algebra-II-lecture-2019-02-08.md 12.2 KB
 velleto committed Mar 13, 2019 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 --- header-includes: | \usepackage{fancyhdr} \pagestyle{fancy} \renewcommand{\footrulewidth}{0.4pt} \fancyhead[CO,CE]{ETH Z\"urich - Lecture notes 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 2019.02.08* ## RECAP In **Lecture 1**, we looked at finite dimensional field extensions. In **Lecture 2**, we looked at the automorphism group that arises from a extension field and an associated fixed field. We used characters to investigate this behaviour. We are now in posession of two "Linear Algebra" arguments: * *Argument **1** *. Distinct characters are linearly independent. * *Argument **2** *. The dimension-cardinality formula for the automorphism group \begin{align*} \mathrm{dim}(E/F) \geq |\mathrm{Aut}(E/F)| \end{align*} --- ## LECTURE 3 ### Some Examples ##### $\mathbb{C}$ over $\mathbb{R}$ Let's take the extension $\mathbb{C}$ over $\mathbb{R}$. What automorphisms of this extension do we know? Well, we know the identity and complex conjugation. Now we can ask ourselves, *can there be any more?*. Well ..., no. We know that the size of the automorphism group is bounded by the dimension of the extension: 2. ##### $\mathbb{Q}(\sqrt[3]{5})$ over $\mathbb{Q}$ We claim that the only automorphism that fixes $\mathbb{Q}$ is $\{\mathrm{id}\}$. Why? Well, a field isomorphism is uniqely defined by the image of the basis. The the minimal polynomial of this extension is $X^3 - 5$. Let $\alpha \in \mathbb{C}$ be a root of this polynomial. We know that $\{1, \alpha, \alpha ^2, \alpha ^3, \ldots \}$ generates $\mathbb{Q}(\sqrt[3]{5})$. As $\alpha ^3 = 5 \in \mathbb{Q}$, we know that $\{1, \alpha, \alpha ^2 \}$ is a basis. Obviously $1 \in \mathbb{Q}$ must be fixed, and thus mapped to $1$. Now our only hope is to permute the other two basis vectors. Let's investigate that: if $\sigma(\alpha) = \alpha^2$ and $\sigma(\alpha^2) = \alpha$, we get \begin{align*} 5 = \sigma(5) = \sigma(\alpha ^3) = \sigma (\alpha) ^3 = (\alpha ^2) ^3 = (\alpha ^3) ^2= 5^2 = 25 \end{align*} Nonsense. #### $\mathbb{Q}(\sqrt{2}, \sqrt{3})$ over $\mathbb{Q}(\sqrt{2})$, $\mathbb{Q}(\sqrt{3})$ and $\mathbb{Q}$ We know that each extension is 2 dimensional: * $\mathbb{Q}(\sqrt{2})$ over $\mathbb{Q}$ * $\mathbb{Q}(\sqrt{3})$ over $\mathbb{Q}$ * $\mathbb{Q}(\sqrt{2}, \sqrt{3})$ over $\mathbb{Q}(\sqrt{2})$ * $\mathbb{Q}(\sqrt{2}, \sqrt{3})$ over $\mathbb{Q}(\sqrt{3})$ We can easily prove that $\mathbb{Q}(\sqrt{2}, \sqrt{3})$ over $\mathbb{Q}$ is 4 dimensional. *Proof.* The proof would go over the dimension formula, yet we do not now *a priori* that $\mathbb{Q}(\sqrt{3}) \not\subseteq \mathbb{Q}(\sqrt{2})$. Assume this to be the case. Then we would have for $x, y \in \mathbb{Q}$ \begin{align*} \sqrt{3} & = x + y\sqrt{2} \\ \implies 3 & = x^2 + 2y^2 + \sqrt{2}\underset{= 0}{\underbrace{xy}} \end{align*} We know that $xy = 0$ as otherwise the RHS contains irrationals whilst the LHS does not. If, however $xy = 0$, then either $x = 0 \implies y = \sqrt{\frac{3}{2}} \notin \mathbb{Q}$ or $y = 0 \implies x = \sqrt{3} \notin \mathbb{Q}$. Both contradictions. We can do the same for $\mathbb{Q}(\sqrt{2}) \not\subseteq \mathbb{Q}(\sqrt{3})$. Hence we can state that, $\mathbb{Q}(\sqrt{3}, \sqrt{2})$ is an extension over $\mathbb{Q}(\sqrt{2}$, which in turn is an extension over $\mathbb{Q}$, to which we can apply the dimension formula to arrive at our result. $\blacksquare$ We now want to use this information to think about the autmorphism group of the extension of $\mathbb{Q}(\sqrt{2}, \sqrt{3}) / \mathbb{Q}$. Let's go via the extension of the adjoin of only $\sqrt{2}$: * We now know that there exists an automorphism $\sigma \in \mathrm{Aut}(\mathbb{Q}(\sqrt{2}, \sqrt{3})/\mathbb{Q}(\sqrt{2}))$. * This automorphism is given by: * $\sigma (\sqrt{3}) = -\sqrt{3}$ * $\sigma |_{\mathbb{Q}(\sqrt{2})} = \mathrm{id}$ The same for the extension of the adjoin of only $\sqrt{3}$ we have analogously * There exists an automorphism $\tau \in \mathrm{Aut}(\mathbb{Q}(\sqrt{2}, \sqrt{3})/\mathbb{Q}(\sqrt{3}))$. * This automorphism is given by: * $\tau (\sqrt{2}) = -\sqrt{2}$ * $\tau |_{\mathbb{Q}(\sqrt{3})} = \mathrm{id}$ We can now compose $\sigma$ and $\tau$ to find a new automorphism in $\mathrm{Aut}(\mathbb{Q}(\sqrt{3}, \sqrt{2})/\mathbb{Q})$. With the identity, we now have 4 automorphisms in total, and therefore, we have all of the possible automorphisms of the extension. *Remark.