# Tuesday September 10th

## Vector Spaces

**Definition:**
Let $\FF$ be a field.
A **vector space** is an abelian group $V$ with a map $\FF \cross V \to V$ such that

- $\alpha(\beta \vector v) = (\alpha \beta) \vector v$

- $(\alpha + \beta)\vector v = \alpha \vector v + \beta \vector v$,

- $\alpha(\vector v + \vector w) = \alpha \vector v + \alpha \vector w$

- $1\vector v = \vector v$

*Examples:*
$\RR^n, \CC^n , F[x] = \mathrm{span}(\theset{1, x, x^2, \cdots}), L^2(\RR)$

**Definition:**
Let $V$ be a vector space over $\FF$; then a set $W \subseteq V$ *spans* $V$ iff for every $\vector v\in V$, one can write $\vector v = \sum \alpha_i \vector w_i$ where $\alpha_i \in \FF,~\vector w_i \in W$.

**Definition:**
$V$ is *finite dimensional* if there exists a finite spanning set.

**Definition:**
A set $W \subseteq V$ is *linearly independent* iff
$$
\sum \alpha_i \vector w_i = \vector 0 \implies \alpha_i = 0 \text{ for all } i
.$$

**Definition:**
A *basis* for $V$ is a set $W \subseteq V$ such that

1. $W$ is linearly independent, and

2. $W$ spans $V$.

A basis is a midpoint between a spanning set and a linearly independent set.

We can add vectors to a set until it is spanning, and we can throw out vectors until the remaining set is linearly independent.
This is encapsulated in the following theorems:

**Theorem:**
If $W$ spans $V$, then some subset of $W$ spans $V$.

**Theorem:**
If $W$ is a set of linearly independent vectors, then some superset of $W$ is a basis for $V$.

*Fact:*
Any finite-dimensional vector spaces has a finite basis.

**Theorem:**
If $W$ is a linearly independent set and $B$ is a basis, then $\abs{B} \leq \abs W$.

**Corollary:**
Any two bases have the same number of elements.

> So we define the dimension of $V$ to be the number of elements in any basis, which is a unique number.

## Algebraic Extensions

**Definition:**
$E \geq F$ is an algebraic extension iff every $\alpha \in E$ is algebraic of $F$.

**Definition:**
$E \geq F$ is a *finite extension* iff $E$ is finite-dimensional as an $F\dash$vector space.

*Notation:*
$[E: F] = \dim_F E$, the dimension of $E$ as an $F\dash$vector space.

*Observation:*
If $E = F(\alpha)$ where $\alpha$ is algebraic over $F$, then $E$ is an algebraic extension of $F$.

*Observation:*
If $E\geq F$ and $[E: F] = 1$, then $E=F$.

**Theorem:**
If $E \geq F$ is a finite extension, then $E$ is algebraic over $F$.

*Proof:*
Let $\beta \in E$.
Then the set $\theset{1, \beta, \beta^2, \cdots}$ is not linearly independent.
So $\sum_{i=0}^n c_i \beta^i = 0$ for some $n$ and some $c_i$.
But then $\beta$ is algebraic.
$\qed$

> Note that the converse is not true in general. 
*Example*: Let $E = \overline \RR$ be the algebraic numbers. Then $E \geq \QQ$ is algebraic, but $[E : \QQ] = \infty$.

**Theorem:**
Let $K \geq E \geq F$, then $[K: F] = [K: E] [E: F]$.

*Proof:*
Let $\theset{\alpha_i}^m$ be a basis for $E/F$
Let $\theset{\beta_i}^n$ be a basis for $K / E$.
Then the RHS is $mn$.

*Claim:*
$\theset{\alpha_i \beta_j}^{m, n}$ is a basis for $K/ F$.

