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Suppose I have an irreducible affine variety $X \subseteq \mathbb{A}^n_k$. Let us denote $X = \{ x \in k^n : f_j(x) = 0 \ (1 \le j \le M) \}$. $k$ is an algebraically closed field. Let $a_i \in k$, $m_i \ge 1$. Then I want to apply change of variables and consider the following
$Y = \{ y \in k^n : f_j( a_1 y_1^{m_1}, ..., a_n y_n^{m_n} ) = 0 \ (1 \le j \le M) \}$.

By making this change of variables, what I want to know is what happens to

1) dimension of $Y$

2) irreducibility of $Y$ (if not irreducible, is it possible to bound the number of irreducible components?)

Any information would be appreciated. Thank you very much!

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    $\begingroup$ Provided all the $a_i$ are nonzero, the map $(y_1,\dots,y_n)\mapsto (a_1y_1^{m_1},\dots,a_ny_n^{m_n})$ defines a finite surjective morphism from $Y$ to $X$. As a result, $\dim Y=\dim X$. Moreover, on the open subvariety $X':=X\cap \{x_1\cdots x_n\neq 0\}$, the map $Y\to X$ restricts to a torsor under $\mu_{m_1}\times\dots\times \mu_{m_n}$ ($\mu_n$ denotes the variety of $n$-th roots of unity), so if $X'$ is dense in $X$, I expect that $Y$ could not have more than $\prod_i m_i$ irreducible components. This argument is not complete, though. $\endgroup$
    – Uriya First
    Commented Dec 16, 2018 at 7:48
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    $\begingroup$ @UriyaFirst. You are correct about the number of irreducible components. If $X$ is a point, for instance, then $Y$ consists of $\prod_i m_i$ points. In general, since the finite morphism is also flat, every generic point of $Y$ maps to the generic point of $X$. Thus, the number of generic points of $Y$ is bounded above by the degree of the finite map. $\endgroup$
    – Jason Starr
    Commented Dec 16, 2018 at 10:10
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    $\begingroup$ @JasonStarr Thanks! This is a neat explanation. $\endgroup$
    – Uriya First
    Commented Dec 16, 2018 at 11:41
  • $\begingroup$ Thank you very much for the comments. The finite surjective morphism from $Y$ to $X$ you mentioned, suppose I restrict the map to an irreducible component of $Y$. I was wondering would we expect this to be surjective as well? or not in general? I'm trying to understand the irreducible components of $Y$ as much as possible. Thank you. $\endgroup$
    – Johnny T.
    Commented Dec 17, 2018 at 23:01

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