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Let $P_{1}, P_{2},\cdots, P_{k}\in \Bbb{C}^n$. Now define a homomorphism $\phi: \Bbb{C}[x_{1},x_{2},\cdots,x_{n}]\rightarrow \Bbb{C}^k$ which sends $f$ to $(f(P_{1}),f(P_{2}),\cdots,f(P_{k})).$ Then how to show that $\phi$ is surjective?

I saw it here, but don't know how to prove it.

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    $\begingroup$ You can take $f_{i}$ such that $\phi(f_{i}) =e_{i}$, where $e_{i}$s are elements of the standard basis of $\mathbb{C}^{k}$. Then, linear combination of $f_{i}$ gives you desired answer. $\endgroup$
    – user124697
    Mar 18, 2021 at 4:42
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    $\begingroup$ @user124697 that looks like an answer to me - would you care to record it as such below? $\endgroup$
    – KReiser
    Mar 18, 2021 at 5:26
  • $\begingroup$ @MarianoSuárez-Álvarez Ok I opened it and wrote an answer based on your comment $\endgroup$
    – MathEric
    Dec 22, 2022 at 10:24

2 Answers 2

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Fix $P_i$, for every $j\neq i$, you can always find a polynomial $Q_j$ which vanishes on $P_j$ and not on $P_i$ (take an hyperplane passing through $P_j$ and not through $P_i$).

Then $R_i:=\prod_{\j\neq i}Q_j$ is a polynomial which vanishes on all of the $P_j$ but not on $P_i$.

The restriction of $\phi$ to the vector space generated by the $R_i$ is surjective.

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In general proving that functions are surjective is hard but inyectivity is much easier because we have tools, for example the Jacobian that tell us when a function is inyective.

Fortunely for us, the Ax's thoerem states that any polinomial map that is inyective is also surjective, precisley

Let $f:\mathbb{C}^n\rightarrow \mathbb{C}^n$, $f=(f_1,...,f_n), f_i\in \mathbb{C}[x_1,...,x_n]$ if $f$ is inyective then, it also is surjective. (Hils, M., & Loeser, F. (2019). A first journey through logic. Theorem 3.6.3)

I think that Ax's theorem also generalizes to any algebraic variety.

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