User:CaritasUbi/LinearAlgebra

Definition 1.1
A field is a set F with two binary operations + and · such that all the following hold:
 * (a) (Closure) For all x, y $$\in F$$
 * x + y $$\in F$$ and
 * x · y $$\in F$$.
 * (b) (Associativity) For all x, y, and z $$\in F$$,
 * x + (y + z) = (x + y) + z and
 * x · (y · z) = (x · y) · z.
 * (c) (Commutativity) For all x, y $$\in F$$,
 * x + y = y + x and
 * x · y = y · x
 * (d) (Identity) There are elements $$0 \in F$$ and $$1 \in F$$ such that for any element $$x \in F$$,
 * x + 0 = x and
 * x · 1 = x.
 * (e) (Inverses) For any $$x \in F$$, there is an element $$-x \in F$$ and an element $$x^{-1} \in F$$ such that
 * x + (-x) = 0
 * x · x-1 = 1
 * (f) (Distributivity) For any elements $$a, x, y \in F$$,
 * a·(x + y) = a·x + a·y.

Theorem 1.1
''If $$F_1$$ is a subfield of $$F_2$$, and $$F_2$$ is a subfield of $$F_3$$, then $$F_1$$ is a subfield of $$F_3$$.

Proof
The proof follows immediately from the definition of a field, since any property of elements in $$F_1$$ holds in $$F_2$$ and therefore in $$F_3$$ as well.

Theorem 1.2
Every subfield of $$\mathbb{C}$$ contains all of $$\mathbb{Q}$$.

Proof
Let $$F$$ be a subfield of $$\mathbb{C}$$. We need to show that $$F$$ contains every element of $$\mathbb{Q}$$. By definition, for any $$x \in \mathbb{Q}$$, there are integers $$p$$ and $$q$$ such that $$x = \frac{p}{q}$$. Since $$F$$ is a field, it contains an element 1 (the multiplicative identity element of the field), and
 * $$p = \underbrace{1 + 1 + \ldots + 1}_\text{p times} \in F$$.

Similarly,
 * $$q = \underbrace{1 + 1 + \ldots + 1}_\text{q times} \in F$$

and so is its multiplicative inverse $$q^{-1}$$. Then because $$F$$ is closed under multiplication,
 * $$pq^{-1} = \frac{p}{q} \in F$$.

Thus, every $$x \in \mathbb{Q}$$ is in $$F$$. QED

Corollary 1
$$\mathbb{Q}$$ is the smallest subfield of $$\mathbb{C}$$

Proof
Since $$\mathbb{Q}$$ is a subfield of $$\mathbb{C}$$, any subfield of $$\mathbb{Q}$$ is also a subfield of $$\mathbb{C}$$ and therefore by the theorem contains all of $$\mathbb{Q}$$. QED