 года назад +2

    Thank you very much for your videos !

  • @tapioms
    @tapioms 8 месяцев назад

    🎉

  • @HERŞEYKÖTÜ
    @HERŞEYKÖTÜ 2 года назад +6

    I dont understand how people love this series, I for myself, find it very confusing and not clear

    • @guiwenluo1774
      @guiwenluo1774 2 года назад +6

      Second. I found it not newcomer friendly. Probably those who love this series are already learnt Algebraic geometry for at least one time.

    • @mbgdemon
      @mbgdemon 2 года назад +19

      @@guiwenluo1774 That is why the series is called Algebraic Geometry II.

    • @bennoarchimboldi6245
      @bennoarchimboldi6245 4 месяца назад +1

      People love him because he’s a fields medalist not because he’s a good teacher.