Vtyjkyuk

Let $FsubsetmathbbC$ 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 $FinmathbbQ, mathbbR,mathbbC$.

Is the following right?
For $F=mathbbC$ we have $K=mathbbC$, so the Galois Group is trivial.

For $F=mathbbR$ we have $K=mathbbR[i]=mathbbC$ as $sqrt2$ is already in $mathbbR$? Then the Galois Group has 2 elements: The identity and the map sending $ito-i$?

How do I proceed for $F=mathbbQ$?

As you've noted for $F = mathbbC$ the Galois group is the trivial one.

If $F=mathbbR$, then we get that $K = mathbbR(isqrt2)$ (in fact it's equal to $ mathbbC$). We then have $[K:F] = deg (x^2 + 2) = 2$ and so the Galois group is $mathbbZ/2mathbbZ$, as the only group of order 2.

Similarly if $F = mathbbQ$ we get $K = mathbbQ(isqrt2)$ and again the Galois group is $mathbbZ/2mathbbZ$.

If $F= mathbbC$ we see that the Galois group is trivial.

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

If $F = mathbbQ$, we see that the splitting field $K = mathbbQ[isqrt2]$ so the order of the extension is 2 again, and $Gal(K/F) = mathbbZ/2$.