### Lecture 3: Formatting.

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 ... ... @@ -11,7 +11,7 @@ header-includes: | *Lecture 2019.03.08* ## RECAP # RECAP In **Lecture 1**, we looked at finite dimensional field extensions. ... ... @@ -28,14 +28,14 @@ We are now in posession of two "Linear Algebra" arguments: --- ## LECTURE 3 # LECTURE 3 ### Some Examples ## Some Examples ##### $\mathbb{C}$ over $\mathbb{R}$ #### $\mathbb{C}$ 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. ##### $\mathbb{Q}(\sqrt{5})$ over $\mathbb{Q}$ #### $\mathbb{Q}(\sqrt{5})$ 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 $\mathbb{Q}(\sqrt{5})$. 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*} ... ... @@ -43,7 +43,7 @@ We claim that the only automorphism that fixes $\mathbb{Q}$ is $\{\mathrm{id}\}$ \end{align*} Nonsense. #### $\mathbb{Q}(\sqrt{2}, \sqrt{3})$ over $\mathbb{Q}(\sqrt{2})$, $\mathbb{Q}(\sqrt{3})$ and $\mathbb{Q}$ ### $\mathbb{Q}(\sqrt{2}, \sqrt{3})$ over $\mathbb{Q}(\sqrt{2})$, $\mathbb{Q}(\sqrt{3})$ and $\mathbb{Q}$ We know that each extension is 2 dimensional: ... ... @@ -85,7 +85,7 @@ We can now look at a subset of the automorphism group $\{\mathrm{id}, \sigma\} \ Let's consider a further example. #### Elementary symmetric functions ### Elementary symmetric functions Let$E = \mathbb{C}(x_1, x_2)$. ... ... @@ -118,7 +118,7 @@ as both$x_1$and$x_2$fulfill the quadratic equation. *A priori*, now, we say In summary, we have exactly two automorphisms. As we have found two, and there are maximally$\mathrm{dim}(E/F) = 2$. #### Finding a tight bound for the dimension formula ### Finding a tight bound for the dimension formula Suppose$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$is a **group**? ... ... @@ -192,7 +192,7 @@ These Galois extensions are very important as they have the most symmetries. *$\mathbb{Q}(\sqrt{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}$. --- ## SUMMARY # SUMMARY **Theorem.** (Lecture 3) *Let$E$be a field, and$H$a subfield fixed by a group of automorphisms$\Sigma\$. Then we have * \begin{align*} ... ...
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