# $\sqrt{3} \notin \mathbb{Q}(i)$

I'm trying to show that the irreducible polynomial of $\sqrt{3}$ over $\mathbb{Q}(i)$ is $x^2-3$, where $i \in \mathbb{C}, i^2=1$. But I'm having issues proving $\sqrt{3} \notin \mathbb{Q}(i)$. So far I've said that if $\sqrt{3} \in \mathbb{Q}(i)$, then $\exists a,b \in \mathbb{Q}$ such that $\sqrt{3}=a+bi$, so $3=a^2+2abi-b^2$ and $i=\frac{3-(a^2-b^2)}{2ab} \in \mathbb{Q}$, which is a contradiction. I'm not sure if this argument works though. I applied it forward from a similar problem with $\sqrt{2} \notin \mathbb{Q}(\sqrt{3})$.

Your argument is basically OK, with a small gap: You need to worry about the possibility that $a=0$ or $b=0$ (else you can't divide by $2ab$).

But simpler is this . . .

Since $\sqrt{3}$ is real, $\sqrt{3} = a + bi$ implies $b=0$, and $a=\sqrt{3}$.

But then, $a$ is not rational.

Alternatively, you can argue this way . . .

Since $\sqrt{3}$ is real, $\mathbb{Q}(\sqrt{3}) \subseteq \mathbb{R}$, but since $i \notin \mathbb{R}$, it follows that $\mathbb{Q}(\sqrt{3}) \ne \mathbb{Q}(i)$.

You're doing too much work. ;-) You can directly work in $\mathbb{C}$.

Two complex numbers are equal if and only if they have the same real and imaginary parts: thus $\sqrt{3}=a+bi$ with $a,b\in\mathbb{Q}$ is impossible, because $\sqrt{3}\notin\mathbb{Q}$.