# Finding $[\mathbb{Q}(\sqrt{3}, i\sqrt{2}):\mathbb{Q}(\sqrt{3})]$

I am very new to field extensions, so I am very sorry for this basic question.

I want to determine the degree of the field extension $$[\mathbb{Q}(\sqrt{3}, i\sqrt{2}):\mathbb{Q}(\sqrt{3})]$$

So my problem now is that I dont know what the minimal polynomial could look like. I have only dealt with field extensions of a single element over a "normal" field such as $$\mathbb{Q}$$.

Help is very much appreciated!

The extension is generated by a single element $$i\sqrt{2}$$, so the extension degree is just the degree of the minimal polynomial of $$i\sqrt{2}$$ over $$\mathbb{Q}(\sqrt{3})$$. First of all note that $$i\sqrt{2}\notin\mathbb{Q}(\sqrt{3})$$ because $$\mathbb{Q}(\sqrt{3})$$ is a field which contains only real elements. So the extension degree can't be $$1$$, it is at least $$2$$. Also, it is easy to see that $$i\sqrt{2}$$ is the root of the polynomial $$x^2+2\in\mathbb{Q}(\sqrt{3})[x]$$, so the extension degree is also at most $$2$$. Hence it is exactly $$2$$.