TITLE: Ring homomorphism between $R[x]$ and $R$
QUESTION [7 upvotes]: I am trying to show that, if $\phi : R[x] \rightarrow R$ is a ring homomorphism, such that $\phi$ restricted to $R$ is identity map, then $\phi=\phi_a$ for some $a$, where $\phi_a$ is the substitution homomorphism.
What I'm strugglin with is: what does $\phi|_R$ mean exactly? I am assuming that we want to restrict $\phi$ on only constant polynomials, and there, $\phi = id$, but is that correct?
But from there on, I don't manage to proceed. Can someone give a hint?
-marie

REPLY [3 votes]: HINT $\ $ Since $\phi$ is a ring hom $\ \phi(f+g)\: =\: \phi(f)+\phi(g),\ \ \phi(f\:g)\: =\: \phi(f)\:\phi(g)\:.\:$ Hence by induction
$$\phi(x)\: =\:a\ \ \Rightarrow\ \ \phi(r_0 +\: r_1\ x + \:\cdots\: + r_n\ x^n)\ =\ r_0 +\: r_1\ a +\:\cdots\: + r_n\ a^n$$ 
Generally algebra hom's are determined by their values on generators, and $R[x]$ is generated as a ring by $R$ and $x$.