Show that the ideal $\displaystyle (2 + \sqrt{2}) \subseteq \mathbb{Z}[ \sqrt{2} ]$ is not prime.
Also, could you explain how to how show that any given ideal is not prime.
Recall that an ideal $\displaystyle P$ of a ring $\displaystyle R$ is prime if
* it doesn't equal the whole ring
* If $\displaystyle a,b\in R$ and $\displaystyle ab\in P$, then $\displaystyle a\in P$ or $\displaystyle b\in P$.
Now, consider the following example:
Clearly, $\displaystyle \sqrt{2},1+\sqrt{2}\in\mathbb{Z}[\sqrt{2}]$
So it follows that $\displaystyle \sqrt{2}\cdot (1+\sqrt{2}) = 2+\sqrt{2}\in (2+\sqrt{2})$, but $\displaystyle \sqrt{2}\notin(2+\sqrt{2})$ and $\displaystyle 1+\sqrt{2}\notin(2+\sqrt{2})$. If the ideal was prime, one of the factors $\displaystyle \sqrt{2}$ or $\displaystyle 1+\sqrt{2}$ must be contained in the ideal $\displaystyle (2+\sqrt{2})$, but that's not the case here.
Therefore, $\displaystyle (2+\sqrt{2})$ is not a prime ideal in $\displaystyle \mathbb{Z}[\sqrt{2}]$.
Does this make sense?