Prove whether $(5, X^2 + 3) \subseteq \mathbb{Z}[X]$ is a maximum/prime ideal.
My attempt:
I have a hunch that it is prime I will try to use the following theorem $$ I \subseteq R \text{ is an ideal} \iff R/I \text{ is a domain} $$ with $R$ being commutative. From the third the third isomorphism theorem we can conclude that \begin{align*} \mathbb{Z}[X]/(5,X^2 + 3) &\cong \frac{\mathbb{Z}[X]/(X^2 + 3)}{(5,X^2 + 3)/(X^2 + 3)} \\ &\cong \frac{\mathbb{Z}[X]/(X^2 + 3)}{(5)}. \end{align*} From here I would like to prove that $\mathbb{Z}[X]/(X^2 + 3) \cong \mathbb{Z}[i\sqrt{3}]$. Define $z = i\sqrt{3}$. The evaluation homomorphism is now given by $$ \Phi_z: \mathbb{Z}[X] \to \mathbb{Z}[i\sqrt{3}], f \mapsto f(z) $$ It is clear that $\Phi_z$ is surjective, because $f(aX + b) = ai\sqrt{3} + b$ for arbitrary $a,b \in \mathbb{Z}$. From here I would like to prove that $\text{Ker}(\Phi_z) = (X^2 + 3)$. It is clear that $\text{Ker}(\Phi_z) \supseteq (X^2 + 3)$ because $\Phi_z(X^2 + 3) = (-3 + 3) = 0$.
Now take $f \in \text{Ker}(\Phi_z)$. We will devide with remainders with X^2 + 3. We see that $f = q(X^2 + 3) + r$ with $q,r \in \mathbb{Z}[X]$ and $r = aX + b$ or $r = 0$. If $r=0$, then we're finished. If $r = aX + b$ we see \begin{align*} 0 =\Phi_z(f) &= \Phi_z(q)\Phi_z(X^2 + 3) + \Phi_z(r) \\ &= \Phi_z(r) \\ &= ai\sqrt{3} + b. \end{align*} Since $a,b \in \mathbb{Z}$, we must have $a = b = 0$. So $f \in (X^2 + 3)$. Now use the first isomorphism theorem $$ \mathbb{Z}[X]/(X^2 + 3) \cong \mathbb{Z}[i\sqrt{3}]. $$ The only thing left to do is, proving whether $\mathbb{Z}[i\sqrt{3}]/(5)$ is a domain. From here I have no clue how to pick things up. Any hints/tips?
It's probably easier to do it the other way $$\mathbb Z[x]/(x^2+3,5)\simeq \mathbb Z_5[x]/(x^2+3) $$
where $\mathbb Z_5$ is the integers modulo $5$. Since $x^2+3$ is irreducible over $\mathbb Z_5[x]$ (just check that it has no roots), $(x^2+3)$ is a prime ideal and in fact, maximal ideal in $\mathbb Z_5[x]$ since it is a principal ideal domain. Therefore $\mathbb Z_5[x]/(x^2+3) $ is a field and therefore $(x^2+3,5)$ is a maximal ideal in $\mathbb Z[x]$ (and so prime as well).