Prove that $x^2 − 2$ is irreducible over $\mathbb Q (\sqrt 3)$ Prove that $x^2 − 2$ is irreducible over $\Bbb Q(\sqrt 3)$.
I was originally trying to use the fact that if $K=\Bbb Q(\sqrt 3)[x]/(x^2-2)$ and $[K:\Bbb Q(\sqrt 3)]=2$ then $x^2-2$ is irreducible over $\Bbb Q(\sqrt 3)$. But I'm not sure how to do this. Any help would be great, thank you in advance!
 A: Note that the roots of $f(x)=x^2-2$ are $\pm \sqrt{2}$. Since $f(x)$ has degree $2$, if it were to be reducible then it would be the product of two linear factors, namely $f(x)=(x+\sqrt{2})(x-\sqrt{2})$. 
So suppose that $\sqrt{2}\in\mathbb{Q}(\sqrt{3}) \implies \exists a,b\in \mathbb{Q}: \sqrt{2}=a+b\sqrt{3}\implies 2=a^2+2ab\sqrt{3}+3b^2$ $\implies 2-a^2-3b^2 = 2ab\sqrt{3}$ but the L.H.S. is rational and the R.H.S. is irrational*, a contradiction. As such, $\sqrt{2}\notin \mathbb{Q}(\sqrt{3})$ and so $f(x)$ is irreducible.
*Edit: egreg makes a crucial point in the comments, and this argument is incomplete without taking it into consideration. As such, note that if $b=0, a\neq 0$ then we have $\sqrt{2}\in\mathbb{Q}$, a contradiction. Otherwise, if $a=0, b\neq 0$ then $\sqrt{2}=b\sqrt{3} \implies 2=3b^2 \implies \sqrt{\frac{2}{3}} \in \mathbb{Q}$, a contradiction as well. 
A: If $x^2-2$ is reducible, then it has a root; this means there exist $a,b\in\mathbb{Q}$ such that $(a+b\sqrt{3})^2=2$. This becomes
$$
a^2+2ab\sqrt{3}+3b^2=2
$$
Since $\sqrt{3}$ has degree $2$ over $\mathbb{Q}$, we know that $1$ and $\sqrt{3}$ are linearly independent over $\mathbb{Q}$. Hence the above equality becomes
\begin{cases}
a^2+3b^2-2=0 \\[4px]
2ab=0
\end{cases}
Hence either $b=0$, which is impossible because $a^2\ne2$, or $a=0$. In this case we obtain
$$
(3b)^2=6
$$
that implies $\sqrt{6}$ is rational, which it isn't.
A: Any element of $\Bbb Q(\sqrt 3)$ may be written in the form $s + t\sqrt 3$, where $s, t \in \Bbb Q$.  If $x^2 - 2$ were reducible over $\Bbb Q(\sqrt 3)$, it would be expressible as the product of two linear polynomials
$x^2 - 2 = p(x)q(x), \; p(x), q(x) \in \Bbb Q(\sqrt 3)[x], \; \deg p = \deg q = 1; \tag 1$
that is,
$p(x) = x - a, \; q(x) = x - b; a, b \in \Bbb Q(\sqrt 3); \tag 2$
then 
$x^2 - 2 = (x - a)(x - b) = x^2 - (a + b)x + ab; \tag 3$
it follows that
$b = -a, \; a^2 = 2; \tag 4$
now writing
$a = s + t\sqrt 3, \tag 5$
we find
$a^2 = s^2 + 2st\sqrt 3 + 3t^2; \tag 6$
we see that
$t \ne 0, \tag 7$
lest from (5),
$a \in \Bbb Q, \tag 8$
contradicting (4); likewise, we cannot have 
$s = 0, \tag 9$
lest
$a = t\sqrt 3 \Longrightarrow 2 = a^2 = 3t^2  \Longrightarrow 3 \mid 2, \tag{10}$
impossible.  Thus
$s \ne 0 \ne t, \tag{11}$
and thus from (6),
$\sqrt 3 = \dfrac{2 - s^2 - 3t^2}{2st}; \tag{12}$
but the right-hand side of this equation is rational; thus it cannot be $\sqrt 3$; thus there is no $a \in \Bbb Q(\sqrt 3)$ satisfying (4); hence no such polynomials $p(x), q(x) \in \Bbb Q(\sqrt 3)$; and thus $x^2 - 2$ is irreducible over $\Bbb Q(\sqrt 3)$.
