I want to prove number of prime ideal of $ \Bbb{Z}[ \sqrt{-5}]$ above $(2)$ is 1.
My try: From ramification theory, (number of prime ideal above $(2)$)=[$\Bbb{Q}[ \sqrt{-5}:\Bbb{Q}]/$(order of decomposition group at $(2)$). Now, decomposition group is isom to $Gal(\Bbb{Q}_2(\sqrt{-5})/\Bbb{Q}_2)$ which is order $2$, so there is only one prime above $(2)$.
But I think this is overkill, could you tell me what is the standard way to prove this kind of statement ?
Presumably you know that for a number field $K$ with number ring $\mathcal{O}_K$, $[K : \mathbb{Q}] = n$, and $p$ a prime in $\mathbb{Z}$, the sum of the products of the ramification and inertial degrees of the primes $P_1, \dots, P_r$ of $\mathcal{O}_K$ lying over $p$ is equal to $n$, i.e., $\sum_{i = 1}^r e_if_i = n$. In your case, $n = 2$, so if $p$ ramifies in $\mathbb{Z}[\sqrt{-5}]$ you know it's totally ramified. Thus, it suffices to use the discriminant criterion, which says that $p$ ramifies in a number ring $\mathcal{O}_K$ if and only if it divides the discriminant of $\mathcal{O}_K$.
Since $-5 \equiv 3 \pmod{4}$, the discriminant of your number ring is $-20$. Clearly, $2 \mid -20$, so $2$ is ramified in $\mathbb{Z}[\sqrt{-5}]$.