*Linear independence:*

\begin{align*}
\sum_{i, j} c_{ij} \alpha _i \beta_j &= 0 \\
\implies \sum_j \sum_i c_{ij} \alpha_i \beta_j &= 0 \\
\implies \sum_i c_{ij} \alpha_i &= 0 \quad \text{since $\beta$ form a basis} \\
\implies \sum c_{ij} &= 0 \quad \text{since $\alpha$ form a basis}
.\end{align*}

*Exercise*:
Show this is also a spanning set.

$\qed$

**Corollary:**
Let $E_r \geq E_{r-1} \geq \cdots \geq E_1 \geq F$, then
$$
[E_r: F]= [E_r: E_{r-1}][E_{r-1}:E_{r-2}] \cdots [E_2: E_1][E_1 : F]
.$$

*Observation:*
If $\alpha \in E \geq F$ and $\alpha$ is algebraic over $F$ where $E \geq F(\alpha) \geq F$, then $F(\alpha)$ is algebraic (since $[F(\alpha): F] < \infty$) and $[F(\alpha): F]$ is the degree of the minimal polynomial of $\alpha$ over $F$.

**Corollary:**
Let $E = F(\alpha) \geq F$ where $\alpha$ is algebraic.
Then
$$\beta \in F(\alpha) \implies \deg \min(\beta, F) \divides \deg \min(\alpha, F)
.$$

*Proof:*
Since $F(\alpha) \geq F(\beta) \geq F$, we have $[F(\alpha): F] = [F(\alpha): F(\beta)][F(\beta): F]$.
But just note that
\begin{align*}
[F(\alpha): F]  &= \deg \min (\alpha, F) \text{ and } \\
[F(\beta): F]   &= \deg \min (\beta, F)
.\end{align*}

$\qed$

**Theorem:**
Let $E \geq F$ be algebraic, then
$$
[E: F] < \infty \iff E = F(\alpha_1, \cdots, \alpha_n)
\text{ for some }
\alpha_n \in E
.$$

## Algebraic Closures

**Definition:**
Let $E \geq F$, and define
$$
\overline{F_E} = \theset{\alpha \in E \mid \alpha \text{ is algebraic over } F}
$$
to be the **algebraic closure of $F$ in $E$**.

*Example:*
$\QQ \injects \CC$, while $\overline \QQ = \Bbb{A}$ is the field of algebraic numbers, which is a dense subfield of $\CC$.

**Proposition:**
$\overline{F_E}$ is a always field.

*Proof:*
Let $\alpha, \beta \in \overline{F_E}$, so $[F(\alpha, \beta): F] < \infty$.
Then $F(\alpha, \beta) \subseteq \overline{F_E}$ is algebraic over $F$ and
$$
\alpha\pm \beta, \quad \alpha\beta,\quad \frac \alpha \beta \quad  \in F(\alpha, \beta)
.$$

So $\overline{F_E}$ is a subfield of $E$ and thus a field.

**Definition:**
A field $F$ is **algebraically closed** iff every non-constant polynomial in $F[x]$ is a root in $F$.
Equivalently, every polynomial in $F[x]$ can be factored into linear factors.

If $F$ is algebraically closed and $E\geq F$ and $E$ is algebraic, then $E=F$.

### The Fundamental Theorem of Algebra

**Theorem (Fundamental Theorem of Algebra):**
$\CC$ is an algebraically closed field.

*Proof:*

**Liouville's theorem**:
A bounded entire function $f: \CC \selfmap$ is constant.

- *Bounded* means $\exists M \suchthat z\in \CC \implies \abs{f(z)} \leq M$.

- *Entire* means analytic everywhere.

Let $f(z) \in \CC[z]$ be a polynomial without a zero which is non-constant.

Then $\frac 1 {f(z)}: \CC \selfmap$ is analytic and bounded, and thus constant, and contradiction.

$\qed$

## Geometric Constructions:

Given the tools of a straightedge and compass, what real numbers can be constructed?
Let $\mathcal C$ be the set of such numbers.

**Theorem:**
$C$ is a subfield of $\RR$.