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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
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@Stahl Actually it is sufficient to suppose that the field $F$ is infinite. –  Matemáticos Chibchas Nov 4 '13 at 5:16

1 Answer 1

up vote 1 down vote accepted

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).

This means that the map is not surjective.

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.

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