I am aware of this question but I think the proof in that question uses homomorphisms (I think) while I want to use the Subring Test to prove the following statement.
Let $F$ be a field. An element $\phi$ of $F^F$ is a polynomial function on $F$, if there exists $f(x) \in F[x]$ such that $\phi (a)=f(a)$ for all $a \in F$. For $\phi, \psi \in F^F,$ define the sum $\phi + \psi$ by $(\phi + \psi)(r) = \phi(r) + \psi(r)$ and the product $\phi \cdot \psi$ by $(\phi \cdot\psi)(r) = \phi(r) \cdot \psi(r)$ for all $r \in F$.
Show that the set $P_F$ of all polynomial functions on $F$ forms a subring of $F^F$. Here's my attempt thus far:
Proof. Firstly, we show that $P_F$ is closed under subtraction. Let $\gamma, \beta \in P_F$. Then, there exist $f(x), g(x) \in F[x]$ such that $\gamma (a)=f(a)$ and $\beta (a)=g(a)$ for all $a \in F$. Then, $(\gamma - \beta)(a) = \gamma(a) - \beta(a) = f(a) - g(a) = (f-g)(a)$. Here, we have shown that $(f- g) \in F[x]$ is s.t. $(\gamma - \beta) (a)=(f-g)(a)$. So, by definition, $P_F$ is closed under subtraction.
Secondly, we show that $P_F$ is closed under multiplication. Let $\gamma, \beta \in P_F$. Then, there exist $f(x), g(x) \in F[x]$ such that $\gamma (a)=f(a)$ and $\beta (a)=g(a)$ for all $a \in F$. Then, $(\gamma \cdot \beta)(a) = \gamma(a) \cdot\beta(a) = f(a) \cdot g(a) = (f \cdot g)(a)$ and we are done.
Lastly, we show that $P_F$ is nonempty. This is where I run into trouble. I have no clue as to how this should be proved. I attempted a proof by contradiction to no avail. Can someone please show how this can proved? Also, is my proof for closure under multiplication and subtraction accurate?
I should note that I found an alternative rendition of the Subring Test according to which I need to prove that the multiplicative identity of $F^F$ is in $P_F$ (and won't need to prove that $P_F$ is nonempty) in addition to what I currently have. Can someone show how this can be proved?