algebra-II-lecture-2019-03-15.md 13.7 KB
 velleto committed Mar 18, 2019 1 2 3 4 --- geometry: margin=1in header-includes: | \usepackage{tikz}  rrueger committed Mar 26, 2019 5  \usetikzlibrary{calc,decorations.pathmorphing,shapes}  velleto committed Mar 18, 2019 6 7 8 9 10 11 12 13 14  \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} ---  rrueger committed Mar 26, 2019 15 \newpage  rrueger committed Mar 26, 2019 16   velleto committed Mar 18, 2019 17 18 *Lecture 2019.03.15*  19 # LECTURE 4  velleto committed Mar 18, 2019 20   21 In this lecture we will introduce two equivalences of the definition of a Galois extension that we saw last lecture. From now on we will refer to the inital definition as the **first equivalence**. We continue with the  velleto committed Mar 18, 2019 22   23 ## Second Equivalence  velleto committed Mar 18, 2019 24   25 **Theorem.** (Extension dimensions and Galois) *An extension is Galois if and only if*  velleto committed Mar 18, 2019 26 27 28 29 \begin{align*} \mathrm{dim}(E/F) = |\mathrm{Aut}(E/F)| \end{align*}  30 *Proof.* (Galois to dimension) Let $E/F$ be a Galois extension. We know from Lecture 3 that $\mathrm{dim}(E/F) = |\mathrm{Gal}(E/F)| = |\mathrm{Aut}(E/F)|$, as $\mathrm{Gal}(E/F)$ is a group which fixes exactly $F$.  velleto committed Mar 18, 2019 31   32 *Proof.* (Dimension to Galois) 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  velleto committed Mar 18, 2019 33 34 35 36 \begin{align*} \mathrm{dim}(E/H) = |G| \overset{\mathrm{def}}{=} |\mathrm{Aut}(E/F)| \overset{\mathrm{ass}}{=} \mathrm{dim}(E/F) \end{align*}  37 The first equality is given by the the Lecture 3 Theorem. By the extension dimension formula we can follow that $\mathrm{dim}(H/F) = 1$, and thus $F=H$, meaning that $F$ is the field fixed by $G$. Hence the extension is Galois. $\blacksquare$  velleto committed Mar 18, 2019 38   39 This gives rise to the idea that *"Galois extensions are extensions of maximal symmetry"*. Usually the size of the automorphism group is smaller than dimension of the extension, but in the case of a Galois extension, the automorphism group as large as it can be - namely with cardinality **equal** to the dimension.  velleto committed Mar 18, 2019 40   41 ## Third Equivalence  velleto committed Mar 18, 2019 42 43 44 45 46 47 48  **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.  49 50 51 (i) *(First equivalence, Inital definition)* Fixed fields. (ii) *(Second equivalence)* Galois groups of maximal symmetry, i.e. largest possible Galois groups. (iii) *(Third equivalence)* Splitting fields of some polynomial with distinct roots.  velleto committed Mar 18, 2019 52 53 54  *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 $G$.  55 We're not sure if $G$ acting on $x$ will produce pairwise distinct elements, so we manually pick a maximal subset $\mathcal{G} = \{g_1, \ldots, g_k\} \subseteq G$ that will, whilst still having $\mathrm{Orb} _G (x) = \mathrm{Orb} _{\mathcal{G}} (x)$.  velleto committed Mar 18, 2019 56 57 58 59 60  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*}  61 which has distinct roots, owing to the choice of $\mathcal{G}$.  velleto committed Mar 18, 2019 62   63 We now want to show that $p_x$ actually lies in $F[Z]$, because then we can claim that  velleto committed Mar 18, 2019 64 \begin{align*}  65  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]  velleto committed Mar 18, 2019 66 67 68 \end{align*} is a polynomial with distinct roots for which $E$ is the splitting field, concluding our proof.  69 *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 transitivity of the equivalence relation ensures that 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 are left with distinct roots.  velleto committed Mar 18, 2019 70   71 So lets show that $p \in F[Z]$. We don't know much about the field $F$, except that it is fixed by $G$. So to show that the coefficients of $p$ lie in $F$, we could try to show that they are fixed by $G$.  velleto committed Mar 18, 2019 72 73 74 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)) \\  75  \implies \sum _{i=0} ^k g(c_i) z^i & = \prod _{i=1} ^k (z - (g \circ g_k)(x))).  