# Show that $\sqrt{2} \notin \mathbb{Q}(i)$ using infinite descent

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

• 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. – Mark Bennet Dec 15 '17 at 15:49
• perhaps I could argue that $1+i$ divides $p$ ? And also $1-i$ and therefore again $2$ divides $p$. – cactus314 Dec 15 '17 at 15:52
• 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 $\Bbb Q$. – ÍgjøgnumMeg Dec 15 '17 at 15:59

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 $$\sqrt2\in\Bbb{R}.$$

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)