# 02.6.分解体の存在

## Proposition

### Statement

$F$:Field
$f(x) \in F[x] \quad \text{deg} f(x) = n \gt 0$
then $\exists E/F$ such that
$f(x) = a (x - \alpha_1)\cdots(x - \alpha_n) \quad a \in F, \alpha_1,\cdots\alpha_n \in E$

### Proof

when $n = 1$
$f(x) = ax + b \quad a,b \in F$
$f(x) = a(x - (-b/a))$
$-b/a \in F$
$\therefore E = F$
when $n \gt 2$
Suppose when $\text{deg} f(x) \lt n$, the statement is true.---(*)
Let $\text{deg}f(x) = n$
case $f(x)$ is a reducible polynomial
$\exists p(x),q(x) \in F[x], f(x) = a p(x)q(x), 0 \lt \text{deg}p(x), \text{deg}q(x) \lt n, p(x),q(x)$:monic
then $p(x), q(x)$ fills the condition (*). $\therefore \exists E',E'',p(x) = (x - \alpha_1) \cdots (x - \alpha_i), q(x) = (x - \alpha_{i+1}) \cdots (x - \alpha_n) \quad (\alpha_1, \cdots, \alpha_i \in E', \alpha_{i+1} \cdots \alpha_n \in E'')$ $\therefore E = F(\alpha_1, \cdots \alpha_n)$ follows the statement.
case $f(x)$ is a irreducible polynomial
$F[x]$ is a PID.($\because \text{Prop.多項式環と単項イデアル整域}$)
$\therefore (f(x))$ is a Maximal Ideal.($\because \text{Prop.PIDにおける既約元と極大イデアル}$)
$\therefore F[x]/(f(x))$ is a Field.($\because \text{Prop.極大イデアルと体}$)
$\phi:F \hookrightarrow F[x] \twoheadrightarrow F[x]/(f(x)) = E \quad$($E$:Field)
$\phi$:injective $(\because \text{Prop.体の準同型写像は単射})$
$$\begin{xy} \xymatrix { & F[x] / (f(x)) \ar[r]^{\simeq} & F(\alpha)\\ F \ar[r] \ar[ur]^{\phi} & F[x] \ar[u]_{\psi} \ar[ur] \\ } \end{xy}$$ $\therefore E$ is an expansion field of $\text{Img}\phi$.
$F \simeq \text{Img}\phi \subset E$
$\therefore E$ is an expansion field of $F$.
Let
$\psi:F[x] \twoheadrightarrow F[x]/(f(x)) = E$
$\psi:x \mapsto \alpha \quad(= x + (f(x)))$
$f(x) = a_0 + a_1 x + \cdots + a_n x^n$
$\bar F = \psi(F) \subset E$
then
$\psi(f(x)) = 0_E \because \psi(f(x)) = f(x) + (f(x)) = 0 + (f(x))$
$\psi(f(x)) = \psi(a_0) + \psi(a_1) \psi(x) + \cdots + \psi(a_n)\psi(x)^n$
$= \bar a_0 + \bar a_1 \alpha + \cdots + \bar a_n \alpha^n = \bar f(\alpha)$
$\quad (\bar f(x) = \bar a_0 + \bar a_1 x + \cdots + \bar a_n x^n \in \bar F[x] \subset E[x])$
$\bar f(\alpha) = 0_E$ ---(1)
then
$F \simeq \bar F \subset \bar F(\alpha)\quad$ (adding $\alpha$ to $F$) $\subset E$
$\bar f(x) \in \bar F[x] \subset \bar F(\alpha)[x]$
$\exists g(x) \in \bar F(\alpha)[x]$ and $\exists a \in \bar F(\alpha)$ such that
$\bar f(x) = (x - \alpha)g(x) + a \quad (\because\text{Prop.剰余定理})$
(1) $\Rightarrow a = 0_E \therefore \bar f(x) = (x - \alpha)g(x)$
$\text{deg }g(x) = n-1 \lt n$
$\therefore \exists E'/\bar F(\alpha) \text{ s.t. }g(x) = (x - \alpha_2)\cdots(x - \alpha_n) \quad (\alpha_2 \cdots \alpha_n \in E')$
$\therefore E'' = \bar F(\alpha, \alpha_2, \cdots \alpha_n) \simeq F(\alpha, \alpha_2, \cdots \alpha_n)$ follows the statement.
QED