Let $K=\mathbb{Q}[\sqrt{2}]$. Consider its ring of integers $O_K=\mathbb{Z}[\sqrt{2}]$ and the ideal ideal $(3)$. I want to show that $(3)$ is inert in $O_K$.
By Gauss' reciprocity $\left(\frac{2}{3}\right)=2^{(3-1)/2}=2=-1\bmod{3}$, so $(3)$ is non-splitting in $O_K$. We also have $2=[K/\mathbb{Q}]=ef$, where $e$ is the ramification index and $f=[(O_K/P):(\mathbb{Z}/(3))]$. So, if I prove that $f=2$, then $e=1$ and we are finished. But I am stuck here.
Another way to do this would be to just prove that $O_K/(3)$ is a field. But I also can't seem to think of something. For $O_K/(2)\to\mathbb{F}_2$ there is the isomorphism $a+b\sqrt{2}\mapsto a$. But for this case, it feels like it is not similar to the latter.
I appreciate any hint or suggestion, thanks!
We show that $O_K/P$ is not isomorphic to $\mathbb{Z}/3\mathbb{Z}$, where $P$ is a prime in $O_K$ lying over $(3)$.
Consider the polynomial $X^2-2$. It has a root in $O_K$, but not in $\mathbb{Z}/3\mathbb{Z}$ because $3\not\mid 2$. So $O_K/P$ and $\mathbb{Z}/3\mathbb{Z}$ are not isomorphic.