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Lecture 4: Draft.

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\fancyhead[CO,CE]{ETH Z\"urich - Algebra II - Prof. R. Pandharipande}
\fancyfoot[CO,CE]{\href{http://git.ethz.ch/rrueger/algebra-notes}{git.ethz.ch/rrueger/algebra-notes} - \href{mailto:rrueger@ethz.ch}{rrueger@ethz.ch}}
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*Lecture 2019.03.15*
## LECTURE 4
In this lecture we will introduce two equivalences of the definition of a Galois extension that we saw last lecture.
### First Equivalence
**Proposition.** (Extension dimensions and Galois)(I) *Let $E/F$ be a Galois extension. Then*
\begin{align*}
\mathrm{dim}(E/F) = |\mathrm{Aut}(E/F)|
\end{align*}
*and*
\begin{align*}
\mathrm{Gal}(E/F) := \mathrm{Aut}(E/F)
\end{align*}
*Proof.* We have by Lecture 3
\begin{align*}
\mathrm{dim}(E/F) \geq |\mathrm{Aut}(E/F)| \geq |\mathrm{Gal}(E/F)|
\end{align*}
<!-- \begin{align*} -->
<!-- \mathrm{dim}(E/F) \geq \mathrm{dim}(E/H) = |\mathrm{Aut}(E/H)| \geq |\mathrm{Aut}(E/F)| = |\mathrm{Aut}(E/F)| \geq |\mathrm{Gal}(E/F)|. -->
<!-- \end{align*} -->
<!-- * 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
\begin{align*}
\mathrm{dim}(E/H) = |G| \overset{\mathrm{def}}{=} |\mathrm{Aut}(E/F)| \overset{\mathrm{ass}}{=} \mathrm{dim}(E/F)
\end{align*}
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
\begin{align*}
\mathcal{O} = \{g_1(x), \ldots, g_k(x) = g_{k+1}(x) = \ldots = g_n(x)\}
\end{align*}
We now define some polynomial $p_x \in E[Z]$:
\begin{align*}
p_x (z) = \prod _{i=1} ^k (z - g_i(x))
\end{align*}
which has distinct roots, owing to the choice of $\mathcal{O}$.
We want to show that $p$ **actually** lies in $F[Z]$, because then we can claim that
\begin{align*}
p(x) = \prod _{x \in \mathcal{B}} p_x (z) = \prod _{j=1} ^r \prod _{i=1} ^{k_x} (z - g_i (x_j)) \in F[Z]
\end{align*}
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,
\begin{align*}
\sum _{i=0} ^k c_i z^i & := \prod _{i=1} ^k (z - g_k(x)) \\
\implies \sum _{i=0} ^k g(c_i) z^i & = \prod _{i=1} ^k (z - (g \circ g_k)(x)))
\end{align*}
As $\mathcal{O}$ is a subset of the orbit, applying $g$ once more only serves as a permutation; more formally
\begin{align*}
g(\mathcal{O}) = \{(g \circ g_1)(x), \ldots, (g \circ g_k)(x)\} = \{g_{\sigma (1)} (x), \ldots, g_{\sigma (k)} (x) \} = \mathcal{O}
\end{align*}
for some $\sigma \in S_k$. Thus,
\begin{align*}
\sum _{i=0} ^k g(c_i) z^i = \prod _{i=1} ^k (z - (g \circ g_k)(x))) = \prod _{i=1} ^k (z - g_{\sigma (i)}(x))) = \prod _{i=1} ^k (z - g_i(x))) = \sum _{i=0} ^k c_i z^i.
\end{align*}
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
\begin{align*}
|\mathrm{Aut}(E/F)|
& = |\mathrm{Orb}_G (x)| |\mathrm{Stab}_G (x)| \\
& = |\mathrm{Orb}_G (x)| |\mathrm{Aut}(E/F(x))| \\
\end{align*}
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 $|\mathrm{Orb}_G (r_1)| = \ldots = |\mathrm{Orb}_G (r_{d_1})| = d_1$.
We now have
\begin{align*}
|\mathrm{Aut}(E/F)| = d_1 \cdot |\mathrm{Aut}(E/F(r))|
\end{align*}
This is a powerful insight, because now with $q := p_1/ (x-r)$,
\begin{align*}
\widetilde{p}(z) := q(z) \prod _{j=2} ^k p_j (z) \in F(r_1)[Z]
\end{align*}
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
\begin{align*}
|\mathrm{Aut}(E/F)| = d_1 \cdot |\mathrm{Aut}(E/F(r))| \\
\end{align*}
and that
\begin{align*}
\widetilde{p}(z) := q(z) \prod _{j=2} ^k p_j (z) \in F(r_1)[Z]
\end{align*}
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]
{
& & & E & & \\
Q & F(r_1) & F(r_2) & \ldots & F(r_{k-1}) & F(r_k) \\
F & & & F & & \\
};
\path[-stealth]
(m-2-1) edge [-] (m-3-1)
(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-2-5) edge [-] node [above] {$\sim$} (m-2-6)
(m-3-4) edge [-] (m-2-2)
(m-3-4) edge [-] (m-2-3)
(m-3-4) edge [-] (m-2-4)
(m-3-4) edge [-] (m-2-5)
(m-3-4) edge [-] (m-2-6)
(m-1-4) edge [-] (m-2-2)
(m-1-4) edge [-] (m-2-3)
(m-1-4) edge [-] (m-2-4)
(m-1-4) edge [-] (m-2-5)
(m-1-4) edge [-] (m-2-6)
;
\end{tikzpicture}
\end{figure}
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
* Let p_j be the maximal degree sub poly
* let r_1, ..., r_n be the distinct roots
* G acts on roots
* Use orb stab
* prove orb = d_1
* Claim |aut(e/f)| = |stab(r_1)| |orb(r_1)| = |aut(e/fr1)| d_1
* p_1 without root is of degree one less and is in F(r_1) and E is the splitting field.
= |dim(e/fr1)| d_1 = dim(e/f)
---
## SUMMARY
We now have the following equivalences of a Galois extension:
(i) *An extension is Galois if the field fixed by the automorphism group is the base field.*
(ii) *An extension is Galois if the dimension of the extension is the same as the size of the automorphism group.*
(iii) *An extension is Galois if it is normal.*
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