I'm trying to prove $\mathbb Z [\sqrt{14}]$ is not euclidean regarding the Norm $N(a+b\sqrt{14})=|a^2-14b^2|$ using $2$ and $1+\sqrt{14}$. I know how to prove it for the Gaussian integer case and I know that I have to find an Ideal in $\mathbb Z [\sqrt{14}]$ that is not principal using $2$ and $1+\sqrt{14}$ but I'm stuck here:
MY ATTEMPT: let $I= a + b\sqrt{14}$ be a principal ideal, then $2 = x(a + b\sqrt{14})$ and $1+\sqrt{14} = y(a + b\sqrt{14})$ where $x,y \in \mathbb Z [\sqrt{14}]$. Then $4=N(x)|a^2-14b^2|$ and $13=N(y)|a^2-14b^2|$. This is where I'm stuck. what do i do or what's the way to prove something is not an euclidean domain using two of its elements? any help would be appreciated...
You are asking two different questions: is $\mathbf Z[\sqrt{14}]$ not norm-Euclidean and is $\mathbf Z[\sqrt{14}]$ not Euclidean? The answers are "yes" and "no": it isn't Euclidean with respect to the absolute value of its norm function, but it is Euclidean with respect to another function.
It is hopeless to approach these problems by showing $\mathbf Z[\sqrt{14}]$ is not a PID since it actually is a PID. To prove that, I would use algebraic number theory: class number bounds imply $\mathbf Q(\sqrt{14})$ has class number $1$, so its ring of integers $\mathbf Z[\sqrt{14}]$ is a PID. But this may be a rather advanced method depending on your background.
That $\mathbf Z[\sqrt{14}]$ is not norm-Euclidean is simple to show. See Theorem 5.8 here. It turns out that $\mathbf Z[\sqrt{14}]$ is a Euclidean domain with respect to a function other than the absolute value of the norm, and this is a theorem of Malcolm Harper. See here. This has been discussed on MSE here.