The proof that $1-\sqrt{-5}$ is irreducible is incorrect. In addition to a minor (and alas, common) mistake in writing that makes what you write not what you intended to write, there is an assertion which is just plain false.
Explicilty, you write:
Assume $1-\sqrt{-5}$ is reducible, then there must exist $a,b \in \mathbb{Z}[\sqrt{-5}]$ so that $N(1-\sqrt{-5})=N(a)N(b) \Rightarrow N(a)=N(b)= \pm (1-\sqrt{-5})$
First: the use of $\Rightarrow$ is incorrect. What you have written is that there exist $a$ and $b$ in $\mathbb{Z}[\sqrt{-5}]$ for which the following statement holds:
If $N(1-\sqrt{-5}) = N(a)N(b)$, then $N(a)=N(b) = \pm (1-\sqrt{-5})$.
What you actually wanted to write was that
If there exist $a,b\in\mathbb{Z}[\sqrt{-5}]$ such that $N(1-\sqrt{-5}) = N(a)N(b)$, then $N(a)=N(b)=\pm(1-\sqrt{-5})$.
What's the difference? The first statement will be true if you can find an $a$ and a $b$ for which $N(1-\sqrt{-5})$ is not equal to $N(a)N(b)$! It will be true that the implication holds, because the antecedent will be false. So, exhibiting $a=7$ and $b=10578432$ makes the statement you wrote true. However, they are irrelevant towards the second statement (and towards establishing what you want to establish, namely, that no such $a$ and $b$ exist).
Second: this is incorrect. What you want to assume is that there exist $a$ and $b$ such that $1-\sqrt{-5} = ab$, and neither $a$ nor $b$ are units; you do not simply want to assume that the product of the norms of $a$ and $b$ equals the norm of $1-\sqrt{-5}$.
Third: Even so, you conclusion is nonsense. The norm of any element of $\mathbb{Z}[\sqrt{-5}]$ must be an integer. What you want to conclude is that $N(a)N(b) = N(1-\sqrt{-5}) = 6$, and then get a contradiction. The norm cannot equal $\pm(1-\sqrt{-5})$.
It's also nonsense to say "since $1-\sqrt{-5}$ is not a quadratic remainder modulo $5$". It's not even a remainder modulo $5$, because it's not an integer!
The argument about $2$ not being prime is likewise incorrect in its use of the norm, which quickly reduces to nonsense symbols being strewn around:
But with $a,b \in \mathbb{Z}[\sqrt{-5}]$ it immediately follows that for :
$(1\pm\sqrt{-5})= 2(a+b\sqrt{-5})$ $2b = \pm 1$. So 2 is not a prime in $\mathbb{Z}[\sqrt{-5}]$.
This is nonsensical as written. I suspect you wanted to say something like
But there cannot exist $a,b\in\mathbb{Z}$ (not in $\mathbb{Z}[\sqrt{-5}]$) such that $2(a+b\sqrt{-5}) = 2a+2b\sqrt{-5} = 1-\sqrt{-5}$.
Same issues with the rest of the arguments. They are either terse to the point of nonsense, or contain incorrect or incoherent claims. I strongly urge you to write complete sentences, use words, and don't over rely on symbols. And to read your own arguments with a critical eye after you are done.
In case someone is still interested in an elementary solution: By proceeding analogously to the case of $\mathbb{Z}[i]$ (in Neukirch's book on algebraic number theory) we can give a list that contains all prime elements of $\mathbb{Z}[\sqrt{2}]$. However, the solution is incomplete insofar that associated prime elements might appear more than once.
The corresponding theorem about the representability of integer primes as the norm of elements in $\mathbb{Z}[\sqrt{2}]$ is the following.
Lemma: For a prime number $p>2$, the diophantine equation
\begin{equation*}
p = a^2 - 2b^2
\end{equation*}
is solvable in integers $a$ and $b$ if and only if $p \equiv 1$ or $7 \bmod 8$.
Note that $2 = 2^2-2*1^1$ is representable as well.
Proof: This is analogous to Neukirch's proof that a prime $p>2$ is representable as a sum of two squares if and only if $p \equiv 1 \bmod 4$. We replace Wilson's theorem (Theorem 80) by Theorem 95 in Hardy and Wrights classic book. It says that the congruence $x^2 \equiv 2 \bmod p$ has a solution if $p \equiv 1, 7 \bmod 8$ (two is a quadratic residue mod $p$).
Note that squares are $\equiv 0,1,4 \bmod 8$, hence only odd numbers $\equiv 1,7 \bmod 8$ can be represented by $a^2-2b^2$.
For the converse, by theorem 95, there is an integer $x$ such that $p\vert x^2-2 = (x-\sqrt{2})(x+\sqrt{2})$, so $p$ is no longer prime in $\mathbb{Z}[\sqrt{2}]$, since $p$ divides neither of the factors in $\mathbb{Z}[\sqrt{2}]$. Hence $p$ can be written as the product $\alpha \beta$ of two non-units, such that for the norm $N(p) = p^2 = N(\alpha) N(\beta)$ holds. Since $\alpha$ and $\beta$ are non-units, $N(\alpha) = N(\beta) = p$ (up to a possible minus sign of which we can get rid of by choosing appropriate associate elements). If we write e.g. $\alpha = a+bi$, we found the integers $a$ and $b$ with $N(\alpha) = a^2 - 2b^2 = p$.
Now the prime elements of $\mathbb{Z}[\sqrt{2}]$ (up to, and possibly including) associates are
- $\sqrt{2}$,
- $\alpha \in \mathbb{Z}[\sqrt{2}]$ such that $N(\alpha) = p$ is a prime $p \equiv 1, 7 \bmod 8$,
- and integer primes $p \equiv 3,5 \bmod 8$.
Now, one can use the proof as for the prime elements in $\mathbb{Z}[i]$ almost word by word by first showing that all these elements are in fact prime elements, and then by showing that any prime element must be associated to an element in the list. (Elements of prime norm are prime because they cannot be further factorized in non-units (and using that we work in an UFD here), and elements in (3) with square prime norm are prime because they are not representable) (any prime element has norm in $\mathbb{Z}$ which factorizes there uniquely. Then one can say again that the prime element must have norm $\pm 1$, $\pm p$ or $\pm p^2$, for $p$ being one of the primes dividing the norm in $\mathbb{Z}$. $\pm 1$ cannot be, $\pm p$ and we are in case (1) or (2), and $\pm p^2$ gives case (3) ).
Best Answer
Yes, $\sqrt{-5}$ is irreducible and prime in $\mathbb{Z}[\sqrt{-5}]$. If it wasn't, it would be possible to find numbers in the domain such that $ab = c \sqrt{-5}$ yet $\sqrt{-5}$ divides neither $a$ nor $b$. Since the norm is multiplicative, that would require $N(ab) = 5$ and so the only possibilities are $N(a) = 1$, $N(b) = 5$ or vice-versa.
For the second part of your question, by $\mathbb{Z}[\sqrt{5}]$ do you mean $\mathbb{Z}[\phi]$ where $$\phi = \frac{1 + \sqrt{5}}{2}?$$ I'll assume that you do. Since $x^2 \equiv \pm 2 \pmod 5$ is insoluble in integers, we can be assured that primes ending in $3$ or $7$ are at least irreducible in both domains (and of course prime in $\mathbb{Z}[\phi]$).
So what we're looking for is that $x^2 \equiv -5 \pmod p$ has no solutions but $x^2 \equiv 5 \pmod p$ does. I'd work out the Legendre symbol for you but I'm running late to dinner.