# Ring homomorphism between $R[x]$ and $R$

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

In general, given any sets $X$ and $Y$, a subset $S\subseteq X$, and a function $f:X\to Y$, the function $f|_S:S\to Y$ is the function $f$, where the domain is now considered to be $S$. The subset $S\subset R[x]$ consisting of constant polynomials can be identified with $R$ in the obvious manner, and in fact we usually write $R\subset R[x]$, leaving this identification implicit. So you are correct, $\phi|_R$ means exactly the restriction of $\phi:R[x]\to R$ to the subset of $R[x]$ consisting of constant polynomials.
Hint on how to proceed: show that, for any ring homomorphism $\psi:R[x]\to T$ where $T$ could be any ring whatsoever, the place where any $p\in R[x]$ is sent by $\psi$, namely $\psi(p)$, is determined completely by where elements of $R$ are sent, and where $x$ is sent. That is, if I tell you how $\psi$ acts on elements of $R$, and how $\psi$ acts on $x$, and tell you that $\psi$ is a ring homomorphism, you can figure out what $\psi$ does to any element of $R[x]$.
Note that we are assuming $\phi:R[x]\to R$ doesn't change elements of $R$, and then think about where $x$ is sent under the substitution homomorphism $\phi_a$...
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 $x$.