# How to prove the following evaluation map is surjective?

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.

• 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. Mar 18, 2021 at 4:42
• @user124697 that looks like an answer to me - would you care to record it as such below? Mar 18, 2021 at 5:26
• @MarianoSuárez-Álvarez Ok I opened it and wrote an answer based on your comment Dec 22, 2022 at 10:24

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.

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.