a) Find a field extension $K/\mathbb{Q}$ such that $\text{Gal}(K/\mathbb{Q})\cong Z_5$ ($Z_5$ denotes the cyclic group on $5$ elements).
b) Let $L=K(\sqrt{2})$. Show $L/\mathbb{Q}$ is Galois and determine the cardinality of $\text{Gal}(L/\mathbb{Q})$.
c) Give the isomorphism type of $\text{Gal}(L/\mathbb{Q})$ explicitly and describe the automorphisms explicitly.
My thoughts: For part (a), we already know $\mathbb{Q}(\zeta_{11})/\mathbb{Q}$ is Galois with $\text{Gal}(\mathbb{Q}(\zeta_{11})/\mathbb{Q})\cong Z_{10}$. Also, $Z_{10}$ has $Z_2$ as its unique normal subgroup of index $5$, and thus $Z_2$ corresponds to a Galois extension $K/\mathbb{Q}$ with $\text{Gal}(K/\mathbb{Q})\cong Z_{10}/Z_2\cong Z_5$.
For part (b), I'm not sure if I can just use the fact that $L$ is a quadratic extension of $K$, or if I need to explicitly determine $K$ first and find a polynomial for which $L=K(\sqrt{2})$ is its splitting field. I think I can determine $K$ explicitly (as in finding a suitable generator to adjoin to $\mathbb{Q}$) using my work from part (a), but this might be working too hard.
For part (c), I think this will follow quickly once I have part (b), but I am again unsure.
Any advice/hints?
Best Answer
For part (b): An extension is Galois iff is the splitting field of a separable polynomial.
So $K$ is the splitting field of a separable polynomial $f(x)$.
You know that $K(\sqrt{2})$ is the splitting field of the polynomial $f(x)(x^2-2)$; this is a separable polynomial so $K(\sqrt{2})$ is Galois.
You know that $\sqrt{2}$ can't be in $K$ because the degree of $K$ on $\mathbb{Q}$ is 5, so you know that the degree of the intersection of $K$ and $\mathbb{Q}[\sqrt{2}]$ is 1 and the degree of $K$ is $\frac{5\cdot 2}{1}=10$.
For part (c) you know that that the Galois group of $L$ contains two different normal subgroup, one related to $\mathbb{Q}[\sqrt{2}]$ and the other related to $K$. So you have that $Gal(L/\mathbb{Q})$ is a group of cardinality 10 with two different normal subgroup of cardinality 2 and 5 with trivial intersection. You can say that it's isomorphic to $Z_{10}$.