# Surjective map between polynomial ring and field.

Let $\phi:R[x]\rightarrow C$ defined for a fixed $z\in C$ as $\phi(p(x))=p(z)$ where $p\in R[x]$. Then this map is supposed to be surjective. Could someone explain me please why?

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What is $C$? And what is its relation to $R$? (is $R$ supposed to be a ring by the way) – Nils Matthes Mar 6 '13 at 8:56
Assuming that $R$ and $C$ are intended to be $\mathbb{R}$ and $\mathbb{C}$ respectively, then this is false. For example, let $z=0$; then the image of $\phi$ is just $\mathbb{R}$, not all of $\mathbb{C}$. – Zev Chonoles Mar 6 '13 at 8:57
Please give the full context. I suspect $C$ to be a simple extension of $R$ by $z$, all things being commutative. – Andreas Caranti Mar 6 '13 at 9:13
Link to OP's prior variant where $R = C.\ \$ – Math Gems Mar 6 '13 at 18:24

I assume that $C$ wants to be $\Bbb C$ and $z\in\Bbb C\setminus\Bbb R$. (Else, if $z\in R$, then for all $p\in R[x]$, we have $p(z)\in R$ whatever ring $R$ is.)
Let's assume then that $z=a+bi$ for $a,b\in\Bbb R$ and $b\ne 0$. Then $$g(x):=\frac1b\cdot (x-a)$$ and for arbitrary $r+si\in\Bbb C$, consider $p(x):=r+s\cdot g(x)\ \in \Bbb R[x]$, then, as $g(z)=i$, we have $p(z)=r+s\cdot i$.
Note $\:z\in\Bbb C\setminus\Bbb R\:\Rightarrow\:\Bbb R[z] = \Bbb C\:$ follows immediately from $\rm\,[\Bbb C:\Bbb R] = 2\,$ is prime, see my answer. – Math Gems Mar 6 '13 at 18:30
Let $\rm\,\bar R = \phi R,\,$ Then $\rm\, \phi\,$ onto $\rm\!\!\iff\!\! \phi(R[x]) = \bar R[z] = C.\:$ Thus $\,\phi\,$ is surjective iff $\rm\,C\,$ is the ring generated by adjoining $\,z\,$ to the subring $\rm\,\bar R\subseteq C.\,$ To say more requires knowing further details.
For example, if $\rm\,\phi : R\subset C\,$ are fields then $\,\phi\,$ is onto if $\rm\,[C:R] = p\,$ is prime, and $\rm\,z\not\in R,\:$ since then $\rm\,1 < [R[z]:R]\mid [C:R]=p,\,$ so $\rm\,[R[z]:R] = p.\,$ This is true, e.g. for $\rm\,R = \Bbb R,\ C = \Bbb C$.