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

Question

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.

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 05 '13 at 22:48
Link to OP's generalization, where $\phi:R[x]\rightarrow C.\ \ $ — Math Gems, Mar 06 '13 at 18:25

Clive Newstead · Answer 1 · 2013-03-05T22:38:09.480

5

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.

edited Mar 05 '13 at 22:38

answered Mar 05 '13 at 22:29

Clive Newstead

score 4 · Answer 2 · answered Mar 05 '13 at 22:38

4

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

answered Mar 05 '13 at 22:38

Math Gems

2 Answers2