How do I show this is a homomorphism? Let $F$ be a field, let $R_1 = F[x]$ be the ring of polynomials with coefficients in $F$, and let $R_2$ be the ring of all functions from $F$ to itself, with addition and multiplication defined as the usual operations on functions with values in a ring. Show that the function
$\phi: R_1 \to R_2$ which sends $f \in F[x]$ to the $F$-valued function $a\to f(a)$ on $F$ which it induces, is a homomorphism of rings.
I'm not sure where to start, any help would be appreciated
 A: Start with the definition of homomorphism: what do you have to show in order to demonstrate that $\varphi$ is a homomorphism? You need to show that 
$\quad(1)\quad\varphi(f+g)=\varphi(f)\oplus\varphi(g)$ for all $f,g\in F[x]$, 
and you need to show that 
$\quad(2)\quad\varphi(f\cdot g)=\varphi(f)\varphi(g)$ for all $f,g\in F[x]$,
where for greater clarity I’ve used $\oplus$ and $\cdot$ for the operations in $F[x]$; I’ll use $+$ and juxtaposition for the operations in $R_2$ and $F$. I’ll prove $(1)$ as an illustration and leave $(2)$ for you.
Let $f,g\in F[x]$; to show that $\varphi(f+g)=\varphi(f)+\varphi(g)$, you must show that $$\big(\varphi(f\oplus g)\big)(a)=\big(\varphi(f)+\varphi(g)\big)(a)$$ for each $a\in F$. Since $\big(\varphi(f)+\varphi(g)\big)(a)=\big(\varphi(f)\big)(a)+\big(\varphi(g)\big)(a)$, this boils down to showing that
$$\big(\varphi(f\oplus g)\big)(a)=\big(\varphi(f)\big)(a)+\big(\varphi(g)\big)(a)$$
for each $a\in F$. For this you’ll have to specify $f$ and $g$, say as $f(x)=a_0+a_1x+\ldots+a_mx^m$ and $g(x)=b_0+b_1x+\ldots+b_nx^n$. Without loss of generality assume that $m\le n$ and let $a_k=0_F$ for $k=m+1,\ldots,n$, so that $f(x)=a_0+a_1x+\ldots+a_nx^n$. Then $f\oplus g$ is the polynomial $h(x)=c_0+c_1x+\ldots c_nx^n$, where $c_k=a_k+b_k$ for $k=0,\ldots,n$, and
$$\big(\varphi(f\oplus g)\big)(a)=\sum_{k=0}^n(a_k+b_k)x^k=\sum_{k=0}^na_ka^k+\sum_{k=0}^nb_ka^k=\big(\varphi(f)\big)(a)+\big(\varphi(g)\big)(a)$$
for each $a\in F$, as desired.
The proof of $(2)$ is a bit messier, simply because multiplication of polynomials is a bit messier, but the principle is the same.
