Lecture 14
Galois group of a Galois Extension I
 

13. The Galois Group of a Field Extension


Definition 13.1  

A field extension $ E/F$ is called a Galois extension if it is normal and separable. The Galois group of a Galois extension $ E/F$ denoted by $ G(E/F)$ or $ Gal (E/F)$ is the group of all $ F$-automorphisms of $ E$ under composition of maps.


Proposition 13.2  

The Galois group of the Galois extension $ \mathbb{F}_{q^n}/\mathbb{F}_q$ is a cyclic group of order $ n$ generated by the Frobenious automorphism $ \phi: \mathbb{F}_{q^n} \rightarrow \mathbb{F}_{q^n},$ defined as $ \phi(a)=a^q.$


Proof.
Note that $ \phi $ is an $ \mathbb{F}_{q}$-automorphism since any $ a \in \mathbb{F}_q$ is a root of $ x^q-x.$ Let $ G = \langle \phi \rangle$. Then $ \phi^{n} (x) = x^{q^{n}} = x$. Therefore $ \vert G\vert \leq n$. Suppose $ \vert G\vert= d$. Then $ \phi^{d} = id$, so $ \phi(x) = x^{q^{d}} = x$. But $ x^{q^{d}} - x$ has at most $ q^{d}$ roots. Thus $ d = n$.

We now show that $ G(\mathbb{F}_{q^n} / \mathbb{F}_{q} )= \langle \phi \rangle$. Since $ \mathbb{F}_{q^n} / \mathbb{F}_{q}$ is a separable extension, $ [\mathbb{F}_{q^n} : \mathbb{F}_{q}]_s = n$. Hence the number of $ \mathbb{F}_q$-automorphisms of $ \mathbb{F}_{q^n}$ is $ n$ whence $ \langle\phi\rangle=G(\mathbb{F}_{q^n}/\mathbb{F}_q).$