# showing that $\mathbb Z [\sqrt{14}]$ is not euclidean regarding usual norm using only 2 and 1 + $\sqrt{14}$.

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...

• Which is exactly the question? To show that the map $x\to N(x)$ is not an euclidean norm? Or that the ring $\Bbb Z[\sqrt{14}]$ is not euclidean / is not factorial? When you start with "let $I=(a+b\sqrt{14})$ be a principal ideal" what are $a,b$, and why should $I$ divide / include $2$ and $1+\sqrt{14}$? Is $I$ in fact $(2, 1+\sqrt {14})$ and we suppose that it is principal? Commented Mar 3 at 17:42
• You say you know that you "have to find an ideal in $\mathbb Z[\sqrt{14}]$ that is not principle" but you don't say why that is the appropriate strategy. The definition of a Euclidean ring does not mention ideals. Do you have some reason why directly applying the definition is not the correct strategy? Commented Mar 3 at 18:25
• oh sorry, the problem is "show that Z[√14] is not an euclidean domain". since every euclidean domain is a PID, if i can show that Z[√14] is not a PID then its not euclidean, am I right? Commented Mar 3 at 20:48
• @mmdmxi: $\Bbb Z[\sqrt{14}]$ IS EUCLIDEAN but it is not norm-Euclidean (the function which plays the role of the known numerical function in $\Bbb Z$ is not the absolute value of the norm). You can see this in About Euclidean rings, Pierre Samuel *Journal of Algebra 19 p.282-301 (1971). Commented Mar 4 at 13:14

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.