TITLE: Number of unitary homomorphisms $\phi \ : \ \mathbb{Z}[X]/(X^3+3X+5) \longrightarrow \mathbb{R}$
QUESTION [6 upvotes]: I need some help to solve an exercise: What is the number of unitary homomorphisms $\phi \ : \ \mathbb{Z}[X]/(X^3+3X+5) \longrightarrow \mathbb{R}$?.

Research effort
The zeros of $X^3+3X + 5$ must divide $5$, there are none. Because of its degree, we know that the polynomial is irreducible. The ring $\mathbb{Z}[X]$ is a factorization ring, $(X^3+3X+5)$ is prime and the given quotient ring is an integrity domain. This is what I've got.
Could you please provide me a subtle but fruitful hint to count the number of unitary homomorphisms? Thank you

REPLY [7 votes]: Given a unitary homomorphism
$$
\mathbb{Z}[X]/(X^3 + 3X + 5) \to R
$$
(where $R$ is any ring), once you know where $X$ goes, you know where everything goes (why?)
You can't map $X$ to anything that you please, because since $X^3 + 3X + 5$ is $0$ in the domain, its image must be $0$ in the codomain. Conversely, that is the only obstruction: if you send $X$ to an element of $R$ satisfying this polynomial, you've defined a map. (why?)
Therefore, the unitary homomorphisms are in bijection with the places $X$ can go. In the specific case that $R = \mathbb{R}$, how many of those are there, and how do you know? (Use a little high school math here.)