# Surjective map from polynomial ring over a field to the field.

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 $f(x)=x+r-z$?
 Or, far more simply, consider $f(x)=r$, as egreg points out in the comments.
Or the constant polynomial $f(x)=r$. –  egreg Mar 5 '13 at 22:36
Hint $\,\ \phi\$ restricts to the identity map on $\rm\,R,\:$ i.e. constants evaluate to themselves.