--------              PART  1         Macaulay2   computer  code      ----------------------

-- Original file:   code_handout_001a_galois_theory_macaulay2_code_22mar18.txt


--  Macaulay2 computer algebra  online system 
--                         http://habanero.math.cornell.edu:3690/

--  Macaulay2 about  polynomials taken from 
--                       [Brzezinski, Galois Theory Through Exercises]

R = QQ [x]

factor  (x^4 +4)  -- answer:    factor  (x^4 +4)  

factor  (x^4 +64)  -- answer:    (x^2-4*x+8)*(x^2+4*x+8)

factor (x^3 -2)       -- answer:   (x^3 -2)

factor (x^4 +1 )       -- answer:   (x^4  +1 )

factor  (x^6 + 27)
      -- answer:    (x^2+3)*(x^2-3*x+3)*(x^2+3*x+3)


isPrime ideal (x^2+1)  -- answer  true  ,  hence  x^2+1 is ireducible  in QQ[x]

isPrime ideal (x^3+1)  -- answer  false,  hence  x^3+1 is not irreducible in Q[x]
    factor (x^3 +1)    --  answer:    (x+1)*(x^2-x+1)



R = ZZ/2 [x]  

factor  (x^7 + 1 )   --  answer:  (x+1)*(x^3+x+1)*(x^3+x^2+1)



R = ZZ/3 [x]  

factor  (x^3 + 2 )   --  answer:  (x-1)^3

factor  ( x^4 + x+ 2)  --   answer:      ( x^4 + x -1 )   



R = ZZ/5 [x] 

factor (x^4 + 2 )    --  answer:  (x^4+2)





---  Find all irreducible polynomails in  ZZ/2 [x]  up to degree 5

R = ZZ/2[x]

list1 = {x,x+1}

-- degree 2:    --  polynomials of the form   x^2 + a1*x + a0
list2 = {}  
for  a1 from 0 to 1 do 
  for a0  from 0 to 1  do
      if    isPrime ideal  (1_R*x^2 + a1*x + a0)  then 
                         list2 = list2 | {1_R*x^2 + a1*x + a0} 
list2   -- answer:   {x^2+x+1}


-- degree 3:    --  polynomials of the form   x^3 + a2*x^2 + a1*x + a0
list3 = {}  
for a2 from 0 to 1 do
  for  a1 from 0 to 1 do 
    for a0  from 0 to 1  do
       if    isPrime ideal  (1_R*x^3 + a2*x^2 + a1*x + a0)  then  (
                         list3 = list3 | {1_R*x^3 + a2*x^2 + a1*x + a0} )
list3   -- answer:    {x^3+x+1, x^3+x^2+1}


-- degree 5:    --  polynomials of the form   x^4 + a3*x^3 + a2*x^2 + a1*x + a0
list4 = {}  
for a3 from 0 to 1 do
 for a2 from 0 to 1 do
  for  a1 from 0 to 1 do 
    for a0  from 0 to 1  do
       if    isPrime ideal  (1_R*x^4 +a3*x^3+ a2*x^2 + a1*x + a0)  then  (
                         list4 = list4 | {1_R*x^4 + a3*x^3 +a2*x^2 + a1*x + a0} )
list4   -- answer:     {x^4+x+1, x^4+x^3+1, x^4+x^3+x^2+x+1}



-- degree 5:    --  polynomials of the form  x^5 +a4*x^4+ a3*x^3+a2*x^2+a1*x + a0
list5 = {}  
for a4 from 0 to 1 do
 for a3 from 0 to 1 do
  for a2 from 0 to 1 do
   for  a1 from 0 to 1 do 
     for a0  from 0 to 1  do
        if    isPrime ideal  (1_R*x^5+ a4*x^4 +a3*x^3+ a2*x^2 + a1*x + a0)  then  (
                         list5 = list5 | {1_R*x^5 + a4*x^4+ a3*x^3 +a2*x^2 + a1*x + a0} )
list5   -- answer:   {x^5+x^2+1, x^5+x^3+1, 
         --                 x^5+x^3+x^2+x+1, x^5+x^4+x^2+x+1, x^5+x^4+x^3+x+1, 
         --                 x^5+x^4+x^3+x^2+1}



































































--------     PART  2   Pari/GP  and         Macaulay2   computer  code      --------------

