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

Thanks in advance.

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- Mar 21st 2009, 04:51 PMBiscaimabout finite fields
Let K be a finite feld. Show that for any $\displaystyle x\in K$ there is an $\displaystyle f\in K[X_{1},...,X_{n}]$ with $\displaystyle f(x)=1$ and $\displaystyle f(y)=0$ for all $\displaystyle y \in K\$\{x}.

Thanks in advance. - Mar 21st 2009, 09:41 PMThePerfectHacker
Your question makes no sense. You cannot evaluate $\displaystyle f$ as a single point in $\displaystyle K$ if $\displaystyle n>1$. Thus, I am going to show that for any $\displaystyle x\in K$ there is $\displaystyle f\in K[X]$ with $\displaystyle f(x) = 1 $ and $\displaystyle f(y) = 0$ for all $\displaystyle y\in K - \{x\}$. Since $\displaystyle K$ is a finite field we can define $\displaystyle g(X) = \Pi_{a\in K-\{x\}} (X-a)$. We see that $\displaystyle g(x) \not = 0$ and $\displaystyle g(y) = 0$ for all $\displaystyle y\in K-\{x\}$. Define $\displaystyle f(X) = \tfrac{1}{g(x)}g(X)$.

- Mar 22nd 2009, 02:19 AMBiscaimabout finite fields
Sorry for the typo. The correct question was.

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

Thanks again. - Mar 22nd 2009, 08:54 AMThePerfectHacker