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Let $R[x]$ be a polynomial ring over a ring $R$. Let $A$ be any ring and suppose that $\varphi: R \to A$ is a ring homomorphism. Show that the ring homomorphisms $\psi:R[x] \to A$ extending $\varphi$ are in $1-1$ correspondence with elements of $A$.

What I think is asked is to show a bijection $\{\psi : R[x] \to A \mid \psi \text{ homomorphism} \} \longleftrightarrow \{a \mid a \in A\}$?

I've found out that for any polynomial $p = c_nx^n + \dots + c_0 \in R[x]$ I have that $$\psi(p)=\psi(c_n)\psi(x)^n + \dots + \psi(c_0) = \varphi(c_n)\psi(x)^n + \dots \varphi(c_0)$$ since all $c_i$'s are in $R$ and $\psi$ was defined to extend $\varphi$. What I don't understand is how is this homomorphism supposed to correspond to singular $a \in A$ in any way? It looks as $\psi$ is determined only by where it's sending $x$, but I still don't understand what is being asked here?

Gregori
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    As you say, $\psi$ is determined by the element it's sending $x$ to. That element is $a$. – Vercassivelaunos Apr 21 '22 at 20:06
  • @PedroBrunialti: the statement is not about the "free ring on one generator" (i.e., $\Bbb{Z}[x]$). It is relative to a specific homomorphism of one ring $R$ into another ring $A$ and is true (see other comments and an answer). – Rob Arthan Apr 21 '22 at 20:20
  • Yes, you are right, sorry! I've deleted the comment so not to generate any confusion. Thanks! – HelloDarkness Apr 21 '22 at 20:21
  • What @RobArthan is referring to is mentioned in an answer here: https://math.stackexchange.com/a/2576777/1043280 – PTrivedi Apr 21 '22 at 20:22

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All of maps $\psi$ can be thought of as evaluation maps. That is, each polynomial $p(x) \in R[x]$ is mapped to $p(a)$ where $x \mapsto a \in A$ and the coefficients $c_i$ of $p$ are mapped to $\psi(c_i) = \varphi(c_i)$. So each $\psi$ is determined by the element $a \in A$ that $x$ is mapped to the $\psi$s are in $1-1$ correspondence with $A$. In other words, $Hom(R[x],A) \cong A$.

PTrivedi
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