1. ## Irreducibles

If we define $\displaystyle Z\left[ {\sqrt { - 5} } \right] = \left\{ {a + b\sqrt { - 5} :a,b \in Z} \right\}$, how would I show that 2, 3, $\displaystyle 1 + \sqrt { - 5}$ and $\displaystyle 1 - \sqrt { - 5}$ are all irreducible?

I have shown that none of these are units of $\displaystyle Z\left[ {\sqrt { - 5} } \right]$.

For example, if I assume that $\displaystyle 2 = xy$, I need to show that at least one of x and y is a unit. Obviously if 2 = 2 x 1 = 1 x 2, then 1 is a unit, but how do I show that there are no other factorisations of 2 in $\displaystyle Z\left[ {\sqrt { - 5} } \right]$?

Thanks for help!

2. Originally Posted by BeWareWereWolf
If we define $\displaystyle Z\left[ {\sqrt { - 5} } \right] = \left\{ {a + b\sqrt { - 5} :a,b \in Z} \right\}$, how would I show that 2, 3, $\displaystyle 1 + \sqrt { - 5}$ and $\displaystyle 1 - \sqrt { - 5}$ are all irreducible?

I have shown that none of these are units of $\displaystyle Z\left[ {\sqrt { - 5} } \right]$.

For example, if I assume that $\displaystyle 2 = xy$, I need to show that at least one of x and y is a unit. Obviously if 2 = 2 x 1 = 1 x 2, then 1 is a unit, but how do I show that there are no other factorisations of 2 in $\displaystyle Z\left[ {\sqrt { - 5} } \right]$?

Thanks for help!
Let N be a multiplicative norm such that $\displaystyle N(a + b\sqrt { - 5}) = a^2 + 5b^2$ (verify this is an indeed multiplicative norm).

We have $\displaystyle N(\alpha \beta)=N(\alpha)N(\beta)$ for all $\displaystyle \alpha, \beta$ in $\displaystyle \mathbb{Z}\left[ {\sqrt { - 5} } \right]$, and $\displaystyle N(\alpha)=1$ if $\displaystyle \alpha$ is a unit.

Now consider 2 and suppose 2=$\displaystyle \alpha \beta$.
Then $\displaystyle 4=N(2)=N(\alpha)N(\beta)$. If 2 is not irreducible,$\displaystyle N(\alpha)$ should be 2, but there is no integer satisfyng $\displaystyle 2= a^2 + 5b^2$. Contradiction! Thus 2 is irreducible in $\displaystyle \mathbb{Z}\left[ {\sqrt { - 5} } \right]$.

Now consider $\displaystyle 1 + \sqrt { - 5}$ and suppose $\displaystyle 1 + \sqrt { - 5} = \alpha \beta$.
Then $\displaystyle 6=N(1 + \sqrt { - 5})=N(\alpha)N(\beta)$. If $\displaystyle 1 + \sqrt { - 5}$ is not irreducible, then $\displaystyle N(\alpha)$ should be either 2 or 3. But there is no integer satisfyng $\displaystyle a^2 + 5b^2=2$ or 3. Contradiction! Thus $\displaystyle 1 + \sqrt { - 5}$ is irreducible in $\displaystyle \mathbb{Z}\left[ {\sqrt { - 5} } \right]$.

I leave it to you to show the remainder elements are irreducible in $\displaystyle \mathbb{Z}\left[ {\sqrt { - 5} } \right]$.