Prime ideals in quadratic ring $\mathbb{Z}[\sqrt{-5}]$ Consider quadratic ring $\mathbb{Z}[\sqrt{-5}]$. For each of the following elements tell whether or not the principal ideal $\langle x\rangle$ generated by $x$ is  a prime ideal.
$x=29,11.$
My approach: Firstly, I proved that $29$ is reducible element and $11$ is irreducible element in this ring. 
1) Consider the quotient-ring $\mathbb{Z}[\sqrt{-5}]/\langle 29\rangle$ and we can show that $\mathbb{Z}[\sqrt{-5}]/\langle 29\rangle \cong \mathbb{Z}_{29}[\sqrt{-5}]$. After trial-error I've found that $\mathbb{Z}_{29}[\sqrt{-5}]$ is not integral domain because $(3+2\sqrt{-5})(3-2\sqrt{-5})=0.$ Hence, an ideal $\langle 29\rangle$ is NOT prime.
2) Consider the quotient-ring $\mathbb{Z}[\sqrt{-5}]/\langle 11\rangle$ and we can show that $\mathbb{Z}[\sqrt{-5}]/\langle 11\rangle \cong \mathbb{Z}_{11}[\sqrt{-5}]$. After trial-error I've found that $\mathbb{Z}_{11}[\sqrt{-5}]$ is integral domain. 
How did I show that it is an integral domain? Suppose $(a+b\sqrt{-5})(c+d\sqrt{-5})=0$, where $a,b,c,d\in \mathbb{Z}_{11}$. Multiplying by conjugates we get that $(a^2+5b^2)(c^2+5d^2)=0$ in $\mathbb{Z}_{29}$. But since it is field then at least one of them is zero. WLOG suppose that $a^2+5b^2=0$. And after some trial-error method with $\pmod {29}$ I have found that $a=b=0$ in $\mathbb{Z}_{11}$.
1) Is my solution and approach correct?
2) Is there is another one more shorter I'd be happy to see it.
3) Also how to apply such method for ideal generated by $1+2\sqrt{-5}$?
 A: Summary
Note that below answer is more extensive than what you need: it also finds the exact prime ideal factorization.
Prime ideals in the ring of integers of a quadratic extension of $\mathbb{Q}$ have norm $p$ or $p^2$ where $p$ is an integer prime; for checking whether a principal ideal is prime, it suffices to take the norm of the ideal; if that is a product of two (or more) primes, then the ideal is not prime; if it is a prime then the ideal is prime; if it is the square of a prime then you have to check the reducibility of $X^2+5$ in $\mathbb{F}_p[X]$.
End summary
Ideal factorization is best done by applying Kummer's theorem, that roughly states that the prime ideals in the ring of integers $A$ of an algebraic number field $\mathbb{Q}(\alpha)$ with minimum polynomial $f = f^\alpha_{\mathbb{Q}}$ always extend $p\mathbb{Z}$, and for primes $p$ not dividing $index[A:\mathbb{Z}[\alpha]]$, relate directly to the way that $\bar{f} = f \mod p$ splits in irreducible factors:
if $\bar{f} = \prod_{j=1}^{k}{\bar{g_j}^{n_j}}$, then $(p) = Ap = \prod_{j=1}^{k}{P_j^{n_j}}$ where $P_j = Ap + Ag_j(\alpha) = (p,g_j(\alpha))$
with ideal norm $N(P_j) = p^{deg(g_j)}$
For $\alpha = \sqrt{-5}$, the ring of integers of $\mathbb{Q}(\alpha)$ is $A = \mathbb{Z}[\alpha]$ as $-5 \equiv 3 \mod 4$; finding the irreducible factors of $f = X^2 + 5$ in $\mathbb{F}_p[X]$ is easy by checking whether $1,2,...,\frac{p-1}{2}$ are zeroes of $f$ (no trial and error required; obviously it helps to first check whether $(\frac{-5}{p}) = -1$ as then you can save yourself the trouble of looking for zeroes; note that this works for all primes as $index[A:\mathbb{Z}[\alpha]] = 1$)
$\mod 29: X^2+5 = (X-13)(X+13)$, hence $(29) = (29,\alpha-13)(29,\alpha+13)$.
$\mod 11: X^2+5$ is irreducible, hence $(11)$ is prime in $A$.
Factorizing the principal ideal $I = (1+2\alpha)$ is a bit more intricate: the norm of the ideal  equals the norm of the element $1+2\alpha$, which is $21 = 3\cdot 7$, implying that  the ideal factors of $I$ are prime ideals of norm $3$ and $7$. This fact implies that $(3)$ and $(7)$ are not prime ideals (as then they would have norms $3^2$ and $7^2$); indeed
$\mod 3: X^2+5 = (X-1)(X+1)$, so $(3) = (3,\alpha-1)(3,\alpha+1) = P_3Q_3$
and
$\mod 7: X^2+5 = (X-3)(X+3)$, so $(7) = (7,\alpha-3)(7,\alpha+3) = P_7Q_7$
Finding the right prime ideal factors of $I$ is possible by evaluating the four possible combinations of these prime ideals of norm $3$ and $7$; note that all these combinations will lead to principal ideals, as $N(1 \pm 2\alpha) = 21$ and also $N(4 \pm \alpha) = 21$.
For example,
\begin{equation}
\begin{split}
P_3Q_7 &= (3,\alpha-1)(7,\alpha+3) \\&= (21,3\alpha+9,7\alpha-7,2\alpha-8) \\&= (21,3\alpha+9,\alpha-25,2\alpha-8) \\&= (21,3\alpha+9,\alpha-4) \\&= (21,\alpha-4) \\&= (\alpha-4)
\end{split}
\end{equation}
because $21 \in (\alpha-4)$.
And
\begin{equation}
\begin{split}
P_3P_7 &= (3,\alpha-1)(7,\alpha-3) \\&= (21,3\alpha-9,7\alpha-7,-4\alpha-2) \\&= (21,3\alpha-9,\alpha+11,2\alpha-20) \\&= (21,3\alpha-9,\alpha+11,2\alpha+1) \\&= (21,\alpha+11,2\alpha+1) \\&= (2\alpha+1)
\end{split}
\end{equation}
because $\alpha+11 = (2\alpha+1)(1-\alpha) \in (2\alpha+1)$ and $21 \in (2\alpha+1)$.
Hence $I = P_3P_7$.
