<!-- * The first inequality arises, as $F \subseteq H$. -->

<!-- * The equality is given as the Galois group is a group, along with the theorem from Lecture 3. Finally, the Galois group is a subgroup of the Automorphism group, and thus, the -->

On the other hand, Lecture 3 *also* gives us that $\mathrm{dim}(E/F) = |\mathrm{Gal}(E/F)|$, as $\mathrm{Gal}(E/F)$ is a group which fixes exactly $F$. Thus the claim follows.

**Proposition.** (Extension dimensions and Galois)(II) *Suppose $\mathrm{dim}(E/F) = \mathrm{Aut}(E/F)$. Then $E/F$ is Galois.*

*Proof.* Let $G = \mathrm{Aut}(E/F)$ and $H$ be the field fixed by $G$. Then we have a tower of extensions $E/H/F$. We want to prove that the fixed field $H$ is $F$. We know that

The first equality is given by the the Lecture 3 Theorem. We can follow, that $\mathrm{dim}(H/F) = 1$, by the extension dimension formula, and thus $F=H$, meaning that $F$ is the field fixed by $G$. Hence the extension is Galois. $\blacksquare$

**Corollary.***$E/F$ is Galois if and only if $\mathrm{dim}(E/F) = |\mathrm{Aut}(E/F)|$.*

This gives rise to the idea that *"Galois extensions are extensions with galois groups of maximal symmetry"*. Usually the Galois group is only a subgroup of the automorphism group, but in the case of a Galois extension, the Galois groups *is* the whole automorphism group.

### Second Equivalence

**Definition.** (Normal Field Extension) A field extension $E/F$ is **normal**, if $E$ is the splitting field over $F$ of some polynomial $p \in F[Z]$ with distinct roots.

**Theorem.***A field extension is Galois if and only if it is normal.*

*Remark.* Once we have proven this theorem, we will have 3 different ways to think about Galois extensions.

(i) *(Inital definition)* Fixed fields.

(ii) *(First equivalence)* Galois groups of maximal symmetry, i.e. largest possible Galois groups.

(iii) *(Second equivalence)* Splitting fields of some polynomial with distinct roots.

*Proof. (Galois to Normal) (Construction)* We begin with a Galois extension, and show that it is a normal extension. To that end, let $E/F$ be the Galois extension and $G = \mathrm{Gal}(E/F) = (g_1, \ldots, g_n)$. Now we have to find a polynomial $p \in F[Z]$ with distinct roots, for which $E$ is the splitting field. We know that $E/F$ is finite, and thus generated by some basis $\mathcal{B} = (x_1, \ldots, x_r)$. Lets choose some $x \in \mathcal{B}$ and investigate it. Specifically, as $G$ acts on $\mathcal{B}$, lets look at the orbit of $x$ under $G$.

We can immediately note, that as $r = \mathrm{dim}(E/F) \geq |\mathrm{Aut}(E/F)| \geq |\mathrm{Gal}(E/F)| = n$,

We're not sure if $\mathrm{Orb}_G (x)$ is pairwise distinct, so we manually pick the maximal subset $\mathcal{O} \subseteq \mathrm{Orb}_G (x)$ that is, and without loss of generality claim that

is a polynomial with distinct roots for which $E$ is the splitting field, concluding our proof.

*Remark.* We have just stated that each $p_x$ having distinct roots implies that the product over each $p_x$ also has distinct roots. This is not actually the case, but this is more of a technicality which can easily be circumvented. We remind ourselves that membership in orbits form an equivalence relation, and thus if two $p_x$ and $p_y$ share a root, then the orbits of $x$ and $y$ under $G$ are the same. It follows that $p_x = p_y$. In the product over all $p_x$ we can then simply discard $p_y$ to ensure we have distinct roots.

So lets show that $p \in F[Z]$. We don't know much about the field $F$, except that it is fixed by elements in $G$. So to show that the coefficients lie in $F$, we could try to show that they are fixed by $G$.

We can easily see that, due to $g$ being a homomorphism,

Hence $g$ fixes each $c_i$ meaning each $c_i$ lies in $F$ and thus $p \in F[Z]$.

*Proof. (Normal to Galois) (Induction)*. We now want to show that a normal extension is a Galois extension. Let $G = \mathrm{Aut}(E/F)$.

*Remark.* (Idea behind this proof) Ultimately, to show that the extension is Galois we will use the equivalence proven earlier, and show that $\mathrm{dim}(E/F) = |\mathrm{Aut}(E/F)|$. As we are dealing with a group action ($G$ acting on $E$), and want to know something about the size of the group, it seems close at hand to use the orbit stabilizer theorem.

As $g \in G = \mathrm{Aut}(E/F)$ already fixes $F$, $\mathrm{Stab}_G (x) = \mathrm{Aut}(E/F(x))$ and thanks to the orbit stabilizer theorem, we have

for some $x \in E$. However we don't know much about the orbit of $x$ in general. If $x in F$, then the orbit is trivial, and we haven't learned anything. For $x \in E \setminus F$ we have little idea what could happen. We have seen previously that the automorphism group of a splitting field only permutes the roots. If we pick $x$ to be a root of the polynomial delived by the normailty of the extension, we have a chance of figuring out the orbit. Lets explore that thought.

