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I didn't understand this proof:

What exactly is $f^{-1}(Z)$? and what are these $h_i$ in $Z$?

This proof is easy, I need just a little bit of clarification in these points.

Thanks a lot.

user42912
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1 Answers1

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Well, $f^{-1}(Z) = \{ v \in V : f(v) \in Z \}$ is the preimage of $Z$. The elements $h_i$ are the polynomials that define $Z$ as a closed subset of $W$. They should be viewed as elements of the coordinate ring of $W$. Since the composition of two polynomials is still a polynomial, each $h_i \circ f$ is a polynomial, which means that $f^{-1}(Z)$ is a closed subset of $V$.

Michael Joyce
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  • Why $f^{-1}(Z) = h_i\circ f$? that's my doubt. – user42912 Sep 25 '13 at 21:29
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    Your question, as written, does not make sense, because $f^{-1}(Z)$ is a set while $h_i \circ f$ is a function. What you need to know is that $v \in f^{-1}(Z)$ if and only if $f(v) \in Z$, which occurs if and only if $h_1(f(z)) = \cdots h_r(f(z)) = 0$, which is equivalent to $(h_1 \circ f)(z) = \cdots (h_r \circ f)(z) = 0$. (cont.) – Michael Joyce Sep 25 '13 at 21:42
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    You may also be confused by the definition of $Z$. A clearer way of presenting it is that $Z = { v \in V : h_1(v) = \cdots h_r(v) = 0 }$. A similar comment applies to the definition of $f^{-1}(Z)$ given in the text you cited. – Michael Joyce Sep 25 '13 at 21:43
  • yes yes, of course, now it's clear, thank you very much for your help. – user42912 Sep 25 '13 at 22:02
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    User Michael Joyce, when you define $Z$ as $Z = {{v \in V: h_1(v)=...=h_r(v)=0}}$, shouldn't $v$ be an element of $W$ since $Z \subset W$?. Similarly should the $h_i$ be define as elements in $K[w_1,...,w_m]$? rather than in $K[V] = K[v_1,..,v_n]$. – Mahidevran Feb 26 '17 at 04:00
  • Thanks Mahidevran. You're right and I've edited to fix the errors. – Michael Joyce Feb 26 '17 at 05:48