# Thursday September 12th

## Geometric Constructions

**Definition:**
A real number $\alpha$ is said to be **constructible** iff $\abs \alpha$ is constructible using a ruler and compass.
Let $\mathcal C$ be the set of constructible numbers.

Note that $\pm 1$ is constructible, and thus so is $\ZZ$.

**Theorem:**
$\mathcal{C}$ is a field.

*Proof:*
It suffices to construct $\alpha \pm \beta,~~ \alpha\beta,~~ \alpha / \beta$.

*Showing $\pm$ and inverses:*
Relatively easy.

*Showing closure under products:*

![Image](figures/2019-09-17-09:48.png)\

**Corollary:**
$\QQ \leq \mathcal C$ is a subfield.

Can we get all of $\RR$ with $\mathcal C$? The operations we have are

1. Intersect 2 lines (gives nothing new)

2. Intersect a line and a circle

3. Intersect 2 circles

Operation (3) reduces to (2) by subtracting two equations of a circle ($x^2 + y^2 + ax + by + c$) to get an equation of a line.

Operation (2) reduces to solving quadratic equations.

**Theorem:**
$\mathcal C$ contains precisely the real numbers obtained by adjoining finitely many square roots of elements in $\QQ$.

*Proof:*
Need to show that $\alpha \in \mathcal C \implies \sqrt \alpha \in \mathcal C$.

- Bisect $PA$ to get $B$.

- Draw a circle centered at $B$.

- Let $Q$ be intersection of circle with $y$ axis and $O$ be the origin.

- Note triangles 1 and 2 are similar, so
$$
\frac{OQ}{OA} = \frac{PO}{OQ} \implies (OQ)^2 = (PO)(OA) = 1\alpha
.$$

$\qed$

*Corollary:*
Let $\gamma \in \mathcal{C}$ be constructible.
Then there exist $\theset{\alpha_i}_{i=1}^n$ such that
$$
\gamma = \prod_{i=1}^n \alpha_i
\quad\text{and}\quad
[\QQ(\alpha_1, \cdots, \alpha_j): \QQ(\alpha_1, \cdots, \alpha_{j-1})] = 2
,$$
and $[\QQ(\alpha): \QQ] = 2^d$ for some $d$.

**Applications:**

**Doubling the cube:**
Given a cube of size 1, can we construct one of size 2?
To do this, we'd need $x^3 = 2$.
But note that $\min(\sqrt[3]{2}, \QQ) = x^3 - 2 = f(x)$ is irreducible over $\QQ$.
So $[\QQ(\sqrt[3]{2}): \QQ] = 3 \neq 2^d$ for any $d$, so this can not be constructible.


**Trisections of angles:**
We want to construct regular polygons, so we'll need to construct angles.
We can get some by bisecting known angles, but can we get all of them?

*Example:*
Attempt to construct $20^\circ$ by trisecting the known angle $60^\circ$, which is constructible using a triangle of side lengths $1,2,\sqrt 3$.

If $20^\circ$ were constructible, $\cos 20^\circ$ would be as well.
There is an identity
$$
\cos 3\theta = 4\cos^3 \theta - 3\cos \theta
.$$

Letting $\theta = 20^\circ$ so $3\theta = 60^\circ$, we obtain
$$
\frac 1 2 = 4(\cos 20^\circ)^3 - 3\cos 20^\circ,
$$

so if we let $x = \cos 20^\circ$ then $x$ satisfies the polynomial $f(x) = 8x^3 - 6x - 1$, which is irreducible.
But then $[\QQ(20^\circ):\QQ] = 3 \neq 2^d$, so $\cos 20^\circ \not\in\mathcal C$.

## Finite Fields

**Definition:**
The *characteristic* of $F$ is the smallest $n\geq 0$ such that $n1 = 0$, or $0$ if such an $n$ does not exist.

*Exercise*:
For a field $F$, show that $\ch F = 0$ or $p$ a prime.

Note that if $\ch F = 0$, then $\ZZ \in F$ since $1,~ 1+1,~ 1+1+1, \cdots$ are all in $F$.
Since inverses must also exist in $F$, we must have $\QQ \in F$ as well.
So $\ch F = 0 \iff F$ is infinite.

If $\ch F = p$, it follows that $\ZZ_p \subset F$.

**Theorem:**
$$
\text{For } E \geq F \text{ where } [E: F] = n \text{ and } F \text{ finite }, \quad
\abs F = q \implies \abs E = q^n
.$$

*Proof:*
$E$ is a vector space over $F$.
Let $\theset{v_i}^n$ be a basis.
Then $\alpha \in E \implies \alpha = \sum_{i=1}^n a_i v_i$ where each $a_i \in F$.
There are $q$ choices for each $a_i$, and $n$ coefficients, yielding $q^n$ distinct elements.

$\qed$

**Corollary:**
Let $E$ be a finite field where $\ch E = p$.
Then $\abs E = p^n$ for some $n$.

**Theorem:**
Let $\ZZ_p \leq E$ with $\abs E = p^n$.
If $\alpha \in E$, then $\alpha$ satisfies
$$
x^{p^n} - x \in \ZZ_p[x].
$$

*Proof:*
If $\alpha = 0$, we're done.
So suppose $\alpha \neq 0$, then $\alpha \in E\units$, which is a group of order $p^n - 1$.
So $\alpha^{p^n - 1} = 1$, and thus $\alpha \alpha^{p^n - 1} = \alpha 1 \implies \alpha^{p^n} = \alpha$.

$\qed$

**Definition:**
$\alpha \in F$ is an *$n$th root of unity* iff $\alpha^n = 1$.
It is a *primitive* root of unity of $n$ iff $k\leq n \implies \alpha^k \neq 1$ (so $n$ is the smallest power for which this holds).

**Fact:**
If $F$ is a finite field, then $F\units$ is a cyclic group.

**Corollary:**
If $E \geq F$ with $[E: F] = n$, then $E = F(\alpha)$ for just a single element $\alpha$.

*Proof:*
Choose $\alpha \in E\units$ such that $\generators \alpha = E\units$.
Then $E = F(\alpha)$.

$\qed$

Next time:
Showing the existence of a field with $p^n$ elements.

For now: derivatives.

Let $f(x) \in F[x]$ by a polynomial with a multiple zero $\alpha \in E$ for some $E \geq F$.

If it has multiplicity $m \geq 2$, then note that
$$
f(x) = (x-\alpha)^m g(x) \implies f'(x) m(x-\alpha)^{m-1}g(x) + g'(x)(x-\alpha)^m \implies f'(\alpha) = 0.
$$
So
$$
\alpha \text{ a multiple zero of } f \implies f'(\alpha) = 0
.$$

The converse is also useful.

*Application:*
Let $f(x) = x^{p^n} - x$, then $f'(x) = p^n x^{p^n - 1} - 1 = -1 \neq 0$, so all of the roots are distinct.