# 代數導論 Week 10 - Uniqueness of Splitting Field (Part 3)

[TOC]

## Thm 3

:::danger

Let

$$
\phi : F \to \tilde F
$$

be a field isomorphism, and let

$$
f(x) = \sum_{i = 0}^{a_i x^i}
$$

$$
\tilde f (x) = \sum_{i = 0}^{n}\phi(a_i) x^i
$$

Let $E$ be a splitting field for $f(x)$ over $F$, and $\tilde E$ be a splitting field for $\tilde f(x)$ over $\tilde f$. Then exists a isomorphism

$$
\sigma : E \to \tilde E
$$

such that

$$
\sigma|_{F} = \phi
$$

:::

We prove by induction on $\deg f$

LEt

$$
\tilde \phi : F[x] \to \tilde F[x]
$$

be the ring homomorphism extending $\phi$. i,e,

$$
\tilde \phi(\sum_{i = 0}^{n}a_ix^i) = \sum_{i = 0}^{n}\phi(a_i)x^i
$$

Then

1. $\tilde \phi$ is a ring isomorphism
2. If $p(x)$ is an irreducible factor polynomial of $f$, then $\tilde \phi(p)$ is an irreducible factor of $\tilde f$

Case 1: $f(x)$ splits completely in $F[x]$

Then

$$
E = F
$$

and $\tilde f$ splits completely in $\tilde F[x]$, thus

$$
\tilde E = \tilde F
$$

will do the work

Case 2: $f(x)$ has an irreducible factor $p(x)$ with $\deg p > 1$

Then let

$$
F_1 = \frac {F[x]}{(p)}
$$

and

$$
\tilde F_1 = \frac {\tilde F[x]}{(\tilde \phi(p))}
$$

Then observe that

$$
\phi_1 : F_1 = \frac {F[x]}{(p)} \simeq \frac {\tilde F[x]}{(\tilde \phi(p))} = \tilde F
$$

where

$$
\phi_1 |_{F} = \phi
$$

> Quotient 前就 *surjective*, 因為 *kernel* 送到 *kernel* 所以 *well-defined*; 定義域是 *field* 且非零所以 *injective*

And $p(x)$ has a root in $F_1$. So

$$
p(x) = (x - \alpha)p_1(x) \quad \alpha \in F_1
$$

So

$$
f(x) = (x - \alpha)f_1(x) \quad \alpha \in F_1
$$

So

$$
\tilde f(x) = (x - \tilde \alpha)\tilde f_1(x)
$$

where

$$
\tilde \alpha = \phi(\alpha) \in \tilde F_1
$$

$$
\tilde \phi_1(x - \alpha) = (x - \tilde \alpha)
$$

so

$$
\tilde \phi_1(f_1) = \tilde \phi(f_1)= \tilde f_1
$$

By induction hypothesis, exists splitting field for $f_1(x)/F_1$, splitting field $\tilde E_1$ for $\tilde f_1(x)/\tilde F_1$, and exists isomorphism $\sigma_1 : E_1 \to \tilde E_1$ such that $\sigma_1|_{F_1} = \phi_1$.

Then $E_1 \supseteq F_1$, so $E_1$ contains all roots of $f_1(x)$. Hence $E_1$ contains all roots of $f(x)$.

Suppose that there is a field $K$ such that $F \subseteq K \subseteq E$ containing all roots of $f(x)$, then

$$
\alpha\in K \Rightarrow F(\alpha) = F_1 \subseteq K
$$

and $K$ contains all roots of $f_1(x)$.

### Corollary 4

:::danger

Any two splittings for $f(x)/F[x]$ are isomorphic

:::

Let

$$
F = \tilde F
$$

and

$$
\phi = \text{id}
$$

and apply theorem 3

## Def: Algebraic Closure

:::warning

Let $F$ be a field. The field $\bar F$ is called an algebraic closure of $F$ is $\bar F$ is algebraic over $F$ and every non-constant polynomial $f(x) \in F[x]$ has a root in $\bar F$.

:::

## Def: Algebraically Closed

:::warning

A field $K$ is said to be algebraic closed if every non-constant polynomial $f(x) \in K[x]$ has all its root in $K$.

:::

> 也就是自己就是自己的 *algebraic closure*

### Proposition 5

:::danger

Let $\bar F$ be an algebraic closure of $F$, then $\bar F$ is algebraic closed

:::

Suppose that

$$
f(x) \in \bar F[x]
$$

and let $\alpha$ be a root of $f(x)$ in some extension of $\bar F$.

$$
\bar F(\alpha)
$$

is an algebraic extension of $\bar F$, and $\bar F$ is algebraic over $F$. So

1. (by algebraic tower) $\bar F(\alpha)$ is algebraic over $F$
2. Thus exists irreducible $f(x) \in F[x]$ such that
$$
f(\alpha) = 0
$$
3. so $\alpha \in \bar F$.