Prime ideals in certain quadratic ring Let's consider the quadratic ring $\mathbb{Z}[\sqrt{-5}]$ and the principal ideals $(29)$ and $(11)$. Tell whether or not these ideals are prime.
My approach: In order to solve this theorem I am using the following fact:
Fact: If $R$ - commutative ring with $1_R$. Ideal $I$ in $R$ is prime iff factor-ring $R/I$ is integral domain.
Using the fact that $\mathbb{Z}[\sqrt{-5}]\cong \mathbb{Z}[x]/(x^2+5)$ I have derived the following the following results:
$$\mathbb{Z}[\sqrt{-5}]/(29)\cong \mathbb{Z}_{29}[x]/(x^2+5) \quad \text{and} \quad \mathbb{Z}[\sqrt{-5}]/(11)\cong \mathbb{Z}_{11}[x]/(x^2+5).$$
How to show are these quotient-rings are integral domain or not?
Is there any method except computational one? 
Would be very grateful for help!
EDIT: For the second example I know that $x^2+5$ is irreducible over $\mathbb{Z}_{11}$. Hence this quotient-ring is field. Hence it is an integral domain.  
But what about the first one?
 A: It will suffice to show whether or note $x^2+5$ is irreducible over $\Bbb Z_{29}$.  As a quadratic this only factors if there is some $x$ such that $x^2 = -5 \mod 29$.  This can be done by brute force (trying $x = 0, 1, \ldots, 29$) as a last resort.
The techniques of number theory can fairly efficiently compute this question too (whether or not $-5$ is what is known as a quadratic residue mod $29$).  However, this requires some familiarity with the Lagrange symbol and quadratic reciprocity.  
An alternate approach might be to try to directly find a primitive root mod $29$, ie some nonzero value of $z$ such that $z^i \ne 1$ unless $i$ is a multiple of 28.  Then the values of $z^i$ will consist of all values of $\Bbb Z_{29}$.  Then you can see if the congruence class of $-5$ is an even power of $z$ or an odd power ($x^2+5$ would be reducible only if $-5$ is an even power of $z$).  I don't think this will save effort compared to just computing the squares mod $29$ directly, though.
A: To go the next step beyond @RolfHeyer’s answer, I noticed that $6\cdot29=174=13^2+5$. Thus $(13-\sqrt{-5}\,)(13+\sqrt{-5}\,)\in(29)$, so $(29)$ isn’t a prime ideal.
A: First it is easy to check that $-5$ is not a square $\!\bmod{11}$ by Euler's criterion, i.e.
$\!\bmod 11\!:\ \left[a^{\large 2} \equiv -5\right]^{\large 5}\!\Rightarrow\, a^{\large 10}\equiv -5(25)^{\large 2} \equiv -5(3)^{\large 2} \equiv -1\,$ contra little Fermat.  
Therefore $\ x^{\large 2}\equiv -5\,$ is unsolvable so $\,x^2+5\,$ has no root so is irreducible $\bmod {11}.\,$  Otoh
$\!\!\begin{align}
\bmod 29\!:\,\ {-}5\cdot 2^{\large 2} &\equiv 3^{\large 2}\ \ \\[.3em]
 \Rightarrow\ \ \ \ \ \ \ \ \color{#c00}{{-}5}  &\equiv \left(\dfrac{3}2\right)^{\large 2}\!\!\equiv \left(\dfrac{-26}2\right)^{\large 2}\!\! \equiv \color{#c00}{13^{\large 2}}\\[.3em]
\Rightarrow\ \  x^{\large 2}+\color{#c00}{5}&\equiv x^{\large 2}\color{#c00}{-13^2}\equiv (x-13)(x+13)
\end{align}$
Remark $ $ Here we don't actually need to calculate the value of $\,3/2\,$ since we only need to know that $-5$ is a square to infer that $\,x^2+5\,$ is reducible. But it was easy to do so here so we did it.
