$\Bbb Q(\sqrt{12},i)$ is a simple extension of $\Bbb Q $ Is proving $\Bbb Q(\sqrt{12},i) = \Bbb Q(\sqrt{12} +i)$ the same as proving that $\Bbb Q(\sqrt{12},i)$ is a simple extension of $\Bbb Q$, if not could anyone point me in the right direction.
 A: Clearly, $\Bbb Q(\sqrt{12} + i)\subseteq \Bbb Q(\sqrt{12}, i)$. If you next can show that $\sqrt{12}$ and $i$ are both contained in $\Bbb Q(\sqrt{12} + i)$, you're done. This might be a bit of work, but see what you can do with $\sqrt{12} + i, (\sqrt{12} + i)^2$ and $(\sqrt{12} + i)^3$.
Optionally, you can do it using extension degrees. Because $4 = [\Bbb Q(\sqrt{12}, i):\Bbb Q] = [\Bbb Q(\sqrt{12}, i):\Bbb Q(\sqrt{12} + i)]\cdot [\Bbb Q(\sqrt{12} + i):\Bbb Q]$, showing that $[\Bbb Q(\sqrt{12} + i):\Bbb Q] >2$ means that $[\Bbb Q(\sqrt{12}, i):\Bbb Q(\sqrt{12} + i)] = 1$, which is to say they are the same.
A: All you need to do is show that the degree of $\Bbb Q(\sqrt{12}+i)$ has degree $4$ since clearly $\Bbb Q(\sqrt{12}+i)\subseteq \Bbb Q(\sqrt{12},i)$ and the latter has degree $4$.
Now note that since all degree $2$ extensions of $\Bbb Q$ are of the form $\Bbb Q(\sqrt{d})$ for some square-free integer $d$ we write

$$\sqrt{12}+i = 2\sqrt 3 + i = a+b\sqrt d$$

clearly $d<0$ or else there is no imaginary part, but then $bi\sqrt{|d|}=i$ implying $b= {1\over\sqrt{|d|}}$ which is impossible as $b$ is rational and $\sqrt{|d|}$ is not. So $\sqrt{12}+i$ is not contained in any quadratic extension, so it must be that it generates an extension of degree $4$, i.e. the whole field.
A: Note that $(\sqrt{12}+i)^2=11-2i\sqrt{12}$, so $i\sqrt{12}\in\mathbb{Q}(\sqrt{12}+i)$.
Therefore $(\sqrt{12}+i)i\sqrt{12}=12i-\sqrt{12}\in\mathbb{Q}(\sqrt{12}+i)$.
Can you prove from this that $i\in\mathbb{Q}(\sqrt{12}+i)$?
