Over $\Bbb C$, $p(x)=c(x-\alpha_1)\cdots(x-\alpha_n)$

Definition: A field extension is $\phi:F\to K$ where $\phi$ is an injective ring homomorphism and $F\subseteq K$

Exercise: If F is a field and $\phi$ is any ring homomorphism, then $\phi$ is injective

Examples:

1. $\Bbb Q\subseteq \R$

2. $\R \subseteq \Bbb C$

3. $\Bbb Q \subseteq \Bbb Q[\sqrt 2]$

4. If $F$ is a field and $f(x)$ is an irreducible polynomial then $\underbrace{F[x]/f(x)}_{\text{field}} = \{\ \overline{h(x)}: \deg h < \deg f\ \}$

   $\phi: F \hookrightarrow F[x]/\braket{\ f(x)\ } \\ \phi(a)=\overline a = a+\braket{\ f(x)\ }$

   $\phi$ is an injective ring homomorphism

5. $F=\Bbb Q,\ f(x)=x^2-2$

   $\phi:\Bbb Q\hookrightarrow \underbrace{\Bbb Q[x]/\braket{\ x^2-2\ }}_k=\{\ \overline{a+bx}:\ a,b\in \Bbb Q\ \}$

   $\phi(a)=\bar a$ ← Coset represented by $a$

   Consider $f(\lambda)=\lambda ^2-2$

   $f(\lambda)\in \Bbb Q[\lambda],f(\lambda)$ is irreducible

   But, $f(\lambda)\in K[\lambda]$ irreducible and has a root

   $\Bbb Q[x]/\braket{\ x^2-2\ }=K\cong\Bbb Q[\sqrt 2]$

   $\psi:Q[x]\to \Bbb Q[\sqrt 2]$

   $\psi(p(x))=p(\sqrt2)$

   $(\overline x)^2-\overline 2= \overline x\cdot \overline x-\overline 2 = \overline{x^2-2} = \overline 0$

   So, if $f(\lambda)$ is irreducible then in the field $K=F[x]/\braket{\ f(x)\ }$ $,\ f(\lambda)$ has a root $\overline x$

   $\lambda ^2 -2$ factorization over $K$

   $\lambda ^2 -2=(\lambda-\overline x)(\lambda +\overline x) = (\lambda-\sqrt 2)(\lambda +\sqrt x)$

6. $\Bbb Q[ \sqrt[3]{2}]=\{\ a+b\sqrt[3]{2} +c\sqrt[3]{4}:a,b,c\in \Bbb Q\ \}\subseteq \R$ is a field

   $\Bbb Q[\sqrt[3]{2}]\cong \Bbb Q[x]/\braket{\ x^3-2\ }$

   Proof:

   Using the $1^{\text{st}}$ Isomorphism Theorem,

   $\phi:\Bbb Q[x]\to \Bbb Q[\sqrt[3]{2}]\\ \phi(p(x))=p(\sqrt[3]{2})$

   $x^3-2\in \ker \phi$

   $\braket{\ x^3-2\ }\subseteq \ker \phi$

   Claim: they are equal

   Notice that $\sqrt[3]{2} \notin \Bbb Q$ so $x^3-2$ has no roots in $\Bbb Q$.

   It is of $\deg 3$, thus it is irreducible

   Consequently, $\braket{\ x^3-2\ }$ is maximal

   Hence, $\braket{\ x^3-2\ } =\ker \phi$

   So, $\Bbb Q[x]/\braket{\ x^3-2\ } =\Bbb Q[x]/\ker \phi \cong Im \phi = \Bbb Q[\sqrt[3]{2}]$

   $f(\lambda)=\lambda ^3 -2$ irreducible over $\Bbb Q$ but it is reducible over $K$ because it has a root

   $\implies \lambda - \sqrt[3]{2}\mid \lambda ^2-2$ over $K$

   $$ \begin{array}{r} \lambda^2+\sqrt[3]2\lambda\phantom{11111111111} \\ \lambda-\sqrt[3]2\phantom{1}{\overline{\smash{\big)}\,\lambda^3-2\phantom{11111111111111}}} \\ \underline{-(\lambda^3-\sqrt[3]2\lambda^2)} \phantom{1111111111}\\ \sqrt[3]2\lambda^2-2\phantom{1111-1} \\ \underline{-(\sqrt[3]2\lambda^2-\sqrt[3]4\lambda)}\phantom{1111} \\ \sqrt[3]4\lambda-2\phantom{-11}\\ \underline{\sqrt[3]4\lambda-\sqrt[3]4 \sqrt[3]4\phantom{}}\\ 0

   \end{array} $$

   We have a factorization over $K$

   $\lambda ^3-2=(\lambda -\sqrt[3]{2})(\lambda^2+\sqrt[3]{2}\lambda +\sqrt[3]4)$

   $\Bbb Q\to \Bbb Q[x]/\braket{\ x^3-2\ } \cong K\to L=K[y]/\underbrace{\braket{\ y^2+\sqrt[3]2+\sqrt[3]4\ }}_{\text{irreducible over }K}$

   $(\lambda -\sqrt[3]2)(\lambda^2+\sqrt[3]2+\sqrt[3]4)$

Theorem:

Let $f$ be a polynomial over $F$, then $\exists$ field $F\to K$ st. over $K,\ f(\lambda)=c(\lambda-\alpha_1)\cdots (\lambda-\alpha_n)$ splits

In $L,\ \lambda ^3 -2$ splits

Field Extension: $\phi \to K,$ $K$ extends $F$

$K=F\hookrightarrow K \\ \phantom{00111}\phi \mapsto \overline \phi$

Recall from Linear Algebra:

$F-\text{ Field}$

$V-\text{ Vector Space}$: abelian group with scalar multiples

$\underbrace{\alpha}{\in F} \cdot \underbrace{v}{\in V} \in V$