The Galois Extension of $\mathbb Q$ with cyclic group of prime order as its Galois group

Solution 1:

To construct an Galois extension of $\mathbb{Q}$ of order $p$, take an integer $N$ such that $(\mathbb{Z}/N\mathbb{Z})^*$ maps surjectively onto $\mathbb{Z}/p\mathbb{Z}$, then the field $\mathbb{Q}(\zeta_N)$ contains a subfield $K$ such that $[K:\mathbb{Q}]=p$ and $K$ is Galois over $\mathbb{Q}$ (Here $\zeta_N$ is a primitive $N$-th root of unity). For example, to find a galois extension of $\mathbb{Q}$ of degree 5, you may look at the subfield of $\mathbb{Q}(\zeta_{11})$ that's fixed by the complex conjugation. Kronecker-weber Theorem states that every finite abelian extensions of $\mathbb{Q}$ is a subfield of cyclotomic fields, so every Galois extension of $\mathbb{Q}$ of prime order arises this way.

Continue the example with prime $p=5$, to find all of such extensions, we will look at the sequence $5n+1$. By Dirichlet's theorem on primes in arithmetic progressions, there are infinitely many primes $q$ in this arithmetic progression, each $\mathbb{Q}(\zeta_q)$ contains a subfield which is a degree 5 extension of $\mathbb{Q}$. All these subfields are distinct because they ramifies at different primes. Last, note that that there is also a subfield of $\mathbb{Q}(\zeta_{25})$ which is of degree 5 over $\mathbb{Q}$.

Edit: I was wrong about my complete list statement. For example, $\mathbb{Q}(\zeta_{341})$ (composite of $\mathbb{Q}(\zeta_{11})$ and $\mathbb{Q}(\zeta_{31})$) has 6 subfields which are degree 5 abelian extensions of $\mathbb{Q}$. Apologies.