I am required to find a direct proof that any element of $\mathbb{Q}(\sqrt{2},\sqrt{3})$ has the form $a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}$, I was just wondering if this proof would be sufficient or if it needs a little more.
To start off I assumed that $\mathbb{Q}(\sqrt{2},\sqrt{3}) = \mathbb{Q}(\sqrt{2})(\sqrt{3})$
Then I go on to say that $\mathbb{Q}(\sqrt{3}) = \{a+b\sqrt{3} : a,b \in \mathbb{Q}\}$ and $\mathbb{Q}(\sqrt{2}) = \{a+b\sqrt{2} : a,b \in \mathbb{Q}\}$
So, $\mathbb{Q}(\sqrt{2},\sqrt{3}) = \mathbb{Q}(\sqrt{2})(\sqrt{3}) = \{(a+b\sqrt{2}) + (c+d\sqrt{2})\sqrt{3} : a,b,c,d \in \mathbb{Q}\}$ $$ =\{a+b\sqrt{2} +c\sqrt{3} +d\sqrt{6} : a,b,c,d \in \mathbb{Q}\}$$
Any feedback greatly appreciated thanks! :)
Let's call $K= \mathbb{Q}(\sqrt{2},\sqrt{3})$ and $L = \{a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}\}$. To prove that $K=L$, by definition you have to show
By the description of $L$, (2) and (3) are obvious. So (as the comment by Will suggests) you have to show that $L$ is a field. The only non-trivial thing to check is what Will suggested.
Obviously you can do it the explicit way with some true grit but here's a slick way to prove it: Pick $x \in L$. $L$ is a $\mathbb{Q}$-vector space. Consider the map $\phi: L \rightarrow L$ sending $a \mapsto xa$. This is a $\mathbb{Q}$-linear map, and it's injective. So linear algebra says it's surjective, so $1$ is in the image, which means $x$ admits an inverse.