Fields

Preliminaries:

Irreducible:

Remember, in an integral domain $D$ an element $a\in D$ is irreducible if whenever $a=bc$, then either b or c is a unit.

Prime:

Similarly, in an integral domain $D$ an element $p\in D$ is prime if for some $p|ab$ then $p|a$ or $p|b$

Units and associates:

Units are just multiplicative inverse and associates are any place an unit can take an element, ie a is an associate of b if whenever a=bc, c is an unit.

It could be useful to review the first few portions of Half-Factorial Domains.

Field Definition:

Lastly just remember a field is a commutative division ring. Essentially, Is also an integral domain and basically all of the definitions of rings (W/ unity, commutes, if ab=0 then a=0 or b = 0, all units are invertible excluding 0)

Remember, $\mathbb{Q},\mathbb{R},\mathbb{C}$ are all fields, and moreover $\mathbb{Q}\subseteq \mathbb{R}\subseteq \mathbb{C}$. $\mathbb{Z}$ is not a field as not every element has an inverse. ($\frac{1}{2}\notin \mathbb{Z}$)

Extension Fields:

Class notes on this topic can be found in April 28th

Introduction:

It is known that both $\mathbb{Q},\mathbb{R}$ are fields. Even more, we know $\mathbb{Q}\subseteq \mathbb{R}$.

This is important, suppose we want to find roots. It would be difficult to find roots for $\mathbb{Q}$ as we cannot factor into linear factors with elements with $\mathbb{Q}$.

Consider $p(x)\in \mathbb{Q}$:
$p(x)=x^4-5x^2+6$

Now notice the following factorizations:
In $\mathbb{Q}$
$p(x)=(x^2-2)(x^2-3)$
In $\mathbb{R}$
$p(x)=(x-\sqrt{2})(x+\sqrt{2})(x-\sqrt{3})(x+\sqrt{3})$

Now, we can find linear solutions. Additionally, it is possible to find an even smaller field in which p(x) has zero. Namely
$\mathbb{Q}[\sqrt{2}]=\{a+b\sqrt{2}:a,b\in \mathbb{Q}\}$

The goal is to study such fields for arbitrary polynomials over $\mathbb{Q}$
For a deeper guide on extension fields view Extension Fields