# # Symmetry

Definition. A linear transformation $\sigma:\mathbb{R}^2\rightarrow\mathbb{R}^2$ is called orthogonal if it is distance preserving; that is, if $|U-V|$ denotes the distance between points $U$ and $V$, then

$|\sigma(U)-\sigma(V)|=|U-V|$

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

Definition.: Given a figure $F$ in the plane, its symmetry group $\Sigma(F)$ is the family of all orthogonal transformations $\sigma:\mathbb{R}^2\rightarrow\mathbb{R}^2$ for which

$\sigma(F)=F$

The elements of $\Sigma(F)$ are called symmetries.

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

Definition. A cummutative ring with $1$ is a set $R$ equipped with two binary operations, addtion: $(r,r')\rightarrow r+r'$ and multiplication: $(r,r')\rightarrow rr'$ such that:

• $R$ is an abelian group under addition.

• multiplication is commutative and associative.

• there is an element $1\in R$ with $1\neq 0$ and

$\forall r\in R,1r=r$

• the distributive law holds:

$\forall r,s,t\in R,r(s+t)=rs+rt$

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

Definition. A ring $R$ is a domain (or integral domain) if the product of any two nonzero elements in $R$ is itself nonzero.

Theorem: A ring $R$ is a domain if and only if it satisifies the cancellation law:

$\forall r,a,b,ra=rb\wedge r\neq 0\Rightarrow a=b$

Theorem. $\mathbb{Z}_n$ is a domain if and only if $n$ is prime.

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

Definition. An element $u\in R$ is a unit if there exists $v\in R$ with $uv=1$. （乘法可逆）

Definition. A field is a ring $R$ in which every nonzero $r\in R$ is a unit.

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

Theorem: For every domain $R$, there is a field $\textnormal{Frac}(R)$ containing $R$ as a subring. Moreover, every element $q\in\textnormal{Frac}(R)$ has a factorization:

$q=ab^{-1}$

with $a,b\in R,b\neq 0$.

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

Definition. If $R$ and $S$ are rings, then a function $\varphi:R\rightarrow S$ is a ring homomorphism if for all $r,r'\in R$:

$\varphi(r+r') =\varphi(r)+\varphi(r')\\ \varphi(rr')=\varphi(r)\varphi(r')\\ \varphi(1)=1$

A ring homomorphism is an isomorphism if it is a bijection, we writes $R\cong S$.

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}

Definition. The kernel of a ring map is:

$\text{ker}\varphi=\{r\in R:\varphi(r)=0\}$

Definition. An ideal in a ring $R$ is a subset $I$ containing 0 such that:

• $a,b\in I\Rightarrow a-b\in I$.
• $a\in I,r\in R\Rightarrow ra\in I$.

An ideal $I$ in a ring $R$ is a proper ideal if $I\neq R$.

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 $(a)$.

Theorem. If $\varphi:R\rightarrow S$ is a ring homomorphism, then $\text{ker}\varphi$ is a proper ideal in $R$. Moreover, $\varphi$ is an injection if and only if $\text{ker}\varphi=\{0\}$.

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.

Theorem: Let $I$ be a proper ideal in a ring $R$. Then the additive abelian group $R/I$ can be equipped with a multiplication which makes it a ring and which makes the natural map $\pi:R\rightarrow R/I$ a surjective ring homomorphism:

$\pi(r)=r+I$

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

Theorem: (First Isomorphism Theorem) If $\varphi:R\rightarrow S$ is a ring homomorphism with $\text{ker}\varphi=I$, then there is an isomorphism $R/I\rightarrow\text{im}\varphi$ given by $[r]\rightarrow \varphi(r)$.

Theorem: If $F$ is a field, then every ideal in $F[x]$ is a principal ideal.

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 $f(x)=q(x)m(x)\in(m(x))$.

Definition. A ring $R$ is called a principal ideal domain if every ideal in $R$ is principal.

Definition. Let $F$ be a field. A nonzero polynomial $p(x)\in F[x]$ is irreducible over $F$ if $\partial(p)\geq 1$ and there is no factorization $p(x)=f(x)g(x)$ in $F[x]$ with $\partial (f)<\partial(p)$ and $\partial(g)<\partial(p)$.

where $\partial(f)$ means the degree of $f$.

Definition. An ideal $I$ in a ring $R$ is called a prime ideal if it is a proper ideal and $ab\in I\Rightarrow a\in I$ or $b\in I$.

• 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 $a,b\notin p\mathbb{Z}$.

Theorem: If $F$ is a field, then a nonzero polynomial $p(x)\in F[x]$ is irreducible if and only if $(p(x))$ is a prime ideal.

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.

Theorem: A proper ideal $I$ in $R$ is a prime ideal if and only if $R/I$ is a domain.

Definition: An ideal $I$ in a ring $R$ is a maximal ideal if it is a proper ideal and there is no ideal $J$ with $I\subsetneq J\subsetneq R$.

Theorem: A proper ideal $I$ in a ring $R$ is a maximal ideal if and only if $R/I$ is a field.

Theorem: If $R$ is a principal ideal domain, then every nonzero prime ideal $I$ is a maximal ideal.

Definition. A polynomial $f(x)\in F[x]$ splits over $F$ if it is a product of linear factors in $F[x]$. Of course, $f(x)$ splits over $F$ if and only if $F$ contains all the roots of $f(x)$, i.e.:

$f(x)=(x-a_1)(x-a_2)...(x-a_n),a_i\in F$

Theorem: If $F$ is a field and $p(x)\in F[x]$ is irreducible, then the quotient ring $F[x]/(p(x))$ is a field containing (an isomorphism copy of) $F$ and a root of $p$.

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 $f(x)=x^2+1$.

Theorem(Kronecker) Let $f(x)\in F[x]$ where $F$ is a field. There exists a field $E$ containing $F$ over which $f(x)$ splits.

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 $\mathbb{Z}_2/(x^2+1)$.

Definition. A field has character 0 if its prime field is isomorphic to $\mathbb{Q}$, it has character p if it’s isomorphic to $\mathbb{Z}_p$.

Theorem(Galois): For every prime $p$ and every positive integer $n$, there exists a field having exactly $p^n$ elements.

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.

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

Definition. If $E$ is a field, then an automorphism of $E$ is an isomorphism of $E$ with itself. If $E/F$ is a field extension, then an automorphism $\sigma$ of $E$ fixes $F$ pointwise if $\forall c\in F,\sigma(c)=c$.

And we define the Galois Group as:

$\text{Gal}(E/F)=\{\textnormal{automorphisms }\sigma \text{ of } E \text{ fixing } F\text{ pointwise}\}$

Theorem. If $f(x)\in F[x]$ has $n$ distinct roots in its splitting field $E$, then $\text{Gal}(E/F)$ is isomorphic to a subgroup of the symmetric group $S_n$.

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