Working on Galois Theory. Splitting field of f(x) = $x^2 + 11 \in Q[x]$ Working on Galois Theory. I have a polynomial of f(x) = $x^2 $+11$\in$ Q[x] and I am asked to find the splitting field. I know that solutions to f(x) = 0 are $i\sqrt 11$ and $-i\sqrt 11$. I also believe the splitting field is $Q(i\sqrt 11)$ and not $Q(i, \sqrt 11)$. Thus my degree is 2. Am I on the right track? Most examples I have found deal with roots that are not imaginary. Then with the examples given in lecture I am unsure how to proceed using my elements of Galois group to find subgroups and then the fixed fields. The examples given have more elements and I see how to follow with fixed  elements in those, but not in this case, unless I really messed up the splitting field.
 A: The Splitting field is $\mathbb{Q}(i\sqrt{11}),$ which is not $\mathbb{Q}(i,\sqrt{11}).$ To see it is the splitting field you could note, that $x^2+11$ splits in $\mathbb{Q}(i\sqrt{11}),$ so $\mathbb{Q}(i\sqrt{11})$ contains its splitting field. But the splitting field must contain all roots, hence the splitting field must contain $\mathbb{Q}(i\sqrt{11}),$ hence $\mathbb{Q}(i\sqrt{11}).$ In general if we have a polynomials $f$, and we know all its roots $a_1,...,a_n,$ then its splitting field will be $\mathbb{Q}(a_1,...,a_n),$ though, there are often more clever ways to write it out. 
Now to show that the splitting field is not $\mathbb{Q}(i,\sqrt{11})$ note that $x^2+1$ is irreducible, so $[\mathbb{Q}(i):\mathbb{Q}]=2$. By Fermat's Christmas Theorem we deduce that $11$ is a Gaussian prime, so $x^2-11$ is irreducible in $\mathbb{Q}(i)[x]$ by Eisenstein's Criterion with prime $11$, therefore $[\mathbb{Q}(i,\sqrt{11}):\mathbb{Q}(i)]=2.$ So by the Tower Theorem
$$[\mathbb{Q}(i,\sqrt{11}):\mathbb{Q}]=4>2=[\mathbb{Q}(i\sqrt{11}):\mathbb{Q}],$$
hence $\mathbb{Q}(i,\sqrt{11})$ properly contains $\mathbb{Q}(i\sqrt{11})$ and so is not the splitting field. 
To find the Galois Group in this case is easy as it's degrees $2$, So it's just the Automorphisms sending $i\sqrt{11}$ to itself, and sending it to its additive inverse. The fixed fields in this case are almost trivial, you know they correspond to subgroups by the Fundamental Theorem of Galois Theory, and there are only $2$ subgroups, so the fixed fields are just $\mathbb{Q}$ and $\mathbb{Q}(i\sqrt{11})$. For more complicated extensions finding the Galois group can be more tricky. I don't want to go into details, but you can check out 
https://mathoverflow.net/questions/22923/computing-the-galois-group-of-a-polynomial
To find fixed fields there is a simply, but time consuming method that seems to work. You've constructed a finite algebraic extension, the splitting field of a polynomial. Now, you can easily determine a basis. You can then write out an arbitrary element of the splitting field, and simply test it against each element in a subgroup of the Galois Group. I'm sure there are more efficient methods, but this is simple and it works. 