* This group is isomorphic to $\mathbb{Z}/ 2\mathbb{Z} \times \mathbb{Z}/ 2\mathbb{Z}$. We can now look at a subset of the automorphism group $\{\mathrm{id}, \sigma\} \subset \mathrm{Aut}(\mathbb{Q}(\sqrt{3}, \sqrt{2})/\mathbb{Q})$ and ask ourselves what the fixed field of this subset is? It must be $\mathbb{Q}(\sqrt{2})$. If it were some larger field $\mathbb{Q}(\sqrt{2}, \sqrt{3}) \subset F \subset \mathbb{Q}(\sqrt{2})$, we would fit a field extension we could fit another field extension between the two, but we know that $Q(\sqrt{2}, \sqrt{3})$ is an extension of degree 2, and thus nothing can fit in between. Let's consider a further example. #### Elementary symmetric functions Let $E = \mathbb{C}(x_1, x_2)$. **Definition.** *(Elementary symmetric functions)* Let $e_1 = x_1 + x_2$ and $e_2 = x_1 x_2$ Let us now consider $F = \mathbb{C}(e_1, e_2)$, which is a subfield of $E$ and ask ourselves: *what is $\mathrm{Aut}(E/F)$*? * We know that the identity surely an element, and owing to the symmetric nature of $e_1$ and $e_2$, switching $x_1$ and $x_2$ is also one. Two in the bag. * If we try to compose theses two elements we won't get far. * There is, however a little ray of hope is to compose $\sigma$ with itself, but alas, this is the identity. Now we can think back to our dimension theorem. We know that the size of the automorphism group is bounded by the dimension of the field. * So we can ask ourselves what the dimension of the extension $E/F$ is. * What is the minimal polynomial? Let's suggest \begin{align*} Z^2 - e_1 Z + e_2 = 0 \end{align*} * This is somewhat more clear if we frame it differently \begin{align*} (Z - x_1) (Z - x_2) = Z^2 - (x_1 + x_2) Z + x_1 x_2 = Z^2 - e_1 Z + e_2 = 0 \end{align*} This tells us that both \begin{align*} \mathrm{dim}(F[x_1]/F) = \mathrm{dim}(F[x_2]/F) = 2 \end{align*} as both $x_1$ and $x_2$ fulfill the quadratic equation. *A priori*, now, we say that the extension of $(E = F[x_1, x_2])/F$ is of 4th degree. However as $x_2 = e_1 - x_1$, we can follow $E = F[x_1, x_2] = F[x_1]$, we see that the extension is of degree 2. In summary, we have exactly two automorphisms. As we have found two, and there are maximally $\mathrm{dim}(E/F) = 2$. #### Finding a tight bound for the dimension formula Suppose $E$ is a field. Let $\Sigma = \{ \sigma_1, \ldots, \sigma_n \}$ be distinct automorphisms on $E$. Let $H$ be the field fixed by these automorphisms. Lecture 2 told us that $\mathrm{dim}(E/F) \geq |\Sigma|$. What if $\Sigma$ is a **group**? *Remark* We usually write this so that $\sigma_1$ is the identity. Obviously, as $\Sigma$ is a group, its closed under group operations and inverses. We now want to show equality between the size of the automorphism group and the dimension of the extension **Theorem.** (Lecture 3) *Let $\Sigma, E, F$ be as above. Then we have* \begin{align*} \mathrm{dim}(E/H) = |\Sigma| = n \end{align*} We already know that the dimension is greater or equal to the size of the group. When is the dimension stricly greater than the size of the group? We will assume this to be the case, and look for a contradiction. *Proof. (by contradiction)* Assume $r = \mathrm{dim}(E/H) > n$. Then we have a basis $\{w_1, \ldots, w_r\} \in E$. We now define the matrix $M \in \mathrm{Mat}_{r \times n} (E)$ \begin{align*} M_{ij} = \sigma _i (w_j) \end{align*} As $r > n$, $M$ has a non-trivial kernel. Let $0 \neq a \in E$ be such an element. We now claim (i) **Not all $a_i$ are $0$.** By assumption. (ii) **Not all $a_i$ are in $H$.** Let $\sigma _1 = \mathrm{id}$. Then the first row of $M$ is just $(w_1, \ldots, w_r)$, and thus the first entry of $Ma$ would be $\sum _i a_i w_i = 0$ - a trivial representation of $0$, but ${w_1, \ldots, w_r}$ form a basis, a contradiction. (iii) **It can't be that only one $a_i$ is non-zero.