-- Original file:    code_handout_002c_galois_theory_23apr18.txt


Pari/GP: Computer Algebra program for Number Theory

Homepage: https://pari.math.u-bordeaux.fr/

Online platform: http://pari.math.u-bordeaux.fr/gp.html

Online Documentation: https://pari.math.u-bordeaux.fr/dochtml/html-stable/



\\ INDEX


\\ Task 4: keycode: 12145 Find minimal polynomial with Macaulay2 and Pari/GP

\\ Task 3: keycode: 98211 Factoring with Pari/GP over a field extension

\\ Task 2: keycode: 2536 Factoring with Pari/GP over the finite field ZZ/(p)

\\ Task 1: keycode: 8372 Factoring with Pari/GP over the rational numbers





\\ Task 4: keycode: 12145 Find minimal polynomial with Macaulay2 and Pari/GP







-- Example 1, 15apr18, keycode: 156641

-- M2 code related to the discussion at
--
-- https://math.stackexchange.com/questions/
-- 1779204/prove-or-disprove-that-sqrt32-sqrt1-sqrt2-is-aroot-of-a-polynomial
--
-- PROBLEM: Find a nonzero polynomial with integer coefficients that vanishes on 
--
-- u = A + B
--
-- where 
--
-- A = 3rd root of 2
--
-- B = square Root of ( 1+ C )
--
-- and C = square root of 2


-- Macaulay2 code: 

clearAll

R = QQ [A,B,C,u]

I = ideal ( A^3-2, C^2-2 , B^2 - (1+C), u - (A+ B) )

eliminate (I, { A,B,C} )
-- answer: ideal(u^12-6*u^10-8*u^9+9*u^8+28*u^6-144*u^5+
-- 63*u^4+96*u^3-78*u^2-168*u-41)



\\ Pari/gp computation for the minimal polynomial of algebraic integer: 

a=sqrtn(1+sqrt(7),3)
algdep(a,6)
\\ answer: x^6 - 2*x^3 - 6 [M2, p. 42]











\\ Task 3: keycode: 98211 Factoring with Pari/GP over a field extension


\\ Example 1 : Aim: Define t to be a root of pol1 defined below. Find the 
\\ decomposition of pol1 over the field QQ(t)

pol1 = X^4-4*X^3-20*X^2-8*X+4

polisirreducible(pol1) \\answer 1, hence true, so pol1 is irreducible in QQ[t]

print ( lift( factornf(pol1, t^4-4*t^3-20*t^2-8*t+4) ) )
\\answer [X - t, 1; X + (1/2*t^3 - 2*t^2 - 10*t - 4), 1; 
\\ X^2 + (-1/2*t^3 + 2*t^2 + 11*t)*X + 2, 1]



\\ Example 2:
\\ 
\\ Aim: Factor x^4 +1 over QQ[t]/(t^2+1)
pol1 = X^4 +1
polisirreducible(x^4+1) \\answer 1, hence true, so pol1 is irreducible in QQ[t]

print ( lift( factornf( X^4+1 , t^2+1) ) )
\\ answer: [X^2 - t, 1; X^2 + t, 1] 





\\ Task 2: keycode: 2536 Factoring with Pari/GP over the finite field ZZ/(p)

g = Mod ( x^7+1, 2)

polisirreducible (g) \\ answer 0, so g is not irreducible in ZZ/2

fa = factor (g)

print (fa)

\\ answer: 
\\ [ Mod(1, 2)*x + Mod(1, 2), 1;
\\ Mod(1, 2)*x^3 + Mod(1, 2)*x + Mod(1, 2), 1; 
\\ Mod(1,2)*x^3 + Mod(1, 2)*x^2 + Mod(1, 2), 1 ]

\\ Task 1: keycode: 8372 Factoring with Pari/GP over the rational numbers

\\ Example 1, Polynomial is x^4 + 4 \in QQ[x] keycode: 2831124

polisirreducible(x^4+4) 
\\ answer 0, it means false, so x^4+4 is not irreducible in QQ[x]

fa = factor (x^4+ 4) \\ fa has the structure of a 2 \times 2 matrix
print (fa)
\\ answer: [x^2 - 2*x + 2, 1; x^2 + 2*x + 2, 1]
\\ It means fa = (x^2-2*x+2) * (x^2+2*x+2) is the decomposition
\\ of x^4+4 in QQ[x] as product of irreducible polynomials.



