\[ \DeclareMathOperator{\Q}{Q} \DeclareMathOperator{\Aut}{Aut} \]

Nepal Algebra Project (NAP) / नेपाल बीज-गणित परियोजना

COURSE YEAR 2017

Module - IV : July 02, 2017 – July 14, 2017, René Schoof and Laura Geatti
Topics: Computing Galois Groups, When is Gf contained in An? When is Gf transitive? Polynomials of degree at most three, Quartic poly-nomials, Examples of polynomials with Sp as Galois group over Q, Finite fields, Computing Galois groups over Q
[Homeworks]
Monday, July 3, 2017 (also download from here)

  • Construction of finite fields:
  • For any prime $p$ one has $\mathbf{F}_p[x]/(f)$, with $f$ irreducible in $\mathbf{F}_p[x]$.
  • Proposition $1$.
    1. the cardinality of a finite field $\mathbf{K}$ of characteristic $p$ is $q=p^n$.
    2. $\mathbf{K}$ contains $\mathbf{F}_p$
    3. the additive group $(\mathbf{K},+)$ is isomorphic to $(\mathbf{Z}/p\mathbf{Z} $ $ \times \ldots \times $ $\mathbf{Z}/p\mathbf{Z},+)$
    4. the multiplicative group $(\mathbf{K}^*,\cdot )$ of a finite field is cyclic.
  • Theorem $2$. For every prime power $q=p^n$ there exists a field with $q$ elements. It is of the form $\mathbf{F}[x]/(f)$ with $f$ an irreducible polynomial in $\mathbf{F}_p[X]$ of degree $n$.
  • Theorem $3$. Every finite field of $q$ elements is a splitting field of $x^q-x$ over $\mathbf{F}_p$. Therefore all finite fields with $q$ elements are isomorphic. Notation $\mathbf{F}_q$.
  • Examples. $\mathbf{F}_4$, $\mathbf{F}_5[\sqrt 2]$, $\mathbf{F}_9=$ $\mathbf{F}_3[i]=$ $\mathbf{F}_3[x]/(x^2+1)=$ $\mathbf{F}_3[i+1]$
  • Theorem $4$. $Aut(\mathbf{F}_q) =\langle \phi\rangle$, where $\phi$ denotes the Frobenius automorphism $\phi(x)=x^p$.

Tuesday, July 4, 2017 (also download from here)

  • Subfields.
  • Lemma $1$. Let $p$ be a prime. Let $f\in\mathbf{F}_p[x]$ be an irreducible polynomial of degree $n$. Let $\alpha$ be a zero of $f$. Then
    1. $\alpha^{p^n}=\alpha$;
    2. $n$ is the smallest positive integer for which (a) holds.
  • Proposition $2$. Let $f\in\mathbf{F}_p[x]$ be an irreducible polynomial of degree $n$. Let $\alpha$ be a zero of $f$. Then
    $f(x)=$ $(x-\alpha)$ $(x-\alpha^p)$ $\ldots$ $(x-\alpha^{p^{n-1}}).$
  • Corollary $3$. Every finite extension of a finite field is a Galois extension.
  • Theorem $4$. $Aut(\mathbf{F}_{p^n}) $ is a cyclic group isomorphic to $\mathbf{Z}/n\mathbf{Z}$, generated by the Frobenius automorphism $\phi$.
  • Proposition $5$.
    1. If $\mathbf{K}$ is a subfield of $\mathbf{F}_{p^n}$, then the cardinality of $\mathbf{K}$ is equal to $p^d$, for some divisor $d$ of $n$.
    2. For every divisor $d$ of $n$, there exists a unique subfield of cardinality $p^d$.
  • The Galois correspondence in the case of finite fields:
    Fix $p$ prime and the finite field $ \mathbf{F}_{p^n}$.
    $Aut( \mathbf{F}_{p^n})= $ $\langle \phi\rangle= $ $Gal(\mathbf{F}_{p^n}/\mathbf{F}_p)\cong$ $\mathbf{Z}/n\mathbf{Z}$.
    For every $d$ divisor of $n$, there is a unique subfield $\mathbf{F}_{p^d}\subset \mathbf{F}_{p^n}$.
    For every $d$ divisor of $n$, there is a unique subgroup $G_d$ of $Gal(\mathbf{F}_{p^n}/\mathbf{F}_p)$ of index $d$, namely the subgroup generated by $ \phi^d$.
    One has $ \mathbf{F}_{p^d} = $ $\{x\in \mathbf{F}_{p^n} $ $~|~ g(x)=x$, $\forall g\in G_d\}$.
    Conversely $G_d= \{$ $ g\in Gal(\mathbf{F}_{p^n}/\mathbf{F}_p)$ $~|~g(x)=x$, $~\forall x\in \mathbf{F}_{p^d} \}$.
    Example: $\mathbf{F}_{3^4}$.
  • Excercise $1$.
    1. Find an irreducible polynomial $f$ of degree $2$ in $\mathbf{F}_3[x]$. Then $\mathbf{F}_9=\mathbf{F}_3[x]/(f)$.
    2. Which elements of $\mathbf{F}_9$ are generators of its multiplicative group $\mathbf{F}_9^*$?
    3. Which elements of $ \mathbf{F}_9$ have square roots in $\mathbf{F}_9$?
    4. Prove that the product of all elements of $\mathbf{F}_9^*$ is 2.
    5. Show that the additive group of $\mathbf{F}_9$ is not cyclic.
  • Excercise $2$.
  • Draw the Hasse diagrams of the subfields of each $\mathbf{F}_{2^k}$ for $k=1,..,6.$.