velleto committed Mar 18, 2019 76 \end{align*}  77 As $\mathcal{G}$ is a subset of $G$, applying $g$ once more only serves as a permutation; more formally  velleto committed Mar 18, 2019 78 \begin{align*}  79  g(\mathrm{Orb} _G (x)) = \{(g \circ g_1)(x), \ldots, (g \circ g_k)(x)\} = \{g_{\sigma (1)} (x), \ldots, g_{\sigma (k)} (x) \} = \mathrm{Orb} _G (x)  velleto committed Mar 18, 2019 80 81 82 83 84 85 86 87 88 \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 $G = \mathrm{Aut}(E/F)$.  89 *Remark.* (Idea behind this proof) Ultimately, to show that the extension is Galois we will use the second equivalence: that an extension is Galois if $\mathrm{dim}(E/F) = |\mathrm{Aut}(E/F)|$. 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-stabiliser theorem.  velleto committed Mar 18, 2019 90   91 As $g \in G = \mathrm{Aut}(E/F)$ already fixes $F$, the stabiliser of some $x$ is given by $\mathrm{Stab}_G (x) = \mathrm{Aut}(E/F(x))$ and thanks to the orbit-stabiliser theorem, we have  velleto committed Mar 18, 2019 92 \begin{align*}  93  |\mathrm{Aut}(E/F)|  velleto committed Mar 18, 2019 94  & = |\mathrm{Orb}_G (x)| |\mathrm{Stab}_G (x)| \\  95  & = |\mathrm{Orb}_G (x)| |\mathrm{Aut}(E/F(x))|  velleto committed Mar 18, 2019 96 \end{align*}  97 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 delivered by the normailty of the extension, we have a chance of figuring out the orbit. Lets explore that thought.  velleto committed Mar 18, 2019 98   99 To that end, define $p \in F[Z]$ 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  velleto committed Mar 18, 2019 100 \begin{align*}  101  p(z) = \prod _{j=1} ^k p_j(z).  velleto committed Mar 18, 2019 102 \end{align*}  103 If we now look at $p_1$, with degree $d_1$ and roots $\{r_1, \ldots, r_{d_1}\}$ we see that adjoining any root to $F$ gives us some extension of degree $d_1$ over F. We are sure that we have $d_1$ roots, as the degree is $d_1$ and the roots of $p$ are distinct. In fact we get the following tower of extensions.  velleto committed Mar 18, 2019 104 105 106 107 108 109 110  \begin{figure}[!h] \centering \begin{tikzpicture} \matrix (m) [matrix of math nodes,row sep=3em,column sep=4em,minimum width=2em] { & & E & & \\  111  F(r_1) & F(r_2) & \ldots & F(r_{k-1}) & F(r_{d_1}) \\  velleto committed Mar 18, 2019 112 113 114  & & F & & \\ }; \path[-stealth]  115 116 117 118  (m-2-1) edge [-] node [above] {$\quad\cong$} (m-2-2) (m-2-2) edge [-] node [above] {$\cong$} (m-2-3) (m-2-3) edge [-] node [above] {$\cong$} (m-2-4) (m-2-4) edge [-] node [above] {$\cong\quad$} (m-2-5)  velleto committed Mar 18, 2019 119   120  (m-3-3) edge [-] node [below] {degree $d_1 \quad\quad$} (m-2-1)  velleto committed Mar 18, 2019 121 122 123 124 125 126 127 128 129 130 131 132 133 134  (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}  135 Noticably, we know that $E$ is the unique splitting field of all of the roots.  velleto committed Mar 18, 2019 136   137 138 139 *Remark.* The fact that the $F(r_i)$'s are congruent is a result from Algebra I, and is important later in the proof. We can now progress on our original goal: finding the orbit of some root. We claim that $G$ acts trasitively on $\{r_1, \ldots, r_{d_1}\}$, it follows that $\mathrm{Orb}_G (r_1) = \{r_1, \ldots, r_{d_1}\}$. Specifically, we have $|\mathrm{Orb}_G (r_1)| = \ldots = |\mathrm{Orb}_G (r_{d_1})| = d_1$. (We will prove the claim of transitivity after this proof).  velleto committed Mar 18, 2019 140 141 142  We now have \begin{align*}  143  |\mathrm{Aut}(E/F)| = d_1 \cdot |\mathrm{Aut}(E/F(r_1))|  velleto committed Mar 18, 2019 144 \end{align*}  145 This is a powerful insight, because now  velleto committed Mar 18, 2019 146 \begin{align*}  147  \widetilde{p}_1(z) := \frac{p_1}{x-r} \prod _{j=2} ^k p_j (z) \in F(r_1)[Z]  velleto committed Mar 18, 2019 148 \end{align*}  149 is a polynomial for which $E$ is still the splitting field, but it is of degree one less than the $p$ we began with. It now looks like we can finish the proof by induction.  velleto committed Mar 18, 2019 150 151 152 153 154 155 156 157 158  **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.  159 *Inductive Step.