Let K be a finite feld. Show that for any $x\in K$ there is an $f\in K[X_{1},...,X_{n}]$ with $f(x)=1$ and $f(y)=0$ for all $y \in K\$\{x}.

2. Originally Posted by Biscaim
Let K be a finite feld. Show that for any $x\in K$ there is an $f\in K[X_{1},...,X_{n}]$ with $f(x)=1$ and $f(y)=0$ for all $y \in K\$\{x}.

Your question makes no sense. You cannot evaluate $f$ as a single point in $K$ if $n>1$. Thus, I am going to show that for any $x\in K$ there is $f\in K[X]$ with $f(x) = 1$ and $f(y) = 0$ for all $y\in K - \{x\}$. Since $K$ is a finite field we can define $g(X) = \Pi_{a\in K-\{x\}} (X-a)$. We see that $g(x) \not = 0$ and $g(y) = 0$ for all $y\in K-\{x\}$. Define $f(X) = \tfrac{1}{g(x)}g(X)$.

Sorry for the typo. The correct question was.

Let K be a finite feld. Show that for any $x\in K^{n}$ there is an $f\in K[X_{1},...,X_{n}]$ with $f(x)=1$ and $f(y)=0$ for all $y \in K^{n}-\{x\}$.

Thanks again.

4. Originally Posted by Biscaim
Sorry for the typo. The correct question was.

Let K be a finite feld. Show that for any $x\in K^{n}$ there is an $f\in K[X_{1},...,X_{n}]$ with $f(x)=1$ and $f(y)=0$ for all $y \in K^{n}-\{x\}$.

Thanks again.
It is the same idea! Try to modify what I first wrote to $K^n$.