眼里没有噙泪，那份悲伤已被浓黑地固定成型，深嵌在他的瞳孔里。

# Symmetry

The set $O(2,\mathbb{R})$ of all orthogonal transformations is a group under composition, called the real orthogonal group.

The wonderful idea of Galois was to associate to each polynomial $f(x)$ a group, nowadays called its Galois group, whose properties reflect the behavior of $f(x)$. Our aim in this section is to set up an analogy between the symmetry group of a polygon and the Galois group of a polynomial.

# Rings, Domains and Fields

From now on, we will write ring instead of “commutative ring with 1.”

Proof: $[a][b]=0\Rightarrow [ab]=0\Rightarrow ab\equiv 0\mod p$。

• If $p$ is prime, then $\mathbb{Z}_p$ is a field.

Proof: Just like $\textnormal{Frac}(\mathbb{Z})=\mathbb{Q}$, define:

$$\textnormal{Frac}(R)=\{a/b\;|\;a,b\in R,b\neq 0\}\\$$

• Addition: $a/b+c/d=(ad+bc)/bd$.
• Multiplication: $(a/b)(c/d)=ac/bd$.

where $a/b=ab^{-1}$.

We call $\textnormal{Frac}(R)$ is $R$'s fraction field. And we denote $R[x]$ the ring of polynomials over $R$, and $\textnormal{Frac}(R[x])$ the field of rational functions over $R$, whose elements are of the form $f(x)/g(x)$.

# Homomorphism and Ideals

We can derive $\varphi(0)=0$ immediately:

$$\begin{aligned} &\because \varphi(a)=\varphi(a+0)=\varphi(a)+\varphi(0)\\ &\therefore 0=\varphi(0) \end{aligned}$$

An ideal is a sub additive group of the ring.

If $a\in R$, $\{ra:r\in R\}$ is the ideal generated by $a$, which is called the principal ideal generated by $a$, denoted by $(a)$.

Proof: $\text{ker}\varphi$ contains $0$ is self-evident, and:

$$\forall a\in\text{ker}\varphi,r\in R,\varphi(ra)=\varphi(r)\varphi(a)=\varphi(r)0=0$$

so $ra\in\text{ker}\varphi$. and $\varphi(a-b)=\varphi(a)-\varphi(b)=0-0=0$, so $\varphi(a-b)\in\text{ker}\varphi$.

If $\varphi$ is an injection, then for $r\neq 0,\varphi(r)\neq\varphi(0)=0$, so $\text{ker}\varphi=\{0\}$. Conversely, if $\text{ker}\varphi=\{0\}$, and exists $r\neq r',\varphi(r)=\varphi(r')$, then $\varphi(r-r')=\varphi(r)-\varphi(r')=0$, so $0\neq r-r'\in\text{ker}\varphi$, contradicts.

Proof:

$$R/I=\{r+I\;|\;r\in R\}\\$$

• Addition: $[r_1]+[r_2]=[r_1+r_2]$.
• Multiplitcation: $[r_1][r_2]=[r_1][r_2]$.

where $[r]=\{r'\;|\;r+I=r'+I\}$.

Proof: If $I=\{0\}$, then $I=(0)$. Otherwise, let $m(x)\in I$ be the polynomial of least degree in $I$, then we prove $I=(m(x))$.

$(m(x))\subseteq I$ is obvious since $m(x)\in I$. For the other direction, for $f(x)\in I$, we have:

$$f(x)=q(x)m(x)+r(x)$$

by polynomial modulo, where $r(x)=0$ or $\deg r(x)<\deg m(x)$. Now $r(x)=f(x)-q(x)m(x)\in I$, if $r(x)\neq 0$ then we have contradicted $m(x)$ having the smallest degree. So $f(x)=q(x)m(x)\in(m(x))$.

• Example: for $p\geq 2$, then the ideal $(p)$ in $\mathbb{Z}$ is a prime ideal if and only if $p$ is prime.

  If $ab\in(p)$, then $p\;|\;ab$, so $p\;|\;a$ or $p\;|\;b$.
  Otherwise, if $p=ab$ is a factorization, then $a,b\notin p\mathbb{Z}$.

Proof:

Assume $p(x)$ is a prime ideal. If $p(x)$ is not irreducible, i.e. there is a factorization $p(x)=a(x)b(x)$ and $\partial(a),\partial(b)<\partial(p)$. Since every non-zero element in $(p(x))$ should have degree $\geq\partial(p)$, so contradicts.

On the other direction, If $p(x)$ is irreducible and $ab\in(p)$, then $p\;|\;ab$, then $p\;|\;a$ or $p\;|\;b$, thus $a\in (p)$ or $b\in (p)$. And we need to prove $(p)$ is a proper ideal. If $R=(p)$, then $1\in R=(p)$, so we have $1=p(x)f(x)$, which is impossible.

Where the isomorphism is: $a\rightarrow a+I$. And the root is $\theta(x) \rightarrow t(x)+I$, $t(x)=x$.

$$\begin{aligned} p(x)&=a_0+a_1x+...+a_nx^n\\ p(\theta(x))&=(a_0+I)+(a_1+I)(t(x)+I)+...+(a_n+I)(t(x)+I)^n\\ &=(a_0+I)+(a_1t(x) + I)+...+(a_nt(x)^n+I)\\ &=(a_0+a_1t(x)+...+a_nt(x)^n)+I\\ &=(a_0+a_1x+...+a_nx^n)+I\\ &=p(x)+I\\ &=I \end{aligned}$$

Since $I=(p(x))$, so $p(\theta(x))=I=0+I$. So in $F[x]/(p(x))$, we have a root: $t(x)+I$.

