The ring structure of the set of all functions from a finite field $K$ to itself

Question

$K$ is a finite field and $F$ is the set of all functions of $K$ in $K$. In $F$ we define $(f+g)(a)=f(a)+g(a)$ and $(fg)(a)=f(a)g(a)$.

Show that $F$ is isomorphic a $k[x]/I$ for a some ideal $I$ of $k[x]$ , and find this $I$.

Is correct my solution and $I$ is only this $\ker(\Phi)$? What influences $K$ be field and not just ring.

$\Phi$ is homomorphism , because $(f+g)(a)=f(a)+g(a)$ and $(fg)(a)=f(a)g(a)$.

if $k[x]/I$ is isomorphic a $\Phi$ => $\ker(\Phi)= I = \{p(x) \in k[x] \mid \Phi(p(x))=0\}$ and $\Phi$ is homomorphism

So $k[x]/I$ is isomorphic a $\Phi$ , and $I = \ker(\Phi)$

Answer 1

Let $F = \{f : k \to k\}$, then, under the operations above, $F$ is a ring.

1. Define $\varphi : k[x] \to F$ given by $\varphi(f) = f$, where the $f$ on the RHS is thought of as a function on $k$. This is clearly a homomorphism, so we would be done by the first isomorphism theorem if we could only show that it is surjective.

2. Now write $F = \{x_0, x_1, \ldots, x_n\}$. Given $g \in F$, write $y_i = g(x_i)$, and consider the interpolating polynomial $$ p(x) = \sum_{i=0}^n y_i \prod_{j\neq i} \frac{x-x_j}{x_i - x_j} $$ Then, $p \in k[x]$ (Note : $p$ makes sense because $k$ is a field), and $$ \varphi(p)(x_i) = y_i = g(x_i) \quad\forall i \in \{0, 1,\ldots, n\} $$ Hence, $\varphi(p) = g$ (ie., as a function on $k$, $p = g$).

This completes the proof.