Standard exercise is to show $\sqrt{2} \notin \mathbb{Q}$ (e.g. Wikipedia). There are examples on Math.SE such as [1, 2, 3, 4].
If we adjoin an element to $\mathbb{Q}$ does the same proof by contradiction work? I'd like to show $\sqrt{2} \notin \mathbb{Q}(i)$ where $x = i$ is a solution to $x^2 + 1 = 0$. Perhaps I could begin the same way. We are solving:
$$ p^2 = 2 q^2 $$
where $p, q \in \mathbb{Z}[i]$. We certainly have $2 = (1+i)(1-i)$ so that $2$ does factor in this new ring, but these two numbers are relatively prime. We have:
$$ \sqrt{1 \pm i } \notin \mathbb{Q}(i)$$
And therefore, their product does not belong in that field as well. Does that look correct? Is has correct ideas, but I don't think the logic is presented correctly.
There are may solutions to this problem, so I've also indicated a certain line of proof I'm trying to follow, using infinite descent. I'm asking, Does the descent argument we typically use over $\mathbb{Z}$ carries over to $\mathbb{Z}[i]$ ?
Alternatively, we can write any element of $\mathbb{Q}(i)$ as $a + bi$ with $a,b \in \mathbb{Q}$. If we have $(a+bi)^2 = 2$ Then $a^2 - b^2 = 2$ and $2ab = 0$. Then necessarily $b = 0$ and $a^2 = 2$. This only uses descent over $\mathbb{Q}$, and we never use that $2$ factorizes into $(1+i) \times (1-i)$.
In contrast: Proving $\sqrt{2}\in\mathbb{Q_7}$?
We want to show that $\mathbb Q(i)$ does not contain a root to $x^2+2=0$.
To do this ione possible route is to this we work in $\mathbb C$ and notice that $\mathbb Q(i)$ must contain $\{a+bi| a,b\in \mathbb Q\}$.
If we can show $\{a+bi| a,b\in \mathbb Q\}$ is a field then clearly we must have $\mathbb Q(i)=\{a+bi| a,b\in \mathbb Q\}$.
Clearly it is closed under addition,multiplication and additive inverses. Recall that $(a+bi)^{-1}=\frac{a-bi}{a^2+b^2}$ which is also of the desired form.
Thus all we have to prove is that $(a+bi)^2\neq -2$ for $a,b\in \mathbb Q$.
Clearly $\sqrt{2}i$ and $-\sqrt{2}i$ are not of this form. (One can also prove that $(a+bi)^2=-2$ has no solution by hand)