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.
Using Fraleigh’s seventh edition of A First Course in Abstract Algebra. Need help with problem 22.31c. I’m completely lost. Thank you!
Let $F=\{a_1,a_2,\dots,a_n\}$ and consider the map $$ v\colon F[x]\to F^n \qquad v(f)=(f(a_1),f(a_2),\dots,f(a_n)) $$ which is a ring homomorphism. Its kernel is generated by $$ g(x)=(x-a_1)(x-a_2)\dots(x-a_n) $$ so $v$ is surjective by the Chinese remainder theorem.
Can you finish?