I'm working through Dummit and Foote's algebra textbook. An exercise asks to determine the splitting field and degree over $\mathbb{Q}$ of $x^{2}+2$.
I have what I think is an accurate solution, but I must have made a mistake somewhere because it does not line up with the true answer. Here's my solution:
First, it's easy to see that $x^{2}+2$ is irreducible over $\mathbb{Q}$ (it's a quadratic with no root in $\mathbb{Q}$). So the splitting field of $x^{2} + 2$ is not $\mathbb{Q}$ itself, and so the splitting field has degree $\geq 2$.
Noting that $x^{2} + 2 = (x - i \sqrt{2})(x + i \sqrt{2})$, I determined that the splitting field must contain $\pm i \sqrt{2}$. So I thought to check $\mathbb{Q}[i \sqrt{2}]$, which definitely contains these two roots. Then $x^{2}+2$ definitely splits over $\mathbb{Q}[i \sqrt{2}]$.
Since $i \sqrt{2}$ is algebraic over $\mathbb{Q}$ (as we've seen, it's a root of $x^{2}+2 \in \mathbb{Q}[x]$), we know $\mathbb{Q}[i \sqrt{2}]$ is a field (and so an extension of $\mathbb{Q}$).
So then $\mathbb{Q}[i \sqrt{2}] = \{a + b i \sqrt{2} : a, b \in \mathbb{Q}\}$, and it has the basis $\{1, i \sqrt{2}\}$ when viewed as a vector space over $\mathbb{Q}$. So it's a vector space of dimension 2, and so the extension has degree $[\mathbb{Q}(i \sqrt{2}) : \mathbb{Q}] = 2$.
But this doesn't line up with the true solution, which says that the splitting field is $\mathbb{Q}(i, \sqrt{2})$, which can be seen to be of degree $4$. Can anyone see an error in this solution?