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Math Help - Algebraic Geometry

  1. #1
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    Algebraic Geometry

    Let  V(I) = {v \in V | g(v) = 0 \forall g \in G}, for I \subset k[x], and k an algebraically closed field, Prove that V(I) = {0} \iff I=k[x]
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by terr13 View Post
    Let  V(I) = {v \in V | g(v) = 0 \forall g \in G}, for I \subset k[x], and k an algebraically closed field, Prove that V(I) = {0} \iff I=k[x]
    i'll fix your latex for you..

    V(I) = \{v \in V \ | \ g(v) = 0 \ \forall g \in G\}

    V(I) = \{v \in V \ | \ g(v) = 0 \ \forall g \in G\}

    and i think, V(I) = \emptyset \iff I=k[x] and not V(I) =0.

    and another thing, i think, it should be V(I) = \{v \in k \ | \ g(v) = 0 \ \forall g \in I\}
    (actually, i am assuming you want to prove a version of the weak nullstelensatz.. )

    now, the proof.. i'll prove one direction only.

    (<=) Suppose I=k[x].

    By definition, V(I) = \{v \in k \ | \ g(v) = 0 \ \forall g \in I\}. Since I contains nonzero constant functions, say p(x)=c for a constant c, then there is no v\in k such that p(v)=0. Thus, there is no v\in k such that g(v)=0 for all g\in k[x]=I. Hence, V(I) = \emptyset.

    (=>) To prove this, Let V(I) = \emptyset, I\subset k[x]. show that I contains a unit, since if it does, then I is the entire k[x] (exclude the case that I is empty.. it should be trivial that V(\emptyset)=k).

    or you can prove it by contrapostive: if I is a proper subset of k[x], show that V(I) is not empty. (in particular, let I be a maximal ideal of k[x])

    can you do it?
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