** If this were the case (with $a_i$ non-zero), $Ma = a_i \cdot Me_i = a_i \cdot M^{(i)} \neq 0$. Let's look for elements of the non-trivial kernel. We can characterise this by counting the number of zero entries in the vector. From point (iii), we know that this number can't be 1. Let $s \geq 2$ be the minimal number of non-zero coefficients of any non-zero kernel vector. We can reorder the basis so that the first $s$ entries of $a$ are non-zero. We will get a contradiction by finding a solution with even less non-zero elements. We now write $a = (a_i, \ldots, a_{s-1}, 1, 0, \ldots, 0)^{\mathrm{T}}$. We can force a $1$ in the $s$'th entry by dividing by $a_s$. We can now consider the equations that arise from $Ma = 0$. Up until this point, we haven't used the group structure of the $\sigma$'s. We remind that not all $a_i$ are not in $H$. Let's claim that $a_1$ is not in $H$. Now the idea is to take the $a_1 \notin H$. What does it mean for $a_1 \notin H$? Well, we know that it is not fixed by some $\sigma \in \Sigma$. Let $\sigma_k$ be the automorphism that moves it. As the set of equations $Ma = 0$ are all in $E$, we can apply $\sigma _k$. Let's do exactly that: \begin{align*} 0 = \sigma _k ((Ma)_i) = \sigma_k ( \sum _j a_j \sigma _i (w_j)) = \sum _j \sigma _k (a_j \sigma _i (w_j)) = \sum _j \sigma _k (a_j) \cdot (\sigma _k \circ \sigma _i)(w_j) \end{align*} We can now use the group structure of $\Sigma$. As in any group, composition of $\sigma$'s are again in the group. Thus each $\sigma _k \circ \sigma _i =: \sigma _{i'} \in \Sigma$. This amounts to nothing more than a permutation of the $\sigma$'s. Thus we end up with a new kernel vector $a'$. \begin{align*} a' := (\sigma _k(a_1), \ldots, \sigma _k(a_{s-1}), 1, 0, \ldots, 0)^{\mathrm{T}} \end{align*} To convince ourselves that this is true, we can multiply back: \begin{align*} (Ma')_i & = \sum _j a'_j \sigma _i (w_j) \\ & = \sum _j \sigma _k (a_j) \sigma _i (w_j) \\ & = (\sigma _k \circ \sigma _k ^{-1})(\sum _j \sigma _k (a_j) \sigma _i (w_j)) \\ & = \sigma _k (\sum _j a_j (\sigma _k ^{-1} \circ \sigma _i) (w_j)) \\ velleto committed Mar 14, 2019 166 167 168 & = \sigma _k (\sum _j a_j \sigma _{i'} (w_j)) \\ & = \sigma _k ((Ma)_{i'}) \\ & = \sigma _k (0) \\ velleto committed Mar 13, 2019 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 & = 0 \end{align*} As the kernel is a subspace, we can take the difference $a - a' \in \mathrm{Ker}(M)$. \begin{align*} a' - a & = (\sigma _k(a_1) - a_1, \ldots, \sigma _k(a_{s-1}) - a_{s-1}, 1 - 1, 0, \ldots, 0) \\ & = (\underset{\neq 0}{\underbrace {\sigma _k(a_1) - a_1}}, \ldots, \sigma _k(a_{s-1}) - a_{s-1}, \mathbf{0}, 0, \ldots, 0) \\ & \neq 0 \end{align*} This non-zero vector has at most $s-1$ non-zero entries. This is a contradiction to the statement that $s$ was the minimal number of non-zero entries of non-zero kernel vectors. This means there cannot be a minimal number of non-zero entries. This in turn means that there is no non-zero kernel vector. We have shown by contradiction, that if $r > n$, then we only have a trivial kernel, even though there should be a non-trivial one. $\blacksquare$ **Definition** *(Galois extension)* An extension in which the automorphism group that fixes the base field, only fixes the base field is called a **Galois Extension**. We say \begin{align*} \mathrm{Gal}(E/H) := \mathrm{Aut}(E/H) = G. \end{align*} These Galois extensions are very important as they have the most symmetries. **Examples of Galois extensions** * $\mathbb{C}$ on $\mathbb{R}$. This is Galois. The automorphisms $\{\mathrm{id}, \overline{\cdot}\}$ have $\mathbb{R}$ as a fixed field. * $\mathbb{Q}(\sqrt[3]{5})$. Is not Galois. It's a 3 dimensional extension, but we only know of one automorphism $\{\mathrm{id}\}$. This doesn't only fix $\mathbb{Q}$. --- ## SUMMARY **Theorem.** (Lecture 3) *Let $E$ be a field, and $H$ a subfield fixed by a group of automorphisms $\Sigma$. Then we have * \begin{align*} \mathrm{dim}(E/H) = |\Sigma| \end{align*}