### Notation

In this article, we use following notation.

- $G$: Group, $R$: Ring, $F$: Field

### To be continued until we understand Galois theory

#### Characteristic

*.*

**Prime ring**In other words, prime ring is a subring generated by $1$.

Consider the following map from $\mathbb{Z}$ to a ring $R$,

\begin{align}

n\mapsto 1_{R}.

\end{align}

This map is a homomorphism and the image is the prime ring of $R$.

Similarly, * prime field* of a field $F$ is defined as the smallest subfield of $F$.

- If characteristic of $F$ is $p\neq 0$, $\text{"prime ring of $F$"}\cong\mathbb{Z}/(p)$ is a subfield.

(Note that the prime ring of a domain has characteristic $0$ or a prime $p$.) - If characteristic of $F$ is $0$, $\text{"Prime field of $F$"}\cong \mathbb{Q}$.

(Since the monomorphism $\mathbb{Z}\ni m\mapsto m1\in \text{"Prime field of $F$"}$ can be extended to a monomorphism $\mathbb{Q}\ni n/m\mapsto n1(m1)^{-1}\in \text{"Prime field of $F$"}$.)

*, that is, the coefficient of the highest degree term is $1$.*

**monic polynomial**An extension field $E/F$ is called a

*over $F$ of $f(x)$ if the following conditions are satisfied:*

**splitting field**- $E=F(r_1,...,r_n)$
- $f(x)=(x-r_1)\cdots (x-r_n)$

*Proof.*

Let $S$ be a generator of a ring $R$ (or a field $F$) .

If two homomorphism $\eta_i:R\rightarrow R^\prime\,(i=1,2)$ are equal on $S$, $\eta_1=\eta_2$.

This is because $S\subset\bigl\{r\in R\bigl| \eta_1(r)=\eta_2(r)\bigr\}$ is a ring (field).

Let $\eta:R\rightarrow R^\prime$ be a homomorphism of commutative ring and $u\in R^\prime$.

Then we get a unique extension $\eta_u:R[x]\rightarrow R^\prime$ such that

\begin{align}

\eta_u(x)=u.

\end{align}

Field isomorphism $\eta:F\rightarrow \bar{F}$ has a unique extension $\tilde{\eta}:F[x]\rightarrow \bar{F}[x]$ such that

\begin{align}

\tilde{\eta}(x)=x.

\end{align}

We write $\bar{g}(x):=\tilde{\eta}\bigl(g(x)\bigr)$ below.

*of $E$ over $F$ is defined by \begin{align} \mathrm{Gal\,} (E/F) :=\mathrm{Aut\,} (E/F) =\bigl\{\eta\,\bigl|\, \eta\text{: isomorphism of $E$ to $E$, }\eta|_F=\mathrm{id}_F\bigr\}. \end{align}*

**Galois group***or*

**the subfield of $G$-invariants***.*

**the $G$-fixed subfield of $E$**### Our (First) Goal ~to be contained above

$\Leftrightarrow$ Galois group is solvable