Determine ${\rm Frac}\;\mathbb{Z}[\sqrt{2} ]$. Prove that $\mathbb{Z}[\sqrt{2}i ]$ is Euclidean.

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(c) Determine the field of fractions of $\mathbb{Z}[\sqrt{2} ]$.

(d) Prove that $\mathbb{Z}[\sqrt{2}i ]$ is a Euclidean domain under the Euclidean valuation $$ν(a + b \sqrt{ 2} i) = a^2 + 2b^2$$

Our prof brushed up on fields of fractions for like 10 minutes & said that $\mathbb{Q}$ is the field of fractions for $\mathbb{Z}$ but I don't know what this question is asking for. Any help would be appreciated, I have a final in two days!

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$\Bbb Q[\sqrt 2]$ is a field because if $\rho =\sqrt 2$, $\rho^2=2$, that is $\rho^{-1}=\rho/2\in \Bbb Q[\sqrt 2]$. Hence

$(1)$ Any polynomial in $\rho$ can be reduced to one of degree $1$.

$(2)$ $(a+b\rho)^{-1}=\dfrac{a-\rho b}{a^2-2b^2}$

And $a^2-2b^2=0\iff $ both $a=b=0$ by the irrationality of $\rho$.

We have an injection $A=\Bbb Z[\sqrt 2]\to \Bbb Q[\sqrt 2]$ that sends $a+b\rho\to a+b\rho$. Now suppose $K$ is any field that contains $\Bbb Z[\sqrt 2]$. Since it already contains $\Bbb Z$, it contains $\Bbb Q$. Since it contains $\sqrt 2$, it contains $\Bbb Q[\sqrt 2]$. This means $\Bbb Q[\sqrt 2]$ is indeed the field of fractions of $A$.

Have you tried anything for the other question?