# Ring homomorphism is injective but not surjective

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. The function

$$\phi: R_1 \rightarrow R_2$$

which send the polynomial $f\in F[x]$ to the $F$-valued function $a \rightarrow f(a)$ on $F$ which induces, is a homomorphism.

When $F=\mathbb Q$ or $\mathbb R$, show that the homomorphism $\phi$ is injective but not surjective.

I would start like so

if $\phi(a) = b$ and $\phi(a') = b$, then $a=a'$. Since $a,a' \in F[x]$...

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Note that you will need the fact that $\Bbb Q$ and $\Bbb R$ have characteristic $0$ to show that the map is injective: if $F = \Bbb Z/p\Bbb Z$, then $x^p - x$ is a nonzero polynomial, but $f(a) = a^p - a = a - a = 0$ for all $a\in F$, so $\phi(0) = \phi(x^p - x)$, which means $\phi$ is not injective in this case. – Stahl Nov 4 '13 at 5:13
@Stahl Actually it is sufficient to suppose that the field $F$ is infinite. – Matemáticos Chibchas Nov 4 '13 at 5:16

Every function in the image of your map is (or extends to) a continuous function $\mathbb R\to\mathbb R$. Yet there are functions $F\to F$ which are not continuous (or, when $F=\mathbb Q$, do not extend to continuous functions).
Alternatively, every function in the image of your maps is identically zero if it has infinitely many zeroes, yet there exist non-zero functions $F\to F$ which have infinitely many zeroes. This again implies non-surjectivity, and if you look at it correctly, also injectivity.