In-/surjectivity of $R[X] \rightarrow \text{Map}(R,R)$ for infinite integral domain $R$.

Question

Let $R$ be a ring and consider \begin{eqnarray*} \phi : &R[X] &\longrightarrow \text{Map}(R,R) \\ &f &\mapsto \,\,\,(r \mapsto f(r)) \end{eqnarray*} I have shown that $\phi$ is a ring homomorphism iff $R$ is commutative. Now assume that $R$ is an infinite integral domain. (Then $R$ is commutative, so $\phi$ is a ring homomorphism.) I want to show that $\phi$ is injective, but not surjective. To do this, suppose that for arbitrary $f,g ∈ R[X]$, we have $\phi(f) = \phi(g)$. Then we have: $$ (\forall r ∈ R)(f(r) = \phi(f)(r) = \phi(g)(r) = g(r)). $$ Now if $f, g$ were functions, then we could conclude that $f =g$, and $\phi$ would be injective. I figured that the same would apply for polynomials (because they are equal at every point). However, we never used that $R$ is an infinite integral domain here. For surjectivity (or actually: non-surjectivity) it would suffice to show an example of a function $f \notin \text{im} \phi$. I have the feeling that here is where the infinity of $R$ comes into play, but as $R$ remains abstract, I haven't found a concrete example of this yet.

No homework question, just an exercise. Any help will be greatly appreciated.

Edit

What happens when instead $R$ is finite? (i.e. when $R$ is a finite field.) Can $\phi$ still be injective? And surjective? The latter is at least possible (if not plausible) on grounds of cardinality: Map$(R,R)$ is finite, while $R[X]$ is not.

Answer 1

For $R$ infinite, the set of all possible functions $f:R \to R$ has cardinality $2^{\vert R \vert}$, whereas (because polynomials can be put into $1-1$ correspondence with finite sequences from $R$) $\vert R[X] \vert = \vert R \vert$. Cantor's theorem tells us that for any set, $\vert R \vert \lt 2^{\vert R \vert}$, so $\phi$ can't be surjective.

If $\forall x \in R~f(x)=g(x)$, then because $R$ is an integral domain, it embeds in a field $F$, and because $R$ is infinite $f-g \in R[X] \subseteq F[X]$ has infinitely many roots, so it must be identically $0$.

Answer 2

I formulate your question into an exercise:

Let $K$ be a field. We consider the ring homomorphism $\varphi: K[X] \rightarrow$ $\operatorname{Map}(K, K)$ $$ \varphi\left(a_{0}+a_{1} X+\cdots+a_{n} X^{n}\right): K \rightarrow K, \quad b \mapsto a_{0}+a_{1} b+\cdots+a_{n} b^{n} . $$ a) If $K$ is infinite, show that $\varphi$ is injective. Is $\varphi$ surjective?
b) Suppose $K$ is finite. Is $\varphi$ injective? Show that $\varphi$ is surjective.

Solution

a) If $\varphi(f)=0$, then $f$ has infinitely many zeros and is the zero polynomial. (Prop. Let $f \in K[X]$ with deg $f = n$. Then $f$ has at most $n$ zeros in $K$.)
We show that $\varphi$ is not surjective.
Let $\chi_0:K\rightarrow K$ be the mapping $$ \chi_0(b)=\begin{cases} 0,&\quad b\neq 0,\\ 1,&\quad b= 0. \end{cases} $$ Then $g$ has infinitely many zeros, but $g$ is not the zero function. Therefore $\varphi(f)\neq g$ holds for all $f\in K[X].$

b) We show that $\varphi$ is not injective. Let $f=\prod_{a\in K}(X-a).$ Then $f$ is a polynomial of degree $|K|$. Since $$ \varphi(f)(b)=\prod_{a\in K}(b-a)=0 $$ holds (if $b=a$, then $b-a=0$), $\varphi$ is not injective.
$\varphi$ is surjective: as before, let $\chi_0:K\rightarrow K$ be the mapping $$ \chi_0(b)=\begin{cases} 0,&\quad b\neq 0,\\ 1,&\quad b= 0. \end{cases} $$ Set $f=\prod_{a\in K\setminus\{0\}}(X-a)(-a)^{-1}$. Then $$ \varphi(f)(b)=\begin{cases} 0,&\quad b\neq 0,\\ 1,&\quad b= 0, \end{cases} $$ thus $\varphi(f)(b)=\chi_0(b)$.
For an arbitrary function $g:K\rightarrow K$ holds $$ \varphi\left(\sum_{a\in K}g(a)f(X-a)\right)=g, $$ since $$ \sum_{a\in K}g(a)f(b-a)=g(b). $$