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All questions below are over $C$,complex number or over $C^n$.

$\textbf{Q1:}$ Given a $u$ holomorphic function not vanishing identically at $0\in C^n$, then one can assume $u$ is non-vanishing along $z_1$ axis(i.e. restriction of $u$ to $z_1$ axis leads to non-trivial series expansion in $z_1$) by a linear transformation. I do not see why this is obvious. I used a technique adopted to prove Noether normalization in infinite field by transforming the basis by linear transformation. Then argue there is a $P^{n-1}$ point along which $u$'s lowest homogeneous degree component is non-vanishing and this gives the axis $z_1$. For general case, I need non-linear transformation. Am I thinking too complicated here or was there a simpler argument?

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Given $u,v$ holomorphic functions vanishing at $0\in C^n$, then we can suppose $u,v$ both non-vanishing along $z_2=\dots=z_n=0$ axis by a linear transformation.

$\textbf{Q2:}$ Is above assertion correct? Suppose $u(z_1,\dots, z_n)=z_1^k+\dots$ and $v(z_1,\dots, z_n)=z_2^k+\dots$ where $\dots$ contains all terms of degree $k$ at least. Why there is a linear transformation to make this happening? I did prove this statement by reducing to $P^1_C$ case by looking at mixing $z_1,z_2$ terms and this is basically reusing the result from $Q1$. Once again, am I thinking too complicated? Should there be simpler answer to above questions?

Ref: Analytic Theory of Abelian Varieties by Swinnerton-Dyer pg 26. It seems that the author find the argument very obvious.

$\textbf{Additional Question:}$ What is the intuitive reason to expect $Q1$ and $Q2$?

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Maybe not obvious but this type of argument shows up a lot in local theory of holomorphic functions. Any book in several complex variables may discuss this right before Weierstrass Preparation Theorem.

Since $u$ does not vanish identically, there is a line $L$ through the origin such that $\left. u\right|_L $ does not vanish identically. Then take a linear change of coordinates that sends $L$ to the $z_1$-axis.

Notice that the same holds for any finite number of functions. If you have $u_1, \dots, u_m$ non-trivial holomorphic functions, then $\{ u_1 =0 \} \cup \dots \cup \{u_m = 0 \} = \{ u_1\dots u_m =0 \}$. You can find a line where the restriction of the product, hence the restriction ofeach $u_i$, does not vanish identically.

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  • $\begingroup$ Why does weierstrass preparation theorem holds for finite number of functions here? The version I have seen is associated to a function. How to deduce that I can find an axis for finite number of functions non-vanishing? Thanks. $\endgroup$ – user45765 May 29 '18 at 19:32
  • $\begingroup$ Any non-trivial holomorphic function can vanish on at most a finite number of these lines. Once you have a finite number of such functions you still have a finite number of lines to avoid. $\endgroup$ – Alan Muniz May 29 '18 at 19:38
  • $\begingroup$ I think the holomorphic function vanishing on the open ball implies that function vanishes on the connected component. Why there must be finite number of lines to guarantee non vanishing here? $\endgroup$ – user45765 May 29 '18 at 19:42
  • $\begingroup$ If I have $z_1z_2$ in $C^5$, I could have a lot of line vanishing on $z_1=0$ hyperplane. $\endgroup$ – user45765 May 29 '18 at 19:44
  • $\begingroup$ Sorry, my mistake. I meant hyperplanes instead of lines there. I will make it precise in a moment. $\endgroup$ – Alan Muniz May 29 '18 at 19:54

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