Let $C$ be the set of all complex numbers of the form $a+ b \sqrt {5}i$, where $a$ and $b$ are integers... Let $C$ be the set of all complex numbers of the form $a+ b \sqrt {5}i$, where $a$ and $b$ are integers. Prove that $7$, $1 + 2\sqrt {5}i$, and $1 - 2\sqrt {5} i$ are all prime in $C$.
-I am really lost in this question, the closest looking thing I could muster up from our lecture notes was the section covering the fundamental theorem of arithmetic, and more specifically covering uniqueness. Really not sure how to approach this problem, any help and jump start is appreciated.
 A: Hint: Use the norm: $N(a+b\sqrt 5i)= a^2+5b^2$.
Suppose $7=ab$. This implies $N(u)N(v) = N(7)=49$. 
$N(u)=1, N(v)=49$, write $u=a+i\sqrt5 b$, $N(u) =a^2+5b^2=1$ implies $b=0, a^2=1$.
$N(u)=7=N(v)$. Write $u=a+ i\sqrt5 b$, $N(u) =a^2+5b^2=7$ we deduce that $\mid b\mid \leq 1$. Suppose that $\mid b\mid =1$, $a^2 +5=7$, $a^2=2$ impossible.
Suppose that $b=0$, $a^2=7$ impossible.
Write $1+2i\sqrt 5 =uv$, $N(u)N(v) =21$. 
$N(u)=1, N(v)=21$ implies as above $u=1$.
$N(u)=3, N(v)=7$. The argument above shows you can't have $N(v)=7$.
A: Well, I'm very hesitant to call 7 prime in this domain, since $$(1 - 2 \sqrt{-5})(1 + 2 \sqrt{-5}) = 3 \times 7 = 21.$$ When you only deal with unique factorization domains like $\mathbb{Z}$, $\mathbb{Z}[i]$, $\mathbb{Z}[\sqrt{-2}]$, etc., it's very easy not to worry about the difference between irreducibles and primes, because it's irrelevant:


*

*A number $n$ is irreducible if in every instance of $n = ab$, either $a$ or $b$ is a unit (like 1 or $-1$).

*A number $p$ is prime if in every instance of $p \mid ab$, either $p \mid a$ or $p \mid b$ holds true (maybe both).


Look it up in your textbook, you should find it. Unless your textbook is from prior to 1970...
So we see that 7 is indeed irreducible, since these are the only ways to express it as a product of two numbers in $\mathbb{Z}[\sqrt{-5}]$: $1 \times 7$, $-1 \times -7$. Both 1 and $-1$ are units.
But 7 is not prime, since, even though $7 \mid 21$, we see that $7 \nmid (1 - 2 \sqrt{-5})$, $7 \nmid (1 + 2 \sqrt{-5})$ either.
The nice thing about domains with complex numbers is that finitely few numbers can have a given norm. If you don't have any better ideas, you can draw a circle or an oval in the complex plane and just test every combination of two numbers. For example, $(\sqrt{-5})(1 + \sqrt{-5}) = -5 + \sqrt{-5}$, which falls outside of the oval bounded by $21i$, $1 + 2 \sqrt{-5}$, $1 - 2 \sqrt{-5}$, $-21i$, $-1 - 2 \sqrt{-5}$, $-1 + 2 \sqrt{-5}$.
There is a better idea, though. Since $N(1 + 2 \sqrt{-5}) = 1^2 + 5 \times 2^2 = 21$, and the norm function is multiplicative, what you can do instead is to search for a number $a$ such that $N(a) = 3$ and a number $b$ such that $N(b) = 7$ (we don't have to worry about negative norms, another nice feature of domains with complex numbers).
But $x^2 + 5y^2 = 3$ obviously has no solutions. Neither does $x^2 + 5y^2 = 7$. This proves that $1 + 2 \sqrt{-5}$ is irreducible. And since the sign of the imaginary part (or the real part, for that matter) is irrelevant for calculating the norm, this also proves that $1 - 2 \sqrt{-5}$ is irreducible.
However, none of these numbers are prime, at least not according to the definition given in almost any modern algebraic number theory textbook. The numbers you were given were chosen specifically to allow you to find failures of $p \mid ab$ means $p \mid a$ or $p \mid b$.
If you're given some other number $p$ that is prime in $\mathbb{Z}$ and irreducible in $\mathbb{Z}[\sqrt{-5}]$, there is an even easier way to test whether it is also prime: if you can solve $x^2 \equiv -5 \pmod p$, then $p$ is irreducible but not prime in $\mathbb{Z}[\sqrt{-5}]$. For example, finding $3^2 \equiv -5 \pmod 7$ leads us to $(3 - \sqrt{-5})(3 + \sqrt{-5}) = 14$, which is a multiple of 7 just like 21.