Solution of Lecture 2 download from here
Thursday, July 6, 2017 (also download from here)

  • Proposition $1$ The group $G_f$ permutes the roots of $f$:
    if $\sigma\in G_f$ and $\alpha_i \in Zeros(f)$, then $\sigma(\alpha_i)= \alpha_j$ $\in Zeros(f).$
    There is a homomorphism $\Theta\colon G_f\to S_n,$ where $S_n$ is the permutation group of $n$ elements. The homomorphism $\Theta$ is injective. Hence $\#G_f$ divides $n!$.
  • Proposition $f\in\mathbf{K}[x]$ separable. Then $G_f\cong H\subset S_n$, with $H$ transitive on $\{1,2,\ldots,n\}$, if and only if $f$ is irreducible over $\mathbf{K}$.
  • Criterion. $f\in\mathbf{K}[x]$ separable, $char(\mathbf{K})\not=2$. Then $G_f\cong A_n$ if and only if $Disc(f)$ is a square in $\mathbf{K}$, where $Disc(f):=$ $\prod_{i < j}$ $(\alpha_i-\alpha_j)^2\in\mathbf{K}$ (it is $G_f$-invariant).
  • Example $1$. $\mathbf{K}=\mathbf{Q}$, $f(x)=$ $x^4-4=$ $(x^2+2)$ $(x^2-2)$;
    $\mathbf{K}_f=\mathbf{Q} (\sqrt 2,i\sqrt 2)$, $G_f\cong \mathbf{Z}/2\mathbf{Z}$$\times \mathbf{Z}/2\mathbf{Z}$.
  • Example $2$. $\mathbf{K}=\mathbf{Q}$, $f(x)=x^4-2$;
    $\mathbf{K}_f=\mathbf{Q}({\root 4 \of 2},i{\root 4 \of 2})$, $G_f\cong D_4$.
  • Example $3$. (degree $2$ case). $f\in\mathbf{K}[x]$ separable of degree 2; $G_f\subset S_2=$$\{id,(12)\}$.
    1. $G_f=id$ $\Leftrightarrow[\mathbf{K}_f:\mathbf{K}]$ $=1$ $\Leftrightarrow \mathbf{K}_f=K$ if and only if $f$ factors in $\mathbf{K}$.
    2. $G_f=S_2 \Leftrightarrow$ $[\mathbf{K}_f:\mathbf{K}]$ $=2$ if and only if $K_f$ is a quadratic extension of $\mathbf{K}$ if and only if $f$ is irreducible over $\mathbf{K}.$
  • Example $4$. (degree $3$ case). $f\in\mathbf{K}[x]$ separable of degree 3; $G_f\subset S_3$ $=$ $\{$$id,(12)$,$(13,(23)$,$(123),(132)$$\}$;
    possible subgroups (up to conjugation): id, $S_3$, $\{id,(12)\}$, $\{ id,(123),(132)\}$.
    1. $G_f=id$ $\Leftrightarrow$ $[\mathbf{K}_f:\mathbf{K}]=1$ if and only if $f$ factors in $\mathbf{K}$.
    2. $G_f=S_3$ $\Rightarrow f$ is irreducible.
    3. $G_f$ has order $2 \Rightarrow f$ is a product of a linear and a quadratic polynomial in $\mathbf{K}[X]$.
    4. $G_f$ has order $3$ and hence $G_f=A_3 \Rightarrow f$ is irreducible.
    If $char(\mathbf{K})\not=2$, we can distinguish between cases (b) and (d) using the discriminant. The Galois group $G_f$ is contained in $A_3$ if and only if $Disc(f)$ is a square in $\mathbf{K}$.
  • We have done few exercises from the file Excercises $1$ ."

