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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 $\sqrt{2}\in\mathbb{Q_7}$?

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  • $\begingroup$ Consider the field $\mathbb R$ which does not contain $\pm i$ but does contain $i\times -i$. So you have made an unjustified assumption in your proof. $\endgroup$ – Mark Bennet Dec 15 '17 at 15:49
  • $\begingroup$ perhaps I could argue that $1+i$ divides $p$ ? And also $1-i$ and therefore again $2$ divides $p$. $\endgroup$ – cactus314 Dec 15 '17 at 15:52
  • $\begingroup$ I suppose you can use the same proof. Suppose $a + bi \in \Bbb Q(i)$ has the property that $(a + bi)^2 = 2$. Then $a^2 + 2abi - b^2 = 2$. Comparing the coefficients, it must be that $a = 0$ or $b = 0$, so either $a^2 = 2$ or $-b^2 = 2$. Now use the usual proof that this can't exist in $\Bbb Q$. $\endgroup$ – ÍgjøgnumMeg Dec 15 '17 at 15:59
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Suppose $\sqrt2=a+ib\in\Bbb{Q}[i].$ Then $a^2+b^2=2$ and clearing denominators we can consider the Diophantine equation $p^2+q^2=2r^2.$ Since RHS is even both $p$ and $q$ has to be same parity. Moreover

$$\left(\dfrac{p-q}{2}\right)^2+\left(\dfrac{p+q}{2}\right)^2=\dfrac{p^2+q^2}{2}=r^2.$$

Then using primitive Pythagorean triples we have $p=(m+n)^2-2n^2,$ $q=(m-n)^2-2n^2$ and $r=m^2+n^2$ for some $m,n\in\Bbb{Z}.$

If $p=0$ or $q=0,$ then we have a contradiction $\sqrt2\in\Bbb{Q}.$
Otherwise $b\neq 0,$ which is also contradiction as $\sqrt2\in\Bbb{R}.$

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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)

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