\\ Example 2, Polynomial is x^2 - 3 \in QQ[x] keycode: 155252

polisirreducible(x^2-3) 
\\ answer 1, it means true, so x^2-3 is irreducible in QQ[x]

































--------     PART  3   Pari/GP  and         Macaulay2   computer  code      --------------

-- Original file:     code_handout_003b_galois_theory_18may18.pdf

\\  Pari/GP: Computer Algebra program for Number Theory


\\     Pari/GP   code for computing Galois groups of field extensions

\\ INDEX

\\ Task 5: keycode: 53134  Computing Galois groups etc. with Pari/GP


\\ 01apr18, 3 Important examples of computation of Galois groups with Pari/GP

nf=nfinit(a^4+1)
polgalois (x^4+1)
\\ answer: [4, 1, 1, "E(4) = 2[x]2"] , this means Z_2 \oplus Z_2
print ( lift(nffactor(nf,x^4+1)) )
\\ answer [x - a, 1; x + a, 1; x - a^3, 1; x + a^3, 1]



nf=nfinit(a^4+2) ;
polgalois (x^4+2)
\\answer: [8, -1, 1, "D(4)"] , this means the Dihedral group of order 8
print (lift(nffactor(nf,x^4+2)))
\\ answer : [x - a, 1; x + a, 1; x^2 + a^2, 1] 

nf=nfinit(a^4+2*a+3)
polgalois (x^4+2*x+3)
\\answer: [24, -1, 1, "S4"]
print ( lift(nffactor(nf,x^4+2*x+3)) )
\\answer: [x - a, 1; x^3 + a*x^2 + a^2*x + (a^3 + 2), 1]



for(i = 5, 9, print (polgalois (x^4+2*x+i) ))
\\ answer: [24, -1, 1, "S4"]
\\ [24, -1, 1, "S4"]
\\ [24, -1, 1, "S4"]
\\ [24, -1, 1, "S4"]
\\ [24, -1, 1, "S4"]







for(i = 5, 30, print
 (i, polgalois (x^4+i) ))
\\ answer: interesting pattern depending on whether i is a sqaure or not 

\\ 5[8, -1, 1, "D(4)"]
\\ 6[8, -1, 1, "D(4)"]
\\ 7[8, -1, 1, "D(4)"]
\\ 8[8, -1, 1, "D(4)"]
\\ 9[4, 1, 1, "E(4) = 2[x]2"]
\\ 10[8, -1, 1, "D(4)"]
\\ 11[8, -1, 1, "D(4)"]
\\ 12[8, -1, 1, "D(4)"]
\\ 13[8, -1, 1, "D(4)"]
\\ 14[8, -1, 1, "D(4)"]
\\ 15[8, -1, 1, "D(4)"]
\\ 16[4, 1, 1, "E(4) = 2[x]2"]
\\ 17[8, -1, 1, "D(4)"]







\\ Example 2, 11apr18, keycode: 2989131
\\ 
\\ Example of working with Galois groups with K the 
\\ splitting field of x^4 + 2 \in QQ[t]

P = x^4 + 2;
print ( lift(factornf ( x^4 + 2, a^4+2 )) ) \\ keycode: 282784
\\ answer: [x - a, 1; x + a, 1; x^2 + a^2, 1]

polisirreducible(P)
\\ answer 1, hence P is irreducible in QQ[t] 

G = galoisinit(P)
\\ answer 0 , problem due perhaps (??) to the fact that P does not factor
\\ completely in QQ[a]/( a^4+2)


K = nfsplitting (P)
\\ answer x^8 - 28*x^4 + 2500
\\ Hence the splitting field of K is isomorphic to QQ[x]/ ( x^8 - 28*x^4 + 2500 )

polgalois ( K)
\\ answer: [8, 1, 4, "D_8(8)=[4]2"]
\\
\\ The Dihedral group of order 8

G = galoisinit(K);
\\
\\ G is the Dihedral group of order 8


galoisexport(G)
\\ answer: "Group((1, 5, 8, 4)(2, 3, 7, 6), (1, 2)(3, 4)(5, 6)(7, 8))"
\\ 
\\ Two elements of S_8 that generate G as a subgroup of S_8.



Galoisexport(G,1)
\\ answer: "PermutationGroup<8|[5, 3, 7, 1, 8, 2, 6, 4], [2, 1, 4, 3, 6, 5, 8, 7]>"
\\ 
\\ The order of the group G and two elements of S_8 that generate G as a group.

galoisidentify(G) \\ answer: [8, 3] Related GAP: First coordinate 8 is the order of the group, 
\\ 3 is the identifier of the group among all groups of order 8

\\ Compare: https://groupprops.subwiki.org/wiki/Groups_of_order_8 and
\\ maths/papadakis_partial_screenshot_groups_of_order_8.png


G.group

u = galoissubgroups(G)

matsize (u)
\\ answer [1, 10] This means G has 10 subgroups


vector(#u, i, galoisisabelian(u[i],1))
\\ answer: [0, 1, 1, 1, 1, 1, 1, 1, 1, 1]
\\
\\ 


vector(#u, i, galoisidentify(u[i]))
\\ answer: [ [8, 3], [4, 1], [4, 2], [4, 2], [2, 1], [2, 1], [2, 1], 
\\ [2, 1], [2, 1], [1, 1]]
\\
\\ [4,1] corresponds to the cyclic order 4 subgroup of G, while
\\ the two [4,2] correspond to the two subgroups of G
\\ which are isomorphic to Z_2 \oplus \Z_2 
\\ 


