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

What is(are) the Galois Extension(s) of $\mathbb{Q}$ whose Galois group is cyclic group of prime order? The fundamental theorem of Galois theory says that the degree of the extension is same as the order of the Galois group.Can we find an explicit polynomial of degree which is an arbitrary prime number?

• Abelian extensions of $\mathbb{Q}$ are classified by class field theory (en.wikipedia.org/wiki/Class_field_theory), although I don't know enough to say anything more precise than what's in this blog post: sbseminar.wordpress.com/2008/04/21/… May 1, 2011 at 13:24
• I assume you know the Euler totient function, Cyclotomic polynomial and Group of units of modular arithmetic. If you don't I will be happy to add details. May 1, 2011 at 20:59
• Yes I do, though I dont understand Cyclotomic polynomials very well. May 1, 2011 at 21:57

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.

• Can you please elaborate on your first sentence? May 1, 2011 at 20:49
• @Dinesh: $\mathbb{Q}(\zeta_N)$ is a Galois extension of degree $\varphi(N)$ with Galois group $(\mathbb{Z}/N\mathbb{Z})^{\ast}$, so intermediate extensions of this extension are classified by the fundamental theorem, and the Kronecker-Weber theorem asserts that all finite abelian extensions of $\mathbb{Q}$ are contained in cyclotomic extensions in this way. May 2, 2011 at 20:00
• @Qiaochu Thanks! Proof of Kronecker Weber Theorem is demanding the background of class field theory..is there any shortest possible route one can suggest for its proof? May 5, 2011 at 16:48
• @Dinesh: I don't think so. You need to know some high class complex analysis as well. Anyway there is this link given in Wikipedia jstor.org/pss/2319208.
– user9413
May 6, 2011 at 3:01
• @Chandru Yes I might have done it as soon as Qiaochu explained. May 6, 2011 at 20:31