This is probably a dumb question.
We have a theorem that states
If $E$ is a finite extension field of a field $F$, and $K$ is a finite extension field of $E$, then $K$ is a finite extension of $F$, and $$ [K:F]=[K:E][E:F], $$ where $[E:F]$ is the degree $n$ of $E$ over $F$.
I don't quite think I understand this fully. Let $K=\mathbb{Q}(i)$, $E=\mathbb{Q}(\sqrt{2})$, and $F=\mathbb{Q}$. Then (according to the theorem)
\begin{equation} [\mathbb{Q}(i):\mathbb{Q}]=[\mathbb{Q}(i):\mathbb{Q}(\sqrt{2})][\mathbb{Q}(\sqrt{2}):\mathbb{Q}]. \end{equation} But $[\mathbb{Q}(i):\mathbb{Q}]=2$ and $[\mathbb{Q}(\sqrt{2}):\mathbb{Q}]=2$, so wouldn't that imply $[\mathbb{Q}(i):\mathbb{Q}(\sqrt{2})]=1$ and so $\mathbb{Q}(i)=\mathbb{Q}(\sqrt{2})$? But this cannot be (right)?
What is the error in my thinking that's leading to this apparent contradiction? As far as I know the fields $\mathbb{Q},\mathbb{Q}(i),$ and $\mathbb{Q}(\sqrt{2})$ satisfy the hyptheses. I'm sure I'm looking over something simple, but I'm pulling my hair out over here. Any help would be greatly appreciated.
Thank you.
The error in this reasoning is that $\mathbb{Q}(i)$ does not contain $\mathbb{Q}(\sqrt{2})$. Any element $x$ of $\mathbb{Q}(i)$ has the form $x=a+bi$ for $a$, $b\in\mathbb{Q}$. Then for $x^2=(a^2-b^2)+2abi$ to be real, either $a$ or $b$ must be zero. If $a$ is zero, $x^2$ is nonpositive and does not equal $2$, and if $b$ is zero, $x^2=a^2$ is the square of a rational and does not equal $2$. Therefore $\sqrt{2}\notin\mathbb{Q}(i)$.