ABSTRACT ALGEBRA HOMEWORK
UsingSchonemann and Einstein’s criterion,
P(x) = anxn+an-1xn-1+……. + anx+ ao:
Ifthere exists a prime number p such that:
P divides each ai for i6 =n
P does not an, and
P2 does not divide a0,
ThenP(X) is irreducible over the rationals.
P(X)= P(X1,…………,Xn) and by the fact that its Irreducible its also equal toP(S1,………..,Sn)the two polynomials are equal to zero.
Notealso that we can use Cohn’s irreducibility coefficient
Forexample for the polynomial P(X) = X7+ 5X4+ 35 can be shown to be irreducible.
Takingthe coefficient of X7,which is 1, this is not divisible 5 (obtained because it’s a factorof 5 and 35). Dividing this factor by 5 and 35, we find that itsdivisible hence using Einstein’s criterion, the polynomial isirreducible and equal to zero.
Fromthe remainder theorem, P(x) = Q(x)D(x) + R(x)
P(s)= Q(s)D(s) + R(s)
Sinceboth P(x) = o and P(s) = 0 0 = Q(x)D(x) + R(x)
O= Q(s)D(s) + R(s)
Sincep(x) =P(s), P(x,……..xn)= 0
2(a) From Perron’s Criterion
GivenP(X) = Xn+ an-1Xn-1+…….+ ao
Polynomialsmod P. An integer polynomial
P(X)= anXn+ an-1Xn-1+ …………..+ aomodP
=(anmodP)Xn+ (an-1modP)Xn-1+ …….. +aomodP, this implies that:
Q(,):Q= [)2+ ,)2+(]= 129
b)Q(+)= Q(+)2= 6 +2
Thefield extension Gal[Q(,j)/Q] is Galois of degree 4, hence its Galois group which has order 4.
Theelements of the Galois group are determined by the values of and j. The Q conjugates of and j are and j.Sowe get four possible automrpisms in the Galois group.
Sincethe Galois group has order 4, there are 4 possible assignment ofvalues to and all exist
Eachnonidentity automorphism in the above table has order 2 sinceGal[Q(,j)/Q] contains 4 elements of order 2 , Gal[Q(,j)/Q] has 4 subfields Kisuch that [Gal[Q(,j):ki]= 2 or equivalently
[Ki:Q]=4/2 =2 and that completes the list. Below is a diagram of allsubfields
In the table above, the subgroup fixing Q() is the first and second row, the subgroup fixing Q(j) is thefirst and third row and the subgroup fixing Q(j)is the first and fourth row.