Note to understand Galois theory


In this article, we use following notation.

  • $G$: Group, $R$: Ring, $F$: Field

To be continued until we understand Galois theory


Prime ring
The smallest subring of a ring is called Prime ring.
In other words, prime ring is a subring generated by $1$.

Consider the following map from $\mathbb{Z}$ to a ring $R$,
n\mapsto 1_{R}.
This map is a homomorphism and the image is the prime ring of $R$.

Characteristic of a ring
\begin{align} \text{Prime ring} \cong \begin{cases} \,\mathbb{Z}/(0)=\mathbb{Z} &\text{: Called characteristic $0$}\\ \,\mathbb{Z}/(k) \quad \text{for some } k>0 &\text{: Called characteristic $k$} \end{cases} \end{align}

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

  1. 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 $p$.)
  2. 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 $F$"}$.)
Splitting Field
Let $f(x)\in F[x]$ be a monic polynomial, that is, the coefficient of the highest degree term is $1$.
An extension field $E/F$ is called a splitting field over $F$ of $f(x)$ if the following conditions are satisfied:
  1. $E=F(r_1,...,r_n)$
  2. $f(x)=(x-r_1)\cdots (x-r_n)$
Existence of a splitting field
Any monic polynomial $f(x)\in F[x]$ of positive degree has a splitting field $E/F$.

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

Field isomorphism $\eta:F\rightarrow \bar{F}$ has a unique extension $\tilde{\eta}:F[x]\rightarrow \bar{F}[x]$ such that
We write $\bar{g}(x):=\tilde{\eta}\bigl(g(x)\bigr)$ below.

Galois group
Galois group 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}
The subfield of $G$-invariants
Let $E$ be a field and $G$ be a subgroup of $\mathrm{Aut\,}E$. We call \begin{align} \mathrm{Inv\,}G:=\bigl\{a\in E\,\bigl|\, \eta(a)=a\quad(\forall \eta\in G)\bigr\} \end{align} the subfield of $G$-invariants or the $G$-fixed subfield of $E$.
Equation solvable by radicals
Let $f(x)\in F[x]$ be a monic of positive degree. Equation $f(x)=0$ is said to be solvable by radicals on $F$ if there exists $K/F$ such that \begin{align} &F=F_1\subset F_2\subset\cdots\subset F_{r+1}=K\quad\text{(: Called "Root tower over $F$")}\\ &\bigl(F_{i+1}=F_i(d_i), \quad d_i^{n_i}\in F_i\bigr) \end{align} and $K$ contains a splitting field over $F$ of $f(x)$.

Our (First) Goal ~to be contained above

Galois Critarion
Equation $\bigl(F[x]\ni\bigr) f(x)=0$ is solvable by radicals over $F$ of characteristic $0$
$\Leftrightarrow$ Galois group is solvable


Basic Algebra I: Second Edition (Dover Books on Mathematics)

Some properties of a cyclic group

Let $G=\langle a\rangle$ be a cyclic group.

  1. Let $\eta\in\mathrm{End}\langle a\rangle$. Then "$\eta\in\mathrm{Aut}\langle a\rangle\Leftrightarrow \eta(a)$ is a generator of $\langle a\rangle$".
    • Proof. If $\left |\langle a\rangle\right |=\infty$, $\mathrm{Aut}\langle a\rangle=\{\mathrm{id}\}$. Suppose $\left |\langle a\rangle\right |=n<\infty$. ($\Rightarrow$) $\eta (a^k)=\left(\eta (a)\right)^k$ and $\eta$ is an epimorphism. ($\Leftarrow$) If $1=\eta (a^k)=\left(\eta (a)\right)^k$ then $n\bigl | k$.
  2. Let $\left |\langle a\rangle\right |=n$. Then $\mathrm{Aut}\langle a\rangle=\left\{\eta\in\mathrm{End}\,\left |\,\eta(a)=a^k,\,(n,k)=1\right\}\right.$
    • Proof. Corollary of the above. (Note: There exists $\alpha,\beta\in\mathbb{Z}$ such that $\alpha n+\beta k=1$. Hence, $\left(a^{k}\right)^\beta=a$.)
  3. Let $\left |\langle a\rangle\right |=n$ and $U\left(\mathbb{Z}/(n)\right)$ be the multiple group of $\mathbb{Z}/(n)$. Then \begin{align*}\begin{array}{ccc}\mathrm{Aut}\langle a\rangle&\cong&U\left(\mathbb{Z}/(n)\right)\\\eta_k&\mapsto&\bar{k}=k+(n)\end{array}\end{align*} where $\eta_k(a)=a^k$, $(n,k)=1$.

Some properties of a field

Properties of a (general) field $F$

Let $F$ be a field and $E/F$ a field extension.

  1. Commutative ring $R (\neq 0)$ is a field $\Leftrightarrow$ Ideals of $R$ are only $0\left(=(0)\right)$ or $R\left(=(1)\right)$
    • Proof. $(\Rightarrow):\,a\in I\Rightarrow 1= aa^{-1}\in I$. $(\Leftarrow):\,0\neq a\in R\Rightarrow 1\in R=(a)$.
    • Cor. Homomorphism from a field to a ring is injective.
  2. $F[x]$ is a p.i.d. (Principal Ideal Domain)
    • Proof. If ideal $I$ of $F[x]$ is not $0$, there exists a minimum (positive) degree polynomial.
  3. Suppose $u\in E$ is algebraic over $F$ and its minimum polynomial is $g(x)$. Then, if $g(x)$ is irreducible in $F[x]\Rightarrow F[u]$ is a field.
    • Proof. $F[x]/(g(x))\cong F[u]$. Any ideals of $F[x]/(g(x))$ can be represented as $I/(g(x))$ where $(g(x))\subset I$ is an ideal of $F[x]$. Hence $I=(f(x))$ and $f(x)\bigl| g(x)$. If $g(x)$ is irreducible, $\left(f(x)\right)=\left(1\right)\mathrm{\,or\,}\left(g(x)\right)$.

Properties of a field $F$ (Characteristic $p\neq 0$)

  1. $F^p=\{a^p\,|\,\,a\in F\}$ is a subfield of $F$.
    • Proof. Let $a,b\in F$. $(ab)^p=a^pb^p,1^p=1$ and $(a+b)^p=a^p+b^p$ since $p\left |\begin{pmatrix}p\\k\end{pmatrix}\right.$. Hence $a\mapsto a^p$ is an endomorphism.
  2. If $F$ is a finite, $F=F^p$
    • Proof. Since $F$ is finite, monomorphism $F\ni a\mapsto a^p\in F^p$ is onto.