* Let, without loss of generality, $d_1 = d_{\mathrm{max}} = k+1$ unique. For the case in which there are two irreducible factors with maximal degree see the remark after the proof. We have already seen that for a root $r_1$ of $p_1$  velleto committed Mar 18, 2019 160 \begin{align*}  161  |\mathrm{Aut}(E/F)| = d_1 \cdot |\mathrm{Aut}(E/F(r_1))| \\  velleto committed Mar 18, 2019 162 163 164 \end{align*} and that \begin{align*}  165  \widetilde{p}(z) := \frac{p_1}{x-r_1} \prod _{j=2} ^k p_j (z) \in F(r_1)[Z]  velleto committed Mar 18, 2019 166 167 168 \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*}  169 170 171  |\mathrm{Aut}(E/F)| & = d_1 \cdot |\mathrm{Aut}(E/F(r_1))| \\ & = d_1 \cdot \mathrm{dim}(E/F(r_1)) \\  velleto committed Mar 18, 2019 172 173 174 175 176  & = \mathrm{dim}(E/F) \\ \end{align*} We may conclude that $E/F$ is Galois.$\blacksquare$  177 178 179 180 181 182 183 184 185 186 187 *Remark.* (Two irreducible factors of maximal degree) When proving that normal implies Galois, we said that the reduction of the degree of the polynomial by one, implied that the maximal degree of the irreducible factors also decreased by one. This is not true in general, but the arguments supplied can be applied multiple times, until the maximum *has* been decreased: Suppose there are two $p_j$'s in our factorisation with the same degree $d_\mathrm{max}$. Without loss of generality, let these be $p_1$ and $p_2$, with $r_1$ and $r_2$ roots respectively. We can begin as before and obtain: \begin{gather*} |\mathrm{Aut}(E/F)| = \underset{=d_\mathrm{max}}{\underbrace{d_1}} \cdot |\mathrm{Aut}(E/F(r_1))| \\ \widetilde{p}_1 (z) := \frac{p_1}{x-r_1} \prod _{j=2} ^k p_j (z) \in F(r_1)[Z] \end{gather*} We can now work on $E/F(r_1)$ and get \begin{gather*} |\mathrm{Aut}(E/F)| =\underset{={d_\mathrm{max}}^2}{\underbrace{d_1 \cdot d_2}} \cdot |\mathrm{Aut}(E/F(r_1, r_2))| \\ \widetilde{p}_2 (z) := \frac{p_1}{x-r_1}\frac{p_2}{x-r_2} \prod _{j=3} ^k p_j (z) \in F(r_1, r_2)[Z] \end{gather*} Notice how we needed to augment the field to $F(r_1, r_2)$ for the second factor. We can now proceed iteratively, adjoining more roots to the field, until we can apply the induction hypothesis.  velleto committed Mar 18, 2019 188   189 190 191 192 *Proof.* (Transitivity of $G$ on $\{r_1, \ldots, r_{d_1}\}$) (This fact is a combination of two facts known from Algebra I) We have $p \in F[Z]$ with distinct roots in $E$. We can factorise this into irreducible factors $p_1, \ldots, p_j$ in $F[Z]$ and look at the first factor $p_1$ with degree $d_1$ and roots $R = \{r_1, \ldots, r_{d_1}\}$. We claim that $G = \mathrm{Aut}(E/F)$ acts transitively on $R$. That is, for every pair $r_i, r_j \in R$, we must find some $g \in G$ such that $g(r_i) = r_j$. * (First Fact) We know from Algebra I that $F(r_i) \cong F(r_j)$, with an isomorphism that maps $r_i$ to $r_j$ and fixes $F$ * (Second Fact) Another result known from Algebra I, is that we can extend an isomorphism $\varphi$ between two fields $k, k'$, to an isomorphism between the splitting field $E$ of some $f \in k[X]$ to the splitting field $E'$ of $f(\varphi) \in k'[X]$, as shown in the following diagram.  velleto committed Mar 18, 2019 193 194 195 196 197 198  \begin{figure}[!h] \centering \begin{tikzpicture} \matrix (m) [matrix of math nodes,row sep=3em,column sep=4em,minimum width=2em] {  199 200  E & E' \\ k & k' \\  velleto committed Mar 18, 2019 201 202  }; \path[-stealth]  203 204 205 206  (m-1-1) edge [<->] node [above] {$\cong$} (m-1-2) (m-1-1) edge [-] node [left] {s.f. $f$} (m-2-1) (m-1-2) edge [-] node [right] {s.f. $f' = \varphi(f)$} (m-2-2) (m-2-1) edge [<->] node [below] {$\overset{\varphi}{\cong}$} (m-2-2)  velleto committed Mar 18, 2019 207 208 209 210  ; \end{tikzpicture} \end{figure}  211 Together, these ideas allow us to construct an isomorphism of $E$ that fixes $F$ and maps $r_i$ to $r_j$. This is exactly what we want.  velleto committed Mar 18, 2019 212   213 # SUMMARY  velleto committed Mar 18, 2019 214 215 216  We now have the following equivalences of a Galois extension:  217 218 219 (i) (**First Equivalence**) *An extension is Galois if the field fixed by the automorphism group is exactly the base field.* (ii) (**Second Equivalence**) *An extension is Galois if the dimension of the extension is the same as the size of the automorphism group.* (iii) (**Third Equivalence**) *An extension is Galois if it is normal.*