Is $\mathbb{Q}(\sqrt3 i) = \mathbb{Q}(\sqrt3, i)$?

Question

Hi I am a little bit confused by extension fields and splitting fields.

I have an exercise that asks me to find the degree of the extension $E/\mathbb{Q}$, where $E$ is the splitting field of the polynomial $x^4 -2x^2 - 3$.

I found $E = \mathbb{Q}(\sqrt3,i)$ because the polynomial has roots $\pm\sqrt3, \pm i$

Now $\mathbb{Q}(\sqrt3)/\mathbb{Q}$ is a degree $2$ extension, since $x^2 - 3$ is irreducible over $\mathbb{Q}$, and since $i \notin \mathbb{Q}(\sqrt3)$ it follows that $\mathbb{Q}(\sqrt3, i)/\mathbb{Q}(\sqrt3)$ is also a degree $2$ extension, since $x^2 + 1$ is irreducible over $\mathbb{Q}(\sqrt3)$.

So by the tower law, we have $[E : \mathbb{Q}] = [ E :\mathbb{Q}(\sqrt3)][\mathbb{Q}(\sqrt3):\mathbb{Q}] = 2 * 2 = 4$

But why can't I say, that $\mathbb{Q}(\sqrt3,i) = \mathbb{Q}(\sqrt3 i)$?

Answer 1

You probably meant $x^4 - 2x^2 - 3$.

As to whether $\mathbb{Q}(\sqrt{3}i) = \mathbb{Q}(\sqrt{3},i)$, they aren't the same. One way to look at this is that $\mathbb{Q}(\sqrt{3},i)$ has a subextension $K = \mathbb{Q}(\sqrt{3})$ which satisfies $\mathbb{Q} \subsetneq K \subsetneq \mathbb{R}$, while $\mathbb{Q}(\sqrt{3}i)$ is an extension of $\mathbb{Q}$ of degree two (it is the splitting field of $x^2 + 3$) and therefore has no subfields containing $\mathbb{Q}$ other than $\mathbb{Q}$ itself.

Answer 2

Naïvely, elements of $\mathbb{Q}(\sqrt{3}i)$ should be written $a+b\sqrt{3}i$ where $a,b\in \mathbb{Q}$. You can check that this is, in fact, closed under multiplication. It also clearly contains neither $\sqrt{3}$ or $i$.

Alternatively, note that $(x-\sqrt{3}i)(x+\sqrt{3}i)=x^2+3,$ so $\mathbb{Q}(\sqrt{3}i)$ is the splitting field for this polynomial (this tells you why all elements herein must have the above form and also, why the above sub-ring of $\mathbb{C}$ forms a field).