    • Solution of Lecture 3: download from here
    Table of Field $\mathbf{F}_{16}:$ download from here
Monday, July 10, 2017 (also download from here)

  • $\mathbf{K}$ a field, $f\in \mathbf{K}[x]$ a separable monic polynomial, $\mathbf{K}_f=$$\mathbf{K}[\alpha_1,\ldots,\alpha_n]$ the splitting field of $f$, $\{\alpha_i\}_i$ zeros of $f$, $G_f:=$$Gal(\mathbf{K}_f/\mathbf{K})$$\subset S_n$ the Galois group of $f$.
  • Criterion. Assume $char(\mathbf{K})\not=2$. Then $G_f\subset A_n$ $\Leftrightarrow$ $Disc(f)$ is a square in $\mathbf{K}$.
  • Consequence. $f\in \mathbf{K}[x]$, separable monic polynomial of degree 3, irreducible over $\mathbf{K}$. Then $G_f=A_3$, if $Disc(f)$ is a square in $\mathbf{K}$, and $G_f=S_3$, if $Disc(f)$ is not a square in $\mathbf{K}$.
  • The degree 4 case.
    $f\in \mathbf{K}[x]$, separable monic polynomial of degree 4.
    1. If $f$ is divisible by a degree $1$ factor, we are back in the degree $3$ case.
    2. If $f$ factors as the product of two irreducible polynomials $g$, $h$ of degree $2$, then either $\mathbf{K}_g=\mathbf{K}_h$$=\mathbf{K}_f$ and $G_f\cong C_2$, or $\mathbf{K}_g\not=\mathbf{K}_h$ , $\mathbf{K}_f=\mathbf{K}_g\mathbf{K}_h$ and $G_f\cong C_2\times C_2$.
    3. If $f$ is irreducible, then $G_f$ is isomorphic to a transitive subgroup of $S_4$. These are $S_4$, $A_4$, $D_4$, $V_4$, $C_4$.
  • The group $S_4$ and its transitive subgroups in detail.
  • The cubic resolvent $g$ of $f$: its zeros $\alpha$, $\beta$, $\gamma$ are fixed by $G_f\cap V_4$, hence $g\in\mathbf{K}[x]$.
  • Excercises. Computation of the Galois group of the following polynomials:
    1. $x^2-101$;
    2. $x^3-5x^2$$+6$;
    3. $x^3-5x$$-5$;
    4. $x^3-3x$$+1$;
    5. $x^3-1$;
    6. $x^3-3$;
    7. $x^3-2x^2$$+3x+5$;

Resultants (download from here)
Tuesday, July 11, 2017 (also download from here)