Notice, $F\cong F'\subseteq F[x]/(p(x))$ is the isomorphism from “numbers” to “a set of polynomials”. And once we have a root of $p(x)$ in $F[x]/(p(x))$, it doesn’t mean that there exists a root for $p(x)$ in $F$. If and only if there exists $a\in F$ such that $t-t'\in I$, $t(x)=x,t'(x)=a$, then $a$ is root for $p(x)$ in $F$.

Example: $\mathbb{Q}[x]/(x^2+1)\cong\mathbb{C}$. Where $\mathbb{Q}[x]/(x^2+1)$ contains a root for $f(x)=x^2+1$.

Proof:
If $\partial(f)=1$, then we choose $E=F$ and $f(x)=f(x)\in E[x]$ which is linear.

If $\partial(f)>1$, without loss of generality, we write $f(x)=p(x)g(x)$ where $p(x)$ is irreducible. Let $E=F[x]/(p(x))$, then there exists a root $\theta(x)$ for $p(x)$ in $E$. So in $E$, we have:

$$f(x) =(x-\theta(x))h(x)g(x)+I$$

So by induction, we can split $h(x)g(x)$.

• Example: $f(x)=x^2+1$, then we compute the splitting field of $f(x)$ over $\mathbb{Z}_2$.

  • We factorize $f(x)$ into irreducible ones, $f(x)=x^2+1$.

  • Compute $\mathbb{Z}_2[x]/(x^2+1)$. Here is a trick, let $I=(x^2+1)$, given $f(x),g(x)\in\mathbb{Z}_2[x]$, then $f+I=g+I$ if and only if $f-g\in I$, i.e. $x^2+1\;|\;f-g$.

    So in $\mathbb{Z}_2[x]/(x^2+1)$, there exists no polynomials with degree $\geq 3$. Because:

    $$x^3=x^3+2x=x(x^2+1)+x\equiv x$$

    So the potential items in $\mathbb{Z}_2[x]/(x^2+1)$ are:

    $$0,1,x,x+1,x^2,x^2+1,x^2+x,x^2+x+1$$

    And we have:

    $$0+I= (x^2+1)+I\\ 1+I=x^2+I\\ x+I=x^2+x+1+I\\ x+1+I=x^2+x+I\\$$

    So $\mathbb{Z}_2[x]/(x^2+1)=\{0+I,1+I,x+I,x+1+I\}=\{[0],[1],[x],[x+1]\}$. And

    $$f([0])=[1]\cdot[0]^2+[1]=[1]\\ f([1])=[1]\cdot[1]^3+[1]=[0]\\ f([x])=[1]\cdot[x]^2+[1]=[0]\\ f([x+1])=[1]\cdot[x+1]^2+[1]=[1]$$

    So in $\mathbb{Z}_2[x]/(x^2+1)$, we have $f([x])=0$, which correspond to the $\theta(x)=x+I=[x]$ in the proof.
    Then:

    $$\begin{aligned} \because f(t) &=[1]t^2+[1]\\ &=([1]t-[x])g(t)\\ \therefore g(t)&=([1]t^2+[1])/([1]t-[x])\\ &=[1]t+[x]\\ \therefore f(t)&=([1]t-[x])([1]t+[x])\\ &=([1]t+[x])^2\\ &=([1]t-[x])^2 \end{aligned}$$

    where $t=[x]$ is a root for $f(t)$ in $\mathbb{Z}_2[x]/(x^2+1)$ and satisfies $t^2+[1]=0$. $f(t)$ splits over $\mathbb{Z}_2/(x^2+1)$.

Proof: let $g(x)=x^{p^n}-x$, Then by Kronecker theorem, there exists a field $E$ containing $\mathbb{Z}_p$ over which $g(x)$ splits, let’s construct $F=\{\alpha\in E\;|\;g(\alpha)=0\}$. Since $g(x)$ splits, so it has $\partial(g)=p^n$ roots. And we need to prove that it has no repeat roots. We have:

$$\begin{aligned} g(x)&=x^{p^n}-x\\ &=x(x^{p^n-1}-1)\\ E&=\mathbb{Z}_p[x]/(x)\\ &=\{[0],[1],...,[p-1]\}\\ &\cong \mathbb{Z}_p\\ g(x)&=(x-0)(x^{p^n-1}-1)\\ E&=\mathbb{Z}_p/(x^{p^n-1}-1)\\ \therefore g'(x)&=p^nx^{p^n-1}-1\\ &=-1\in\mathbb{Z}_p\\ \therefore \gcd(g,g')&=1\in\mathbb{Z}_p[x]/(x^{p^n-1}-1)\\ \end{aligned}$$

And if $\gcd(f,f')=1$ in some field, then $f$ has no repetitive roots in the field.

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Example: Let $q=2,n=2$.

$$\begin{aligned} g(t)&=t^4-t=t(t^3-1)\\ &=t(t-1)(t^2+t+1)\\ \mathbb{Z}_2[x]/(x)&=\mathbb{Z}_2\\ \mathbb{Z}_2[x]/(x-1)&=\mathbb{Z}_2\\ \mathbb{Z}_2[x]/(x^2+x+1)&=\{[ax+ b]:a,b\in\mathbb{Z}_2\}\\ g(t)&=(t-[0])(t-[1])(t^2+[1]t+[1])\\ &=(t-[0])(t-[1])(t-[x])(t+[x+1]) \end{aligned}$$

So there are four roots: $[0],[1],[x],[x+1]$ the field containing $4$ elements is $\mathbb{Z}_2[x]/(x^2+x+1)$. When $q=3$ the case is more complicated since $\mathbb{Z}_3[x]/(x^2+1)\not\cong\mathbb{Z}_9$, and the coefficients would be ugly as $[[[1]x^2+[2]x]x]$ something.

# Galois Group