April 28th

#class
Prior to this please take a look at April 25th,Vector Spaces in Abstract Algebra. This is the introduction to Extension Fields

Warm up

Prove that the following are all fields:

1. $\mathbb{Q}[x]/<x^2-2>$
2. $\mathbb{Q}[x]/<x^3-2>$
3. ${\mathbb{F}}_2[x]/<x^3+x+1>$

How to think about elements in this example?
A1 (easy): (apart from normal equivalence classes, $f(x)+I$)
A2 (another way): You can think of redefined equivalence classes, if $f(x)=q(x)p(x)+r(x)$, then $f(x)=[r(x)]$
A3 (uhhh) = $\mathbb{Q}[x]/<x^2-2>\cong \mathbb{Q}[\sqrt{2}]$

Class

Definition:

E and F are fields, s.t. $E\le F$ as a subfield Note: could be error seams to be the other way around.
E (extension Field)
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F (Base Field)

Extension of fields

Ex:
$\mathbb{Q}[\sqrt{2}]=\{a+b\sqrt{2}\mid a,b\in \mathbb{Q}\}$
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$\mathbb{Q}=\{a+b\sqrt{2}\mid a\in \mathbb{Q}andb=0\}$

Lemma ("it's weird")(Reason for last lecture April 25th)

If E is a field extension, then E is an F-vector space. (E is a vector space over F)
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F
(Note; take the E and F together)
(essentially, from the example above, $\mathbb{Q}[\sqrt{2}]$ is a vector space over $\mathbb{Q}$)

Vector space over $k$: $kXv\to v$

• Scalars: $\alpha ,\beta \in k$, $\alpha v=(\alpha v)$
• Vectors: $v,w,u\in V$ for some abelian group V
  Is 2 a vector or scaler?

Every vector in $\mathbb{Q}[\sqrt{2}]$ is of the form $a1+b\sqrt{2}=a{e}_1+b{e}_2$
Is is true for $\mathbb{Q}[\sqrt{2}]$ over $\mathbb{Q}$ that $1,\sqrt{2}$ is a basis of $\mathbb{Q}[\sqrt{2}]$ as a $\mathbb{Q}$-vector space
Back to 2 from the warm up:

$\mathbb{Q}[x]/<x^3-2>$ is an extension of $\mathbb{Q}[x]$

Def:

E over F is a simple algebraic extension $⟺E=F[\alpha ]$ for some $\alpha \in E$ algebraic element over F

Def 2:

another way of stating the above definition

E over F is a simple algebraic extension $⟺E\cong F[x]/<p(x)>$, p(x) is irreducible.

Prop. 21. 12

$\mathbb{F}[\alpha ]\cong F[x]/<p(x)>$ for irreducible p(x)

After for $\mathbb{Q}[x]/<x^3-2>\cong \mathbb{Q}[\omega ]=\{a+b\omega +c{\omega }^2\mid a,b,c\in \mathbb{Q}and{\omega }^3=2\}$

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