<!-- To that end, let $E/F$ be the normal extension, and $p \in F[Z]$ a polynomial with distinct roots in $E$. Let $r_1, \ldots, r_k$ be a roots of this polynomial. We then get the following tower of extensions: -->

<!-- Specifically, we can now claim that $p/(x-r) \in F[r]$ that splits completely over $E$. -->

Let $p \in F[Z]$ with distinct roots for which $E$ is the splitting field. We know that we can factor $p$ into irreducible $p_j$'s over $F$ and write

\begin{align*}

p(z) = \prod _{j=1} ^k p_j(z).

\end{align*}

If we now look at $p_1$, with it's roots $(r_1, \ldots, r_{d_1})$ and degree $d_1$ we see that adjoining any root $r$ to $F$ gives us some extension of degree $d_1$ over F.

In fact we get the following tower of extensions.

\begin{figure}[!h]

\centering

\begin{tikzpicture}

\matrix (m) [matrix of math nodes,row sep=3em,column sep=4em,minimum width=2em]

{

& & E & & \\

F(r_1) & F(r_2) & \ldots & F(r_{k-1}) & F(r_k) \\

& & F & & \\

};

\path[-stealth]

(m-2-1) edge [-] node [above] {$\sim$} (m-2-2)

(m-2-2) edge [-] node [above] {$\sim$} (m-2-3)

(m-2-3) edge [-] node [above] {$\sim$} (m-2-4)

(m-2-4) edge [-] node [above] {$\sim$} (m-2-5)

(m-3-3) edge [-] (m-2-1)

(m-3-3) edge [-] (m-2-2)

(m-3-3) edge [-] (m-2-3)

(m-3-3) edge [-] (m-2-4)

(m-3-3) edge [-] (m-2-5)

(m-1-3) edge [-] (m-2-1)

(m-1-3) edge [-] (m-2-2)

(m-1-3) edge [-] (m-2-3)

(m-1-3) edge [-] (m-2-4)

(m-1-3) edge [-] (m-2-5)

;

\end{tikzpicture}

\end{figure}

Noticably, we know that $E$ is the unique splitting field of all of the roots.

We can also now investigate the orbit of some root. As each of the $F(r_j)$ are isomorphic and actions in $G$ fix $F$, $G$ acts trasitively on $(r_1, \ldots, r_{d_1})$, and $\mathrm{Orb}_G (r) = (r_1, \ldots, r_{d_1})$. Specifically, we have $|\mathrm{Orb}_G (r_1)| = \ldots = |\mathrm{Orb}_G (r_{d_1})| = d_1$.

is a polynomial for which $E$ is still the splitting field, but it is of lesser degree than the $p$ we began with. It now looks like we can finish the proof by induction.

**The proof over induction**

We let $d_j$ denote the degree of the associated $p_j$ and want to perform a proof by induction over the maximal degree $d_{\mathrm{max}}$ of all irreducible factors.

*Basis.* For $d_{\mathrm{max}} = 1$, $p$ splits completely over $F$ and thus $E=F$ and the extension is Galois.

*Assumption.* Assume that normal extensions whose polynomial can be broken into irreducible factors a with maximal degree of $k$ are Galois.

*Inductive Step.* Let $d = d_{\mathrm{max}} = d_1 = k+1$. We have already seen that

is a polynomial of degree $\mathrm{deg}(p) - 1$. This also means that $\widetilde{d}_{\mathrm{max}} = d_{\mathrm{max}} - 1 = k$. We can use the induction hypothesis here, and have

\begin{align*}

|\mathrm{Aut}(E/F)|

& = d_1 \cdot |\mathrm{Aut}(E/F(r))| \\

& = d_1 \cdot \mathrm{dim}(E/F(r)) \\

& = \mathrm{dim}(E/F) \\

\end{align*}

We may conclude that $E/F$ is Galois.$\blacksquare$

---

*Inductive Step.* Let $d = d_{\mathrm{max}} = d_1 \geq 2$. Let $(r_1, \ldots, r_k)$ be the roots of $p_1$. Let's look at the following field

\begin{align*}

Q := F[Z]/(p_1) \cong F[r_k]

\end{align*}

We now can think about $E$ as having $k$ copies of $Q$, one for each root of $p_1$; we get the following tower of extensions.

\begin{figure}[!h]

\centering

\begin{tikzpicture}

\matrix (m) [matrix of math nodes,row sep=3em,column sep=4em,minimum width=2em]

Specifically, we have that E is the unique field extension of each $F(r_k)$.

* Let $r$

* For each $r_k$

* Show that $\mathrm{Aut}(E/F)$ acts transitively on $(r_1, \ldots, r_k)$, and thus we have, via orbit-stabilizer, that $|\mathrm{Aut}(E/F)| = d_1 * |\mathrm{Stab}(E/F(r_1))|$.

* Use induction assumption to move on.

---

* Take polynomial

* Prove by induction over maximal degree because then we have for all polys