Unverified Commit 18b970c9 authored by velleto's avatar velleto
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Lecture 3: Formatting.

parent f14d4fdc
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*Lecture 2019.03.08*
## RECAP
# RECAP
In **Lecture 1**, we looked at finite dimensional field extensions.
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## 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.
##### $\mathbb{Q}(\sqrt[3]{5})$ over $\mathbb{Q}$
#### $\mathbb{Q}(\sqrt[3]{5})$ 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*}
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\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:
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Let's consider a further example.
#### Elementary symmetric functions
### Elementary symmetric functions
Let $E = \mathbb{C}(x_1, x_2)$.
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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**?
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* $\mathbb{Q}(\sqrt[3]{5})$. Is not Galois. It's a 3 dimensional extension, but we only know of one automorphism $\{\mathrm{id}\}$. This doesn't only fix $\mathbb{Q}$.
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## 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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