### Lecture 4: Draft.

parent c2b12809
 --- geometry: margin=1in header-includes: | \usepackage{tikz} \usetikzlibrary{matrix} \usepackage{fancyhdr} \pagestyle{fancy} \renewcommand{\footrulewidth}{0.4pt} \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}} \fancyfoot[LE,RO]{\thepage} --- *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*} 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. 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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