Lecture 19 - Localization (Continued)

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  • Опубликовано: 18 дек 2024

Комментарии • 5

  • @shantanuD99
    @shantanuD99 3 года назад +1

    Brilliantly explained ! Thanks

  • @dimpigautam6536
    @dimpigautam6536 5 лет назад +4

    Thank you sir so much for clear confusion by this lecture

  • @partha360
    @partha360 5 месяцев назад

    In 23:00, x/s is a unit in S^-1 I , it implies s/x is in S^-1 I , it implies s belongs to I but is S a multiplicative closed set of I ?

  • @yogkorde5066
    @yogkorde5066 3 года назад

    I think there is something incorrect here: take the example of the unit circle in R^2. The function x-y is not in the ideal of that circle, and is therefore inverted in the localization. But it DOES have 2 zeros on the unit circle. So it is not true that the localization is the set of regular functions on the unit circle. What is true is that functions that vanish EVERYWHERE on unit circle are not inverted. Functions like x-y that vanish at some points may be inverted, and due to this the local ring is not same as ring of regular functions. Maybe the professor can clarify. Thank you.

  • @kiwanoish
    @kiwanoish 6 лет назад

    Great lectures! I am just a little confused. Question: at 9:08, you are basically saying that g
    ot\in P g(a_1,...,a_n)
    ot=0 for some (a_1,...,a_n)\in V. But, how can we draw this conclusion? It is certainly true that g\in P -> g(a_1,...,a_n)=0 for all (a_1,...,a_n)\in V, which is logically equivalent to the statement (g
    ot\in P) OR (g(a_1,...a_n)=0 for all (a_1,...,a_n)\in V). So it seems we can have g
    ot\in P but still have g(a_1,...,a_n)=0 for all points in V contrary to the proposed statement. Then at 9:43 you even said g was nowhere vanishing on V, exactly what is meant by this? ; certainly we can have g(a)=0 for some a\in V?