The hint is: Let F has as elements $a_1,...,a_n$. Note that if $$f_i(x)=c(x-a_1)...(x-a_{i-1})(x-a_{i+1})...(x-a_n)$$, then $f_i(a_j)=0$ for $i\neq j$ and the value $f_i(a_j)$ can be controlled by the choice of $c\in F$. Use this to show that every function on F is a polynomial function.
First, I need to construct a polynomial $p(x) \in F[x]$ such that $p_i(a_i)\neg 0$ and $p_i(a_j)= 0$ When $i\neq j$ and explain why such a polynomial exists.
Next I think I need to show for all $a\in F$ that f(a)=p(a).
Using Fraleigh’s seventh edition of A First Course in Abstract Algebra. Need help with problem 22.31c. I’m completely lost. Thank you!