\\ Example 1, 11apr18, keycode: 113421
\\
\\ Example of working with Galois groups with K the 
\\ splitting field of x^6 + 108 \in QQ[t]

P = x^6 + 108;
polisirreducible(P)
\\ answer 1, hence P is irreducible in QQ[t] 
\\ Notice: 108 = 2^2 * 3^3 , using factor (108)


K = nfsplitting (P)
\\ K is the splitting field of P
\\ answer: x^6 + 108 
\\ Hence P splits completely over QQ[a]/(a^6+108)
\\ and K = QQ[a]/(a^6+108)

print ( lift(factornf ( x^6 + 108, a^6+108 )) ) \\ keycode: 282784
\\ answer: [ x - a, 1; x + a, 1; 
\\ x + (-1/12*a^4 - 1/2*a), 1; x + (-1/12*a^4 + 1/2*a), 1; 
\\ x + (1/12*a^4 - 1/2*a), 1; x + (1/12*a^4 + 1/2*a), 1 ]
\\ 
\\ This means, that X^6+ 108 has 6 roots in QQ[a]/(a^6+108) 
\\ and the roors are 
\\
\\ a, -a , root_{e1,e2} = (-1/12* e_1 * a^4 - 1/2*e_2 *a)
\\ 
\\ for all possible e_1, e_2 \in {1, -1} 

G = galoisinit(K);

polgalois ( G.pol)
[6, -1, 2, "D_6(6) = [3]2"] -- it is a group isomorphic to S_3 considered as
-- subgroup of S_6

galoisexport(G)
\\ answer: "Group((1, 2, 3)(4, 5, 6), (1, 4)(2, 6)(3, 5))"
\\ 
\\ Two elements of S_6 that generate G as a subgroup of S_6.

galoisexport(G,1)
\\ answer: "PermutationGroup<6|[2, 3, 1, 5, 6, 4], [4, 6, 5, 1, 3, 2]>"
\\ 
\\ The order of the group G and two elements of S_6 that generate G as a group.

galoisidentify(G) \\ answer: [6, 1] Related GAP: First coordinate 6 is the order of the group, 
\\ 6 is the identifier of the group among all groups of order 1

\\ Compare: https://groupprops.subwiki.org/wiki/Groups_of_order_6 and
\\ maths/papadakis_partial_screenshot_groups_of_order_6.png








G.group
\\ answer: [Vecsmall([1, 2, 3, 4, 5, 6]), Vecsmall([2, 3, 1, 5, 6, 4]), 
\\ Vecsmall([3,1, 2, 6, 4, 5]), Vecsmall([4, 6, 5, 1, 3, 2]), 
\\ Vecsmall([5, 4, 6, 2, 1, 3]), Vecsmall([6, 5, 4, 3, 2, 1])]
\\
\\ Very likely, it means that G is the subgroup of S_4 with 6 elements,
\\ namely s_1 = identity = i \mapsto i for all 1 \leq i \leq 6
\\ s_2 : 1 \mapsto 2, 2 \mapsto 3, 3 \mapsto 1, 
\\ 4 \mapsto 5, 5 \mapsto 6 , 6 \mapsto 4
\\ that is s_2 = (1,2,3) (4,5,6)
\\
\\ s_3 : 1 \mapsto 3, 2 \mapsto 1, 3 \mapsto 2, 
\\ 4 \mapsto 6, 5 \mapsto 4 , 6 \mapsto 5
\\ that is s_3 = (1,3,2) (4,6,5) = s_2 \circ s_2
\\
\\ s_4 : 1 \mapsto 4, 2 \mapsto 6, 3 \mapsto 5, 
\\ 4 \mapsto 1, 5 \mapsto 3 , 6 \mapsto 2
\\ that is s_4 = (1,4) (2,6)(3,5)
\\
\\ s_5 : 1 \mapsto 5, 2 \mapsto 4, 3 \mapsto 6, 
\\ 4 \mapsto 2, 5 \mapsto 1 , 6 \mapsto 3
\\ that is s_5 = (1,5) (2,4) (3,6)
\\
\\ s_6 : 1 \mapsto 6, 2 \mapsto 5, 3 \mapsto 4, 
\\ 4 \mapsto 3, 5 \mapsto 2, 6 \mapsto 1
\\ that is s_6 = (1,6) (2,5) (3,4)


