The title is self-explanatory. I know it's irreducible but is it a prime? How to prove these primality and/or irreducibility of $1+\sqrt{5}$.
Can you just briefly state how a prime is defined under $\mathbb{Z}[{\sqrt{5}}]$? I know that it will only be divisible by its associates and the unity. But please tell about the norm conditions and other properties or the complete rigorous definition of primes.
Definitions are indeed very important. Two answerers on this one question have been led astray thinking you're asking about $\mathbb{Z}[\sqrt{-5}]$, which is a big but understandable mistake, one that many askers on this site have made.
There are at least two different things you might mean by $\mathbb{Z}[\sqrt{5}]$. One is the ring of algebraic integers of the form $a + b \sqrt{5}$, with $a, b \in \mathbb{Z}$; this includes numbers like $1 + \sqrt{5}$ and $-3 + 2 \sqrt{5}$. Or you could mean the ring of all algebraic integers in $\mathbb{Q}(\sqrt{5})$, which could potentially include numbers like $$\frac{3}{4} + \frac{7 \sqrt{5}}{3},$$ but I'm getting ahead of myself.
If you mean the former, the answer is clear: $1 + \sqrt{5}$ is not prime. Since $(1 + \sqrt{5})(1 - \sqrt{5}) = -4$ but $1 + \sqrt{5}$ is not a divisor of $-2$ nor $2$, it follows that $1 + \sqrt{5}$ is irreducible but not prime (I'm taking your word for it that the number is irreducible).
If you mean the latter, then you have overlooked a couple of things, because $1 + \sqrt{5}$ is actually composite. Verify that $$\frac{1 + \sqrt{5}}{2}$$ is an algebraic integer with minimal polynomial $x^2 - x - 1$ (ever hear of a thing called the golden ratio?) Then $$2 \left(\frac{1 + \sqrt{5}}{2}\right) = 1 + \sqrt{5}.$$
In a ring $R$, we say a non-unit, nonzero $p \in R$ is prime if for any $a, b \in R \setminus \{ 0 \}$ such that $p | ab$, we have either $p | a$ or $p | b$. Mathmo123 showed $1 + \sqrt{5}$ wasn't prime.
No, it's not prime. Compare $6$. Is $6$ prime in $\textbf Z$? If it was, we'd see that whenever $6 \mid ab$, either $6 \mid a$ or $6 \mid b$. Yet $6 \mid 3 \times 4$ but $6 \nmid 3$ nor $4$.
Likewise, in $\textbf Z[\sqrt{-5}$, $(1 + \sqrt{-5}) \mid 2 \times 3$ but...
One would usually define a non-zero element $a$ of a ring $R$ to be prime if the ideal it generates is a prime ideal.
Under this definition, $1+\sqrt 5$ is not prime, since $4 =2\times2= (1+\sqrt5)(\sqrt5-1) \in \langle1+\sqrt 5\rangle$, but $2\notin\langle1+\sqrt 5\rangle$.
You should note that $\mathbb Z[\sqrt 5]$ is not the ring of integers of $\mathbb Q(\sqrt 5)$. Rather $\mathbb Z[\frac{1+\sqrt 5}2]$ is, and $1+\sqrt 5$ is prime in this ring.
We need to agree on definitions before giving a meaningful answer. If you asked Ethan Bolker fifty years ago, he'd say yes, $1 + \sqrt{-5}$ is indeed prime in $\mathbb Z[\sqrt{-5}]$. It's there in black and white on page 102 of his classic Elementary Number Theory: An Algebraic Approach:
note that $1 + \sqrt{-5}$ is prime in $\textbf A(-5)$ because its norm is 6, which is prime in $\textbf B(-5)$
Dover reprinted the book in 2007, and Bolker wrote a new preface and corrected some errors, but also suggested he would have written the book very differently today.
Fifty years ago, he was using "prime" to mean what today most of us call "irreducible." That is, if $p$, which is not a unit, is divisible only by units and associates, it is irreducible. For example, 7 is irreducible in $\mathbb Z$ since it's only divisible by $-7, -1, 1, 7$.
In infinite domains, we can narrow down where to look by looking only at numbers that are in some sense "smaller" than $p$. In imaginary rings like $\mathbb Z[\sqrt{-5}]$, we can take "smaller" to mean closer to 0, or having a smaller norm.
To call it prime, we now state the additional requirement that whenever $p \mid ab$, $p$ must also divide one of $a$ or $b$, maybe both. In domains like $\mathbb Z$ or $\mathbb Z[\sqrt{-2}]$, the additional requirement is irrelevant: all irreducibles are prime.
But in domains like $\mathbb Z[\sqrt{-5}]$, it makes for a very important distinction. To go back to our example of 7, we see that it is irreducible but not prime in $\mathbb Z[\sqrt{-5}]$, since $7 \mid (3 - \sqrt{-5})(3 + \sqrt{-5})$ but $7 \nmid (3 - \sqrt{-5})$ and $7 \nmid (3 + \sqrt{-5})$ either.
Likewise, we see that $1 + \sqrt{-5}$ is irreducible, since it's divisible only by $-1 - \sqrt{-5}, -1, 1, 1 + \sqrt{-5}$. But it's not prime, because $(1 + \sqrt{-5}) \mid 2 \times 3$ but $(1 + \sqrt{-5}) \nmid 2$, $(1 + \sqrt{-5}) \nmid 3$. And 2 and 3 are also irreducible but not prime in this domain.
And then the norm is a stronger indication of primality. Given a number $z \in \mathbb Z[\sqrt{-5}]$ with nonzero imaginary part, if $N(z)$ is prime in $\mathbb Z$, then $z$ is prime in $\mathbb Z[\sqrt{-5}]$. Quick exercise: verify $3 + 2 \sqrt{-5}$ is irreducible and prime.
As for a purely real positive odd number $p$ that is prime in $\mathbb Z$ and irreducible in $\mathbb Z[\sqrt{-5}]$, this is how you can tell tell if it's also prime: try to solve $x^2 \equiv p - 5 \pmod p$. If that has no solutions, then $p$ is indeed prime.
For example, with 7 we see that $3^2 \equiv 2 \pmod 7$ and $4^2 \equiv 2 \pmod 7$, so 7 is not prime, readily leading us to find that $N(3 + \sqrt{-5}) = 14$ and $N(4 + \sqrt{-5}) = 21$ (there are more, of course, but these are enough to make the point that 7 is irreducible but not prime). On the other hand, 11 is irreducible and prime since $x^2 \equiv 6 \pmod{11}$ has no solutions.
These remarks can be generalized to other imaginary quadratic integer rings. For instance, you want to see if $x^2 \equiv p - 10 \pmod p$ has solutions to determine primality in $\mathbb Z[\sqrt{-10}]$.
One more thing: it might be worthwhile to say that 5 is neither irreducible nor prime in $\mathbb Z[\sqrt{-5}]$, since $(-1)(\sqrt{-5})^2 = 5$.
