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Let $F\subset\mathbb{C}$ be a subfield and $f=x^2+2$. Let $K$ be the field generated by $F$ and the complex roots of $f$. Find the Galois Group of $K/F$ for $F\in\{\mathbb{Q}, \mathbb{R},\mathbb{C}\}$.

Is the following right? For $F=\mathbb{C}$ we have $K=\mathbb{C}$, so the Galois Group is trivial.
For $F=\mathbb{R}$ we have $K=\mathbb{R}[i]=\mathbb{C}$ as $\sqrt{2}$ is already in $\mathbb{R}$? Then the Galois Group has 2 elements: The identity and the map sending $i\to-i$?
How do I proceed for $F=\mathbb{Q}$?

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As you've noted for $F = \mathbb{C}$ the Galois group is the trivial one.

If $F=\mathbb{R}$, then we get that $K = \mathbb{R}(i\sqrt{2})$ (in fact it's equal to $ \mathbb{C}$). We then have $[K:F] = \deg (x^2 + 2) = 2$ and so the Galois group is $\mathbb{Z}/2\mathbb{Z}$, as the only group of order 2.

Similarly if $F = \mathbb{Q}$ we get $K = \mathbb{Q}(i\sqrt{2})$ and again the Galois group is $\mathbb{Z}/2\mathbb{Z}$.

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If $F= \mathbb{C}$ we see that the Galois group is trivial.

If $F = \mathbb{R}$ if $K$ is the splitting field of $x^2+2$, we see that $K = \mathbb{R}[i]$. In this case $|\mathbb{R}[i]:\mathbb{R}| = 2$ and there is only one group of order 2, so $Gal(K/F) = \mathbb{Z}/2$.

If $F = \mathbb{Q}$, we see that the splitting field $K = \mathbb{Q}[i\sqrt{2}]$ so the order of the extension is 2 again, and $Gal(K/F) = \mathbb{Z}/2$.

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