G.gen \\answer: [Vecsmall([2, 3, 1, 5, 6, 4]), Vecsmall([4, 6, 5, 1, 3, 2])]
\\
\\ Hence G is generated (as a group) by the set { s_2, s_4 }



G.orders \\ answer: Vecsmall([3, 2]) , this means 
\\ order (s_2) = 3 , order (s_4) = 2 
\\ (where order means order of an element of a group )


galoisisabelian (G) \\answer 0, which means false. Expected since S_3 is not abelian

u = galoissubgroups(G)

matsize(u) \\ answer [1, 6]

u[1] 
\\ answer: [[Vecsmall([2, 3, 1, 5, 6, 4]), Vecsmall([4, 6, 5, 1, 3, 2])], Vecsmall([, 2])]
\\ Hence we get S_3 as subgroup of S_3

u[2] 
\\ answer: [[Vecsmall([2, 3, 1, 5, 6, 4])], Vecsmall([3])]
\\ Hence we get the cyclic group of order 3 generated by s_2 

u[3]
\\ answer: [[Vecsmall([4, 6, 5, 1, 3, 2])], Vecsmall([2])]
\\ Hence we get the cyclic group of order 2 generated by s_4


u[4]
\\ answer: [[Vecsmall([6, 5, 4, 3, 2, 1])], Vecsmall([2])]
\\ Hence we get the cyclic group of order 2 generated by s_6


u[5]
\\ answer: [[Vecsmall([5, 4, 6, 2, 1, 3])], Vecsmall([2])]
\\ Hence we get the cyclic group of order 2 generated by s_6


u[6]
\\ answer: [[], Vecsmall([])]
\\ Hence we get the trivial subgroup of S_3 


F1= galoisfixedfield( G, u[1] , 2)
\\ answer: [x, Mod(0, x^6 + 108), [x^6 + 108]]
\\
\\ u[1] is the whole Galois group S_3, hence F1 =QQ
\\ so the meaning is that F_1 = QQ [x] / (x) = QQ \subset K
\\ 
\\ The last component is how x^6 + 108 splits over F1


F2 = galoisfixedfield( G, u[2] , 2 )
\\ answer: [x^2 + 972, Mod(3*x^3, x^6 + 108), [x^3 - 1/3*y, x^3 + 1/3*y]]
\\
\\ Apparently, we have F_2 \iso QQ [x] / (x^2+971) 
\\
\\ The last component is how x^6 + 108 splits over F2, where 
\\ by the second component F2 = Q ( 3*x^3) \subset K
\\ 
\\ where K computed above is the splitting field of P

galoisfixedfield( G, u[3] , 2 )
\\ answer: [x^3 + 54, Mod(1/12*x^4 + 3/2*x, x^6 + 108), 
\\ [x^2 - y*x + 1/3*y^2, x^2 + 1/3*y^2, x^2 + y*x + 1/3*y^2]]



galoisfixedfield( G, u[4] ,2 )
\\ answer: [x^3 + 864, Mod(2*x^2, x^6 + 108), 
\\ [x^2 - 1/2*y, x^2 - 1/24*y^2*x - 1/2*y, x^2 + 1/24*y^2*x - 1/2*y]]


galoisfixedfield( G, u[5] ,2 )
\\ answer: [x^3 - 54, Mod(-1/12*x^4 + 3/2*x, x^6 + 108), 
\\ [x^2 - y*x + 1/3*y^2, x^2 + y*x + 1/3*y^2, x^2 + 1/3*y^2]]


galoisfixedfield( G, u[6] , 2 )
\\ answer: [x^6 + 108, Mod(x, x^6 + 108), 
\\ [x - y, x + (-1/12*y^4 + 1/2*y), 
\\ x + (1/12*y^4 + 1/2*y), x + (-1/12*y^4 - 1/2*y), 
\\ x + (1/12*y^4 - 1/2*y), x + y]]
\\
\\ Hence the subgroup is trivial.


vector(#u, i, galoisisabelian(u[i],1))
\\answer: [0, 1, 1, 1, 1, 1]
\\ Meaning: The first subgroup is not abelian, all the other are


vector(#u, i, galoisidentify(u[i]))
\\ answer: [[6, 1], [3, 1], [2, 1], [2, 1], [2, 1], [1, 1]] 
\\
\\ REMARK: galoisidentify (u[i]) returns a pair [a,b] where a is the order of 
\\ the group u[i] and b is the group index in the GAP4 Small Group 
\\ library, by Hans Ulrich Besche, Bettina Eick and Eamonn O'Brien.