Prove that the $q(x)=X^2+3$ is irreducible over $\Q[\sqrt{2}]$. Is my proof correct?
Proof. Since it is a polynomial of second degree it factors iff it has $2$ polynomials of degree $1$. Since we are looking for a factorization over $F[x]$ where $F$ is a field.The factorization is unique.
Now I know factorization is always possible over $\Bbb C$. So $(X^2-i\sqrt{3})(X^2+i\sqrt{3})$.!!! And this factorization is unique. And it cannot be factored in any other way to any subfields of $\Bbb C$!!. (I use this fact by intuition don't know why???). So only job left is to prove that $$i\sqrt{3}$$ does not belong to $\Q[\sqrt{2}]$. Say for contradiction it does then $$i\sqrt{3}=a+b\sqrt{2}$$ ......$$\sqrt{2} =\frac{a^2+2b^2+3}{ab} $$ where a,b are integers so we have a contradiction. Does my proof miss a step where I should prove something that I have taken for granted? Is it bulletproof and complete? Also. Another way to go is that this polynomial is equal to the minimal polynomial of the extension $\Q[i\sqrt{3}]:\Q[\sqrt{2}]$. But where I stuck is that $$X^2+3$$ is irreducible over $\Q$ and $ i\sqrt{3}$is a root so $\Q[i\sqrt{3}]:\Q$ is of degree 2 but that doesn't make sense since $$[\Q[i\sqrt{3}]:\Q]=[\Q[i\sqrt{3}]:\Q[\sqrt{2}] ][\Q[\sqrt{2}]:\Q]$$ so $2=2 *1$ which means $[\Q[i\sqrt{3}]=\Q[\sqrt{2}]$ which is a contradiction. Am I correct? Any help on how to do a proof with the second method?
A monic quadratic polynomial $f(x)$ over a field $K$ is reducible if and only if it can be written as $(x-a)(x-b)$ with $a,b\in K$. Now take $f(x)=x^2+3$ and $K=\mathbb{Q}[\sqrt{2}]$. We have $a=x_1+y_1\sqrt{2}$ and $b=x_2+y_2\sqrt{2}$. Compute $(x-a)(x-b)$ and compare it with $x^2+3$.
Remark: The degree $[L:K]$ is only defined for $K\subseteq L$. This is not the case with your two fields.