  • $f\in \mathbf{K}[x]$, separable monic polynomial of degree $4$
    $g$ the cubic resolvent of $f$.
    $f$ separable $\Rightarrow$ $g$ separable;
    $disc(f)$$=$$disc(g)$.
  • Proposition. $\mathbf{K}[\alpha,\beta,\gamma]$ is the fixed field of $G_f\cap V_4$.
  • Classification of the Galois groups of irreducible separable monic polynomials of degree 4, by means of the cubic resolvent:
    1. $g$ irred., $disc(g)$ not square, $G_g=S_3 $, then $G_f=S_4$;
    2. $g$ irred., $disc(g)$ square, $G_g=A_3 $, then $G_f=A_4$;
    3. $g=$$h_1\cdot h_2$, with $deg(h_i)$$=i$, $disc(g)$ not square, $G_g=C_2 $, then $G_f=D_4$$\Leftrightarrow f$ irred. in $\mathbf{K}_g$;
    4. $g=$$h_1\cdot h_2$, with $deg(h_i)$$=i$, $disc(g)$ not square, $G_g=C_2 $, then $G_f=C_4$$\Leftrightarrow f$ red. in $\mathbf{K}_g$;
    5. $g $ completely red., $disc(g)$ not square, $G_g=id $, then $G_f=V_4$.
  • Construction of an irreducible polynomial of degree $p$, with Galois group $G_f=S_p$, for every prime $p$.
  • Lemma. Let $H$ be a subgroup of $S_p$. If $H$ contains a $2$-cycle and a $p$-cycle, then $H=S_p$.
  • We will construct a polynomial $f\in\mathbf{Q}[x]$, irreducible of degree $p$, with $p-2$ real roots and $2$ complex conjugate roots.
  • Excercises. Computation of the Galois group of the following irreducible polynomials of degree $4$:
    1. $x^4$$-x$$-1$ (type $S_4$);
    2. $x^4$$+2x$$+2$ (type $S_4$);
    3. $x^4$$+8x$$+12$ (type $A_4$);
    4. $x^4$$+36x$$+63$ (type $V_4$);
    5. $x^4$$-10x^2$$+5$ (type $C_4$);
    6. $x^4$$+3x$$+3$ (type $D_4$);

Thursday, July 13, 2017 (also download from here)

  • Construction of an irreducible polynomial of degree $p$, with Galois group $G_f=S_p$, for every prime~$p$.
  • A method to compute Galois groups $Gal(\mathbf{K}_f/\mathbf{Q}).$ By reducing $f$ modulo various primes $p$, one computes automorphisms $\sigma_p$ in the Galois group $G_f\subset S_n$ up to conjugacy in $G_f$. The cycle decomposition of $\sigma_p$ is determined by the factorization of the polynomial $f$ modulo $p$. This method can be used to show that the Galois group $G_f$ is large. To show that it is small, other method are needed (see Milne, p.55). The method was illustrated by a computer presentation using PARI/GP.

  • Excercises.
    1. Let $\mathbf{K}$ be a field of $char\not=2$. If $u,v\in\mathbf{K}^*$, then $\mathbf{K}(\sqrt u)=$$\mathbf{K}(\sqrt v)$ if and only if $u/v$ is a square in $\mathbf{K}^*$.
    2. Compute the Galois group of $\mathbf{Q}_f$, where $f(x)=$$X^4+5x+5$ (irreducible over $\mathbf{Q}$).
    3. Fix $\alpha=$$\sqrt{2-\sqrt3}$.
      1. Show that $\mathbf{Q}(\alpha)$ is a normal extension of $\mathbf{Q}$.
      2. Compute its Galois group.
    4. Let $f$ be the polynomial $x^4-2$. Compute the Galois group of $\mathbf{K}_f$, when $\mathbf{K}=\mathbf{Q}$, $\mathbf{R}$, $\mathbf{C}$, $\mathbf{Q}(\sqrt 2)$, $\mathbf{Q}(\root4\of2)$.

Solutions: preliminary , problem set 1 , problem set 2