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Let $z\in R$ be fixed then the map $\phi:R[x]\rightarrow R $ defined as $\phi(f(x))=f(z)$ is surjective? Could someone please explain me why it is.

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What if instead of $R$ there is $C$? – user65227 Mar 5 '13 at 22:45
The proofs given by Math Gens, egreg, Clive make no assumption about the field / wirk with any field (in fact, with any ring) – Hagen von Eitzen Mar 5 '13 at 22:48
Link to OP's generalization, where $\phi:R[x]\rightarrow C.\ \ $ – Math Gems Mar 6 '13 at 18:25

Let $r \in R$; where does $\phi$ send the polynomial $f$ given by $f(x)=x+r-z$?

[Edit] Or, far more simply, consider $f(x)=r$, as egreg points out in the comments.

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Or the constant polynomial $f(x)=r$. – egreg Mar 5 '13 at 22:36
Ha! I never see the simple solution. – Clive Newstead Mar 5 '13 at 22:37

Hint $\,\ \phi\ $ restricts to the identity map on $\rm\,R,\:$ i.e. constants evaluate to themselves.

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