A polynomial that vanishes at one point but not at finitely many others. 
Problem: Show that given a finite set of distinct points $p_0,p_1,\ldots,p_k\in\Bbb C^n$, there exists a polynomial $f(x_1,\ldots,x_n)\in\Bbb C[x_1,\ldots,x_n]$ such that $f(p_0)=0$ but $f(p_i)\neq 0$ for all $i=1,\ldots,k$.

I get the intuitive idea: The set $X=\{x\in\Bbb C^n:f(x)=0\}$ is a subset of $\Bbb C^n$ of one dimension lower than $\Bbb C^n$, and hence we must be able to arrange it so that it goes through $p_0$ but misses any finite number of other points $p_1,\ldots,p_k$.
But how to prove this rigorously? 
 A: We definitely have $\bigcup_{j=1}^n(p_j-p_0)^\perp\neq\mathbb C^n$. Choose a vector $h\in\mathbb C^n$, which is not in this union and set $f(x) = \langle x-p_0,h\rangle$. This is a polynomial with the desired property.
A: If $p_0=(a_1,...a_n)$. Let $k$ be the field obtained by adjoining $p_{i_j}-a_j$ to $\mathbb{Q}$ (here $p_i=(p_{i_1},...p_{i_n})$ for $i\geq 1$). Now choose some $x\in \mathbb{C}$ that is transcendental over $k$ ($\mathbb{C}$ is not algebraic over $k$ because $k$ is countable). Then $f=(x_1-a_1)+(x_2-a_2)x+\cdots+(x_n-a_n)x^{n-1}$ should do the job.
A: Let the $j$th co-ordinate of $p_i$ be $p_{i,j}.$ For  $1\leq j\leq n,$ take a polynomial $f_j:C\to C$ such that $f_j(p_{0,j})=1$  and $f_j(p_{i,j})= 0$ whenever $i\ne 0$ and $p_{i,j}\ne p_{0,j}.$
Let $f(x_1,..,x_k)=\sum_{j=1}^n(1-f_j(x_j)).$ Then $f(p_0)=0.$ 
And when $i\ne 0,$ we have $f(p_i)=\sum_{j\in i^*}1$ where $i^*=\{j:p_{0,j}\ne p_{i,j}\}\ne \phi,$ so $f(p_i)\geq 1.$
To obtain $f_j$: For each $j$ let $\{p_{i,j}: 0\leq i\leq  k\}=\{p_{0,j}\} \cup S_j$ where  $p_{0,j}\ne S_j.$ If $S_j=\phi$ let $f_j(x)= 1$ for all $x.$ If $S_j\ne \phi,$ let $f_j(x)=[\;\prod_{t\in S_j}(x-t) \;]/ \prod_{t\in S_J}(p_{0,j}-t).$
