Homomorphisms from $\mathbb{Z}[x] \to R$ 
Problem: Let $R$ be a ring with identity. Prove that the number of ring homomorphisms $\mathbb{Z}[x] \to R$ is equal to the number of elements of $R.$

I can't seem to find any approach for this. There are no generators of $\mathbb{Z}[x]$. Can we try something else? Any hint is highly appreciated.
Thank you.
 A: The evaluation map at $r \in \mathbb{R}$ from $\mathbb{Z}[x] \to \mathbb{R}$ is a ring homomorphism.
Thus for any $r \in \mathbb{R}$, there exists $p(x) \in \mathbb{Z}[x]$ such that $$p(x) \mapsto p(r)$$  is the ring homomorphism. Hence the number of ring homomorphisms is the cardinality of $\mathbb{R}$.
A: 
For each $r \in R$ consider the ring homomorphism $\operatorname{ev}_r : \mathbb Z[x] \to R$ given by $\operatorname{ev}_r(p(x)) = p(r)$. Then the map $\operatorname{ev}_{(-)} : R \to \operatorname{Hom}_{\textsf{Ring}}(\mathbb Z[x],R)$ is bijective.

Proof. It is injective since if $\operatorname{ev}_r = \operatorname{ev}_s$, then $r=\operatorname{ev}_r(x) = \operatorname{ev}_s(x) = s$. It if surjective since if $f : \mathbb Z[x] \to R$ is a ring homomorphism, then for any non-negative integer $n$ and any $a_0,\dots,a_n \in \mathbb Z$ we have
$$\begin{align}
f \bigg( \sum_{k=0}^n a_kx^k \bigg) &= \sum_{k=0}^n f(a_k)f(x)^k \\ &=^{\color{blue}1} \sum_{k=0}^n (a_k1_R)f(x)^k \\ &= \sum_{k=0}^n a_kf(x)^k \\ &= \operatorname{ev}_{f(x)} \bigg( \sum_{k=0}^n a_kx^k \bigg) \end{align}$$
and so $f = \operatorname{ev}_{f(x)}$.

${}^{\color{blue}1}$ Recall that there is only one ring homomorphism $\mathbb Z \to R$, namely $m \mapsto m1_R$.
A: Let $\phi:\mathbb{Z}[X]\to R$ be a ring homomorphism. Since $1\in\mathbb{Z}[X]$ must always be mapped to $1\in R$, we have no choice for what $\phi$ does with $\mathbb{Z}\subset\mathbb{Z}[X]$.
This leaves only what $\phi$ is going to do with $X\in\mathbb{Z}[X]$. What could $\phi(X)$ be? The answer is "anything you want in $R$". And once we know what $\phi(X)$ is, by the rule of ring homomorphisms, we know what $\phi$ does to every element of $\mathbb{Z}[X]$, thus determining $\phi$. There are exactly $|R|$ choices for $\phi(X)$, hence the result.
