about finite fields

**Biscaim** · March 21st 2009, 04:51 PM

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}.

Thanks in advance.

**ThePerfectHacker** · March 21st 2009, 09:41 PM

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}.

Thanks in advance.

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 $f(X) = \tfrac{1}{g(x)}g(X)$ .

**Biscaim** · March 22nd 2009, 02:19 AM

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.

**ThePerfectHacker** · March 22nd 2009, 08:54 AM

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 $K^n$ .

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