# Symmetry

Definition. A linear transformation σ:R2R2\sigma:\mathbb{R}^2\rightarrow\mathbb{R}^2 is called orthogonal if it is distance preserving; that is, if UV|U-V| denotes the distance between points UU and VV, then


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

Definition.: Given a figure FF in the plane, its symmetry group Σ(F)\Sigma(F) is the family of all orthogonal transformations σ:R2R2\sigma:\mathbb{R}^2\rightarrow\mathbb{R}^2 for which


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

The wonderful idea of Galois was to associate to each polynomial f(x)f(x) a group, nowadays called its Galois group, whose properties reflect the behavior of f(x)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 11 is a set RR equipped with two binary operations, addtion: (r,r)r+r(r,r')\rightarrow r+r' and multiplication: (r,r)rr(r,r')\rightarrow rr' such that:

  • RR is an abelian group under addition.

  • multiplication is commutative and associative.

  • there is an element 1R1\in R with 101\neq 0 and

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

  • the distributive law holds:

    r,s,tR,r(s+t)=rs+rt\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 RR is a domain (or integral domain) if the product of any two nonzero elements in RR is itself nonzero.

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

r,a,b,ra=rbr0a=b\forall r,a,b,ra=rb\wedge r\neq 0\Rightarrow a=b

Theorem. Zn\mathbb{Z}_n is a domain if and only if nn is prime.

Proof: [a][b]=0[ab]=0ab0modp[a][b]=0\Rightarrow [ab]=0\Rightarrow ab\equiv 0\mod p

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

Definition. A field is a ring RR in which every nonzero rRr\in R is a unit.

  • If pp is prime, then Zp\mathbb{Z}_p is a field.

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


with a,bR,b0a,b\in R,b\neq 0.

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

Frac(R)={a/b    a,bR,b0}\textnormal{Frac}(R)=\{a/b\;|\;a,b\in R,b\neq 0\}\\

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

where a/b=ab1a/b=ab^{-1}.

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

# Homomorphism and Ideals

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

φ(r+r)=φ(r)+φ(r)φ(rr)=φ(r)φ(r)φ(1)=1\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 RSR\cong S.

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

φ(a)=φ(a+0)=φ(a)+φ(0)0=φ(0)\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:

kerφ={rR:φ(r)=0}\text{ker}\varphi=\{r\in R:\varphi(r)=0\}

Definition. An ideal in a ring RR is a subset II containing 0 such that:

  • a,bIabIa,b\in I\Rightarrow a-b\in I.
  • aI,rRraIa\in I,r\in R\Rightarrow ra\in I.

An ideal II in a ring RR is a proper ideal if IRI\neq R.

An ideal is a sub additive group of the ring.

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

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

Proof: kerφ\text{ker}\varphi contains 00 is self-evident, and:

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

so rakerφra\in\text{ker}\varphi. and φ(ab)=φ(a)φ(b)=00=0\varphi(a-b)=\varphi(a)-\varphi(b)=0-0=0, so φ(ab)kerφ\varphi(a-b)\in\text{ker}\varphi.

If φ\varphi is an injection, then for r0,φ(r)φ(0)=0r\neq 0,\varphi(r)\neq\varphi(0)=0, so kerφ={0}\text{ker}\varphi=\{0\}. Conversely, if kerφ={0}\text{ker}\varphi=\{0\}, and exists rr,φ(r)=φ(r)r\neq r',\varphi(r)=\varphi(r'), then φ(rr)=φ(r)φ(r)=0\varphi(r-r')=\varphi(r)-\varphi(r')=0, so 0rrkerφ0\neq r-r'\in\text{ker}\varphi, contradicts.

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



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

  • Addition: [r1]+[r2]=[r1+r2][r_1]+[r_2]=[r_1+r_2].
  • Multiplitcation: [r1][r2]=[r1][r2][r_1][r_2]=[r_1][r_2].

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

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

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

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

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


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

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

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

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

Definition. An ideal II in a ring RR is called a prime ideal if it is a proper ideal and abIaIab\in I\Rightarrow a\in I or bIb\in I.

  • Example: for p2p\geq 2, then the ideal (p)(p) in Z\mathbb{Z} is a prime ideal if and only if pp is prime.

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

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


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

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

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

Definition: An ideal II in a ring RR is a maximal ideal if it is a proper ideal and there is no ideal JJ with IJRI\subsetneq J\subsetneq R.

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

Theorem: If RR is a principal ideal domain, then every nonzero prime ideal II is a maximal ideal.

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

f(x)=(xa1)(xa2)...(xan),aiFf(x)=(x-a_1)(x-a_2)...(x-a_n),a_i\in F

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

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

p(x)=a0+a1x+...+anxnp(θ(x))=(a0+I)+(a1+I)(t(x)+I)+...+(an+I)(t(x)+I)n=(a0+I)+(a1t(x)+I)+...+(ant(x)n+I)=(a0+a1t(x)+...+ant(x)n)+I=(a0+a1x+...+anxn)+I=p(x)+I=I\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))I=(p(x)), so p(θ(x))=I=0+Ip(\theta(x))=I=0+I. So in F[x]/(p(x))F[x]/(p(x)), we have a root: t(x)+It(x)+I.

Notice, FFF[x]/(p(x))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)p(x) in F[x]/(p(x))F[x]/(p(x)), it doesn’t mean that there exists a root for p(x)p(x) in FF. If and only if there exists aFa\in F such that ttIt-t'\in I, t(x)=x,t(x)=at(x)=x,t'(x)=a, then aa is root for p(x)p(x) in FF.

Example: Q[x]/(x2+1)C\mathbb{Q}[x]/(x^2+1)\cong\mathbb{C}. Where Q[x]/(x2+1)\mathbb{Q}[x]/(x^2+1) contains a root for f(x)=x2+1f(x)=x^2+1.

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

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

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

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

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

  • Example: f(x)=x2+1f(x)=x^2+1, then we compute the splitting field of f(x)f(x) over Z2\mathbb{Z}_2.

    • We factorize f(x)f(x) into irreducible ones, f(x)=x2+1f(x)=x^2+1.

    • Compute Z2[x]/(x2+1)\mathbb{Z}_2[x]/(x^2+1). Here is a trick, let I=(x2+1)I=(x^2+1), given f(x),g(x)Z2[x]f(x),g(x)\in\mathbb{Z}_2[x], then f+I=g+If+I=g+I if and only if fgIf-g\in I, i.e. x2+1    fgx^2+1\;|\;f-g.

      So in Z2[x]/(x2+1)\mathbb{Z}_2[x]/(x^2+1), there exists no polynomials with degree 3\geq 3. Because:

      x3=x3+2x=x(x2+1)+xxx^3=x^3+2x=x(x^2+1)+x\equiv x

      So the potential items in Z2[x]/(x2+1)\mathbb{Z}_2[x]/(x^2+1) are:


      And we have:

      0+I=(x2+1)+I1+I=x2+Ix+I=x2+x+1+Ix+1+I=x2+x+I0+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 Z2[x]/(x2+1)={0+I,1+I,x+I,x+1+I}={[0],[1],[x],[x+1]}\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][0]2+[1]=[1]f([1])=[1][1]3+[1]=[0]f([x])=[1][x]2+[1]=[0]f([x+1])=[1][x+1]2+[1]=[1]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 Z2[x]/(x2+1)\mathbb{Z}_2[x]/(x^2+1), we have f([x])=0f([x])=0, which correspond to the θ(x)=x+I=[x]\theta(x)=x+I=[x] in the proof.

      f(t)=[1]t2+[1]=([1]t[x])g(t)g(t)=([1]t2+[1])/([1]t[x])=[1]t+[x]f(t)=([1]t[x])([1]t+[x])=([1]t+[x])2=([1]t[x])2\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]t=[x] is a root for f(t)f(t) in Z2[x]/(x2+1)\mathbb{Z}_2[x]/(x^2+1) and satisfies t2+[1]=0t^2+[1]=0. f(t)f(t) splits over Z2/(x2+1)\mathbb{Z}_2/(x^2+1).

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

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

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

g(x)=xpnx=x(xpn11)E=Zp[x]/(x)={[0],[1],...,[p1]}Zpg(x)=(x0)(xpn11)E=Zp/(xpn11)g(x)=pnxpn11=1Zpgcd(g,g)=1Zp[x]/(xpn11)\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\gcd(f,f')=1 in some field, then ff has no repetitive roots in the field.

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

g(t)=t4t=t(t31)=t(t1)(t2+t+1)Z2[x]/(x)=Z2Z2[x]/(x1)=Z2Z2[x]/(x2+x+1)={[ax+b]:a,bZ2}g(t)=(t[0])(t[1])(t2+[1]t+[1])=(t[0])(t[1])(t[x])(t+[x+1])\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][0],[1],[x],[x+1] the field containing 44 elements is Z2[x]/(x2+x+1)\mathbb{Z}_2[x]/(x^2+x+1). When q=3q=3 the case is more complicated since Z3[x]/(x2+1)≇Z9\mathbb{Z}_3[x]/(x^2+1)\not\cong\mathbb{Z}_9, and the coefficients would be ugly as [[[1]x2+[2]x]x][[[1]x^2+[2]x]x] something.

# Galois Group

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

And we define the Galois